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University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

PROBLEM SET 1

1. Specify the properties of two vectors a and b

such that

(a.) a + b= c and |a|+|b|= |c|.

(b.) a + b= a b.

(c.) a + b= c and |a|2 +|b|2 =|c|2.

(d.) |a + b|= |a b|.

(e.) |a + b|= |a|= |b|.

2. K&K problem 1.2 Find the cosine ofthe angle between A = (3i+ j+ k) and B =(2i 3j k).

3. The relation between Cartesian (x,y ,z) andspherical polar (r,,) coordinates is:

x= r sin cos

y= r sin sin

z= r cos .

Consider two points on a sphere of radius R:(R, 1, 1) and (R, 2, 2). Use the dot productto find the cosine of the angle 12 between thetwo vectors which point to the origin from these

two points. You should obtain:

cos 12= cos 1cos 2+ sin 1sin 2cos (12).

4. New York has North Latitude (= 90 )= 41 and West Longitude (= 360 ) = 74.Sydney has South Latitude (= 90) = 34

and East Longitude (= ) = 151. Take theearth to be a sphere of radius 6370 km; use theresult of Problem 3.

(a.) Find the length in km of an imaginary

straight tunnel bored between New Yorkand Sydney.

(b.) Find the distance of the shortest possiblelow-altitude flight between the two cities.(Hint: The great circle distance along thesurface of a sphere is just R12, where12 isthe angle between the two points, measuredin radians.)

5. K&K problem 1.6 Prove the law of sines us-

ing the cross product. It should only take a cou-ple of lines. (Hint: Consider the area of a triangleformed by A, B, C, where A+ B+ C= 0.)

6. K&K problem 1.11 Let A be an arbitraryvector and let nbe a unit vector in some fixed di-rection. Show thatA = (A n) n+(n A)n.

7. If the air velocity (velocity with respect tothe air) of an airplane is u, and the wind ve-locity with respect to the ground is w, then theground velocity v of the airplane is

v= u + w.

An airplane files a straight course (with respectto the ground) from P toQ and then back to P,with air speed|u|which is always equal to a con-stant U0, regardless of the wind. Find the timerequired for one round trip, under the followingconditions:

(a.) No wind.

(b.) Wind of speed W0 blowing fromP to Q.

(c.) Wind of speed W0 blowing perpendicular toa line connecting P and Q.

(d.) Wind of speed W0 blowing at an angle from a line connecting P and Q.

(e.) Show that the round trip flying time is al-ways least for part (a.).

(f.) What happens to the answers to (b.)-(d.)when W0 > U0? Interpret this limitingcondition physically.

8. A particle moves along the curve y = Ax2

such that its x position is given by x= Bt (t =time).

(a.) Express the vector positionr(t) of the par-ticle in the form

r(t) =if(t) + jg(t) [orxf(t) +yg(t)]

where i and j [or x andy] are unit vectors,and f(t) and g(t) are functions oft.


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(b.) Find the (vector) velocityv(t) as a functionoft.

(c.) Find the (vector) accelerationa(t) as a func-tion oft.

(d.) Find the (scalar) speed |v(t)| as a function

oft.

(e.) Find the (vector) average velocity v(t0)between t = 0 and t = t0 where t0 is anypositive time.

9. Below are some measurements taken ona stroboscopic photograph of a particle under-going accelerated motion. The distance s ismeasured from a fixed point, but the zero oftime is set to coincide with the first strobe flash:

time (sec) distance (m)

0 0.56

1 0.84

2 1.17

3 1.57

4 2.00

5 2.53

6 3.08

7 3.71

8 4.39

Plot a straight-line graph, based on these data,to show that they are fitted by the equation

s= a(tt0)2/2,

where a and t0 are constants, and extrapolatethe line to evaluate t0.


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1


SOLUTION TO PROBLEM SET 1

Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon

1. You may remember the law of cosines fromtrigonometry. It will be useful for several partsof this problem, so we will state it here. If thelengths of the sides of a triangle are a, b, and c,and the angle opposite the side c is , then

c2 =a2 +b2 2ab cos

(a.) When two vectors add up to a third vector,the three vectors form a triangle. If the anglebetween a and b is , then the angle oppositethe side formed by c is 180

.

The law of cosines then tells us that

|c|2 =

|a|2 +

|b|2

2|a||

b|

cos (180

)

From trigonometry, remember that

cos (180 ) = cos

which gives

|c|2 = |a|2 + |b|2 + 2|a||b| cos

We know that |a| + |b| =|c|. Squaring thisequation, we get

|c|2

= |a|2

+ |b|2

+ 2|a||b|If we compare this with the equation above, wecan see that cos has to be equal to one. Thisonly happens when = 0. What this means isthat the two vectors are parallel to each other,and they point in the same direction. If the an-gle between them were 180, then they would beparallel but point in opposite directions.

(b.) This part is simple. Just subtract the vec-tor a from both sides to see that b=b. Theonly way that this can happen is if b = 0, thezero vector.

(c.) This part can also be done by the law ofcosines. Like part (a.), we have the followingtwo equations

|c|2 = |a|2 + |b|2 + 2|a||b| cos This is just the law of cosines again, where is the angle between|a| and b|. The problemstates that |c|2 = |a|2 + |b|2Comparing this with the equation above, we findthat cos = 0. This happens at = 90. Thismeans that the vectors must be perpendicular toeach other.

(d.) Yet again, we can use the law of cosines. Ifthe angle between a and b is , then the anglebetween a andb is 180 . The lengths ofthe sum and difference are

|a + b|2 = |a|2 + |b|2 + 2|a||b| cos |a b|2 = |a|2 + |b|2 2|a||b| cos

For these to be equal, we need cos = 0, whichhappens when = 90. Again, this means thatthe vectors are perpendicular.

(e.) Guess what? Yup, law of cosines. We knowthat|a| = |b| = |a + b|. Addinga to b is goingto look like two vectors stuck together to formtwo sides of a triangle. If the angle between thevectors is , the law of cosines gives

|a + b|2 = 2|a|2 + 2|a|2 cos

where we have used the fact that a and b havethe same length. We also know that |a+b| = |a|.Using this we get

|a|2 = 2|a|2 + 2|a|2 cos Dividing by |a|2, we find a condition on the angle

cos = 12 = 120


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2

2. K&K problem 1.2

We can use the dot product, also known asthe inner product, of two vectors here. Remem-ber that

A B= |A||B| cos where is the angle between the vectors. We canuse the formula for computing the dot productfrom the vector components

A B= AxBx+AyBy+AzBz

The vectors are given as follows: A= 3i + j + kand B = 2i 3j k. Multiplying, we find thatA B= 10. We need the lengths ofA and B.Remember that|A|2 =A A. This tells us that|A|2 = 11 and|B|2 = 14. Dividing, we find that

cos = 1011 14 = 0.805

3. Using the formulas on the problem set, wecan convert the points on the surface of thesphere to Cartesian coordinates.

x1=R sin 1cos 1

y1=R sin 1sin 1

z1=R cos 1

x2=R sin 2cos 2

y2=R sin 2sin 2

z2=R cos 2

As in problem 2, we need to know the length ofthese vectors in order to calculate the angle be-tween them from the dot product. It is fairlyobvious that the lengths of the vectors are justR, because that is the radius of the sphere; wewill show this explicitly.

|(R,,)|2 =R2(sin2 cos2

+ sin2

sin2

+ cos2

)

Using the fact that sin2 + cos2 = 1, we get

|(R,,)|2 =R2 sin2 + cos2 We can just repeat the previous step for nowand get

|(R,,)| =R

as we expected in the first place. Now we calcu-late the dot product. Let x1 be the vector to thefirst point and x2 be the vector to the secondpoint. We find that

x1x2= R2

(sin 1sin 2cos 1cos 2+ sin 1sin 2sin 1sin 2+ cos 1cos 2)

This can be simplified if we remember the for-mula for the cosine of a sum of two angles.

cos( ) = cos cos sin sin

Using this formula, we get the result

x1

x2 = R

2(sin 1sin 2cos(1

2)

+ cos 1cos 2)

To get the angle we just divide by the lengthsof each vector, which are both R. This gives thefinal result.

cos 12= cos 1cos 2

+ sin 1sin 2cos(1 2)

4. This problem is an application of the results

of problem 3.(a.) A straight tunnel between Sydney and NewYork can be represented by the differenceof thevectors pointing to their locations. To say it an-other way, the distance between the ends of twovectors is the length of the difference of the vec-tors. Adjusting for the fact that latitude andlongitude are not quite the same as the coordi-nates and , we find the polar coordinates ofthe cities.

XNY= (6370km, 49

, 286

)XSydney= (6370km, 124

, 151)

Converting to Cartesian coordinates (x,y ,z) us-ing the formulas from problem 3 we get

XNY= (1325 km,4621 km, 4179km)XSydney = (4619km, 2560 km,3562 km)


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3

The distance between New York and Sydneythrough the earth is just|XNY XSydney|. Theresult of the calculation is

Distance = 12, 117km

(b.) Using the result from problem 3 to calcu-late the angle between Sydney and New York,we find that cos 12= 0.809, thus12= 144.0.To calculate the distance along the earths sur-face we need to express this angle in radians.The conversion formula is

(radians) =

180(degrees)

Thus 12 = 2.513 radians. Multiplying this bythe radius of the earth, we get the great circle

distance between New York and Sydney:

Distance = 16, 010km

5. K&K problem 1.6

This question asks you to prove the law ofsines using the cross product. LetA,B, and Cbe the lengths of the vectors making the threesides of the triangle. Let a, b and c be the an-gle opposite each of those sides. The law of sinesstates that

sin a

A =

sin b

B =

sin c

C

Remember that the length of the cross productof two vectors is equal to the areaof the paral-lelogram defined by them. Remember also thatthe the length of the cross product is equal tothe product of the lengths times the sine of theangle between them:|A B| = |A||B| sin . Wehave three vectors to play with in this problem,and using the cross product we can compute the

area of the triangle from any two of them. Wefind that

Area =ABsin c= BCsin a= ACsin b

We just divide the whole thing by ABCand werecover the law of sines.

6. K&K problem 1.11

Let A be an arbitrary vector and let nbe aunit vector in some fixed direction. Show that

A= (A n) n + (n A) nForm a triangle from the three vectors in this

equation. Let B = (A n) n and let C =(n A) n. Let the angle between A andn be . What this formula does is to break upthe vectorAinto a piece parallel to nand a pieceperpendicular to n. B gives the parallel piece.Its length is just|B|=|A| cos . The length ofthe perpendicular piece must then be|A| sin .

Examining the vector C, we see that inside theparentheses is a vector whose length is|A| sin and is perpendicular to n. This vector is thencrossed into n. Since it is perpendicular to n,the length of the final vector is|A| sin , whichis what we want. Now we are just concerned

with the direction. The first cross product isperpendicular to the plane containing n and A.The second cross product is perpendicular to thefirst, thus it is coplanar with n and A. It isalso perpendicular to n. Thus it represents thecomponent of A that is perpendicular to n. Becareful about the sign here.

A useful vector identity that you will be see-ing again is the so-called BAC-CAB rule. It isan identity for the triple cross product.

A

(B

C) =B(A

C)

C(A

B)

Its fairly obvious why this is called the BAC-CAB rule. Using this rule, we see that

n(n A) = n(n A) A(n n)This immediately gives

A= (A n) n + (n A)n


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Of course we havent derived the BAC-CAB rulehere. Its a mess.

7. The idea in all of the parts of this problemis that the plane must oppose any perpendicularwind speed to maintain its straight path. If the

wind is blowing with a speed v perpendicular tothe path, the planes airspeed must bev per-pendicular to the path. The airspeed is u, thewind speed relative to the ground is w, and theground speed is v= u +w. |u| = U0. Let thetotal distance traveled be D.

(a.) No wind, w= 0 so u= v. T =D/U0.

(b.) Wind of speed W0 blowing parallel to thepath. When the wind is going with the plane,v = W0+U0, when it opposes the plane, theground speed is v =U0

W0. The time for the

first leg is T1 = D/2(U0+ W0). The time forthe second leg is T2 = D/2(U0 W0). The totaltime is the sum

T =D

2

1

U0+W0+

1

U0 W0

This can be simplified, and we get the final an-swer, which agrees with part (a.) when W0= 0.

T = DU0U20 W20

(c.) Wind of speed W0 blowing perpendicular tothe path. This part is a little harder. The planewill not be pointed straight along the path be-cause it has to oppose the wind trying to blow itoff course. The airspeed of the plane in the per-pendicular direction will be W0, and we knowwhat the total airspeed is, so we can calculatethe airspeed along the path.

U20 =U2+U

2 U =

U20 W20

The wind has no component along the path ofmotion, so the airspeed in the parallel directionis the same as the ground speed in the paralleldirection. The ground speed is furthermore thesame on both legs of the trip. The final answeragain agrees with part (a.) when W0= 0

T = DU20 W20

(d.) Wind of speed W0 blowing at an angle to the direction of travel. The plane again needsto cancel the component of the wind blowing inthe perpendicular direction. The perpendicularcomponent of the wind speed isW0 = W0sin .As in part (c.) the airspeed in the parallel direc-tion can be computed

U20 =U2+U

2 U=

U20 W20 sin2

In this case, the wind has a component alongthe direction of travel. This parallel componentis W0 = W0cos . On one leg of the trip, thisadds to the ground velocity. On the other leg, itsubtracts. This gives us the following formula:

T =D

2

1

U20 W20 sin2 +W0cos

+ 1

U20 W20 sin2 W0cos

This can be simplified considerably:

T =D

U20 W20 sin2 U20 W20

If you look at this carefully, you will realize thatit reduces to the correct answer for parts (a.),(b.), and (c.) with the proper values for W0 and. If = 90, the wind blows perpendicular tothe path and we get the result from part (c.). If = 0, the wind blows parallel to the directionof travel and we recover the result from part (b.).

(e.) (f.) This part requires some calculus. Weneed to do a minimization of a function. Whatthis part asks is to study the travel time as afunction of wind speed for an arbitrary angle .We need to consider the result from part (d.) asa function of the wind speed:

T(W0) =D

U20 W20 sin2 U20 W20


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For now we are going to ignore the fact that italso depends on U0 and. Remember that func-tions have maxima and minima at places wherethe derivative vanishes, so we need to take thederivative ofTwith respect to W0:

d

dW0T(W0) =D

2W0U20 W20 sin2 (U20 W20 )2

W0sin2

U20 W20 sin2 (U20 W20 )

The derivative is clearly zero when W0 = 0. Inthis case the travel time T = D/U0 as in part(a.). There is another case we have to worryabout though. We divide out what we can toget an equation for another value where the

derivative vanishes

U20sin2 W20 sin2 = 2U20 2W20 sin2

This gives us the other point where the deriva-tive is zero

W20 =U20

2 sin2 sin2

Notice that this point always occurs when thewind speed is greater than the air speed. Noprogress can be made against the wind if thisis the case, so the trip cannot occur. The fi-nal possibility to consider is the case where thewind speed is the same as the air speed. Look-ing at the formula, the time taken is infinite.The only possibility is that the minimum is atW0 = 0. The final piece of this problem is toobserve what happens to the time taken whenW0 > U0. For one thing, it becomes negative.In some circumstances it can even become imag-inary. There is really no interpretation of thisother than ask a stupid question, get a stupid

answer. The answer doesnt make sense be-cause the question didnt make sense. The tripcannot occur whenW0> U0, so it is meaninglessto ask how long it would take.

8. A particle moves along the curve y = Ax2

and its xposition is given by x = Bt.

(a.) We can just plug thex equation into the yequation to get the y position as a function of

time, y = AB2t2. In vector form, the position isthen

r(t) =xBt+yAB2t2

(b.) The vector velocity is obtained from the vec-tor position by differentiating with respect to t

v(t) = d

dtr(t) =xB+ y2AB2t

(c.) The vector acceleration is obtained fromthe vector velocity by again differentiating withrespect to t

a(t) = d

dtv(t) =y2AB2

(d.) The scalar speed is just the length of thevelocity vector. Remember that|A| = A A.

|v(t)| =

v(t)v(t) =

B2 + 4A2B4t2

(e.) The vector average velocity is the integralof the velocity vector over a time interval, di-vided by the time interval. In general, the (time)average of a quantity Ais given by

A

=

1

(t2 t1) t2

t1

A(t)dt

Applying this formula, we see the integral thatneeds to be evaluated:

v(t0) = 1t0

t00

v(t)dt

= 1

t0

t00

(xB+ y2AB2t)dt

We could also use the fact that the integral of thevelocity is the position to get a simpler looking

formula for the average velocity

v(t0) = 1t0

(r(t0) r(0))

Evaluating this integral, we get an answer thatis not surprising

v(t0) =xB+ yAB2t0


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This is just (r(t0) r(0))/t0! The average veloc-ity is just the distance traveled divided by thetime it took.

9. The idea behind this problem is to make agraph of position vs. time data and show that

they fit the equation s = a(t t0)2

/2. In addi-tion you are supposed to find t0. The way todo this is to plot the square root of the distancevs. time, which will give a straight line graph:

s=

(a/2)(t t0). The slope of this graph isapproximately 0.168, so we can use that to ex-trapolate back to zero. We find that the graphreaches zero at about t =4.45, so this meansthat t0 =4.45 to make the distance traveledequal zero at t = 4.45.


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PROBLEM SET 2

1. Calculate the following centripetal accel-

erations as fractions or multiples of g (= 9.8m/sec2):

(a.) The acceleration toward the earths axis of aperson standing on the earth at 45 latitude.

(b.) The acceleration of the moon toward theearth.

(c.) The acceleration of an electron movingaround a proton at a speed of 2106 m/secin a circular orbit of radius 0.5 Angstroms(1 Angstrom = 1010 m).

(d.) The acceleration of a point on the rim of abicycle wheel of 26 in diameter, traveling ata constant speed of 25 mph.

2. K&K problem 1.17 A particle moves in aplane....

3. K&K problem 1.20 A particle moves out-ward along....

4. Att=0 an object is released from rest at thetop of a tall building. At the time t0 a secondobject is dropped from the same point.

(a.) Ignoring air resistance, show that the timeat which the objects have a vertical separa-tion l is given by

t= l

gt0+

t02

.

How do you interpret this result for l 3

2M

Taking the positive root, the final answer is

= cos1

1

3+

1

3

1 3M

2m

4. We assume that the moon is a uniform sphereof mass M= 7.3

1022 kg and radius R= 1740

km. A straight, frictionless tunnel connects twopoints on the surface. Given the mass andradius, the density is just = 3M/4R3. Weneed to know the acceleration due to gravity ata distancer from the center of the moon. This isalso straightforward. Recall that a spherical shellof mass exerts no force on objects inside it, so ata radius r, the only force we need to consider is


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due to the mass in the moon interior to radius r.This is just the density times the volume interiorto r, or M(r) = M r3/R3. The accelerationdue to gravity is then just g(r) = GM(r)/r2 =GMr/R3. Thus the acceleration due to gravityincreases linearly as one moves away from thecenter of a uniform solid sphere.

(a.) In a spherical polar coordinate system withits z axis at the moons north pole, assumethat the tunnel lies in a straight line between(r, 0, ) = (R, 0, 0) and (R, 0, ), i.e. betweentwo points at the same north latitude /2 0having the largest possible difference in longi-tude. This means that the distance along a greatcircle between the ends of the tunnel is 20, whilethe distance from the center of the moon to thecenter of the tunnel isz0 = R cos 0. Now assume

that the mass makes an angle (2 < < 2 )with a line connecting the center of the moonand the center of the tunnel, i.e.with the zaxis.The distance of the mass from the center of thetunnel is then x = z0tan , while its distancefrom the center of the moon is r = z0/ cos . Wenow need to know the component Fx of the grav-itational force GMmr/R3 which lies in the (x)direction of the tunnel, which makes an angle/2 with the radial direction. This is

Fx = GMmr

R3 cos(/2 )= GMm

R3z0

cos sin

= GMmR3

x

This is like the force from a Hookes law springwith effective spring constant keff = GMm/R

3,yielding simple harmonic oscillation with reso-nant angular frequency

0=

keffm

=

GMR3

(b.) Plugging in values of M and R for themoon, and using T = 2/0, we get for theperiod of oscillation

T= 6536 seconds = 109 minutes

(c.) A satellite traveling in a circular orbitmust have centripetal acceleration provided bygravity, which means that

v2

R

=GM

R2

=2R

From the last equality we see that the angu-lar frequency of a circular orbit of radius Raround the moon is the same as 0 above. Ofcourse the period is the same as well.

5. K&K problem 4.23. Two balls of masses Mand m are dropped from height h and collideelastically. The small ball is on top of the largerball. Conservation of energy for the system givesits speed v right before the balls hit the ground:

(M+ m)gh =1

2(M+ m)v2 v=

2gh

The ball M collides with the ground first. Inorder to conserve energy, it must still havespeed v instantaneously after it bounces fromthe ground. Now it immediately collides withthe small ball. (Think of this problem as if therewere a very small gap between the two balls sothat the first ball to hit the ground has a chanceto bounce before the second one hits it.) We con-

sider the elastic collision between the two balls,each moving at speed v towards the other.

The easiest frame in which to study thiscollision is a comoving (inertial) frame that isinstantaneously at rest with respect to the largeball Mimmediately after it has rebounded withvelocityv=

2gh from its elastic collision with

the ground. In this frame,M is instantaneouslyat rest, andm has (upward) velocity 2v. Whenthe collision occurs, ifm M as stated in theproblem,Mseems to m like a brick wall fromwhich it bounces back elastically with the samespeed. Thus, in the comoving frame immediatelyafter the collision, m has velocity +2v. Finally,transforming back to the lab frame, m acquiresan extra velocity increment v, for a total of 3v.Since the height that m reaches is proportionalto the square of its velocity, this means thatm reaches nine times the height from which itoriginally was dropped.


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A less elegant approach considers the colli-sion between M and m in the lab frame. Hereis it essential not to apply the approximationm M until near the end, since cancellationsoccur which may make nonleading terms moreimportant than would initially be suspected.

In the lab frame, conservation of momentumgives

M v mv= M VM+ mVmWe also have conservation of energy through thiscollision. This condition gives

1

2(M+ m)v2 =

1

2M V2M+

1

2mV2m

These are two equations in the two unknowns

Vm and VM, since we already know v = 2gh.We are interested in Vm, which yields the de-sired final height V2m/2g of m, but we are notinterested inVM. So we plan to eliminate VM bysolving for it using the first equation and thensubstituting for it in the second.

Before proceeding with this algebra, it isconvenient to substitute

= m/M

u= Vm/v

U=VM/v

so that all terms are dimensionless. The twoequations above become

1 = U+ u1 + = U2 + u2

Solving the first equation for U,

U= 1 u

Substituting this value forUin the second equa-tion,

1 + = 1 2 + 2 2u + 22u + 2u2 + u20 =(1 + )u2 2(1 )u (3 )0 =u2 2 1

1 + u 3

1 +

Neglecting with respect to 3 or 1 in bothquotients, the polynomial is

u2 2u 3 = (u 3)(u + 1)

with the physical solution

u= 3

Vm= 3v=

18gh

h =V2m

2g = 9h

as before.

6. This is a collision problem that has differentunknown quantities than those to which you areaccustomed, but it is still solvable. We have twocollisions to study, and the unknowns are theneutron mass and the initial and final speeds ofthe neutrons. The initial speeds are the same, sothere are four unknowns in total. We have twocollisions, each of which yields two equations(one for momentum conservation, one for en-ergy conservation since the collisions are elastic).Therefore the system can be solved uniquely.The directions of the scattered neutrons rela-tive to the incident directions do not representadditional unknowns, since the maximum recoilvelocities of the target nucleii will occur when

the collisions take place head-on, with the incom-ing neutrons bouncing straight back. Thus wecan take this to be a one dimensional problem.

The equations are the following (the energyequations have been multiplied by 2):

mnv= mnv + mHvH mnv

2 =mnv2 + mHv

2H

mnv= mnv + mNvN mnv

2 =mnv2 + mNv

2N

Solving these equations for mn and v requirescareful algebra. We square the first momentum

equation to get a relation between v

,mn, andv

v2 =(mnv mHvH)2

m2n

Now we plug this into the first energy equation

mnv2 =

(mnv mHvH)2mn

+ mHv2H


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Expanding,

m2Hv2H 2mnmHvvH+ mnmHv2H= 0

Writing this as an equation for v, we get

v=1

2

1 +

mHmn

vH

This is fairly simple result. If we perform thesame manipulations on the nitrogen equations,we will get an analogous result

v=1

2

1 +

mNmn

vN

We can now use these to solve for mn and v.

Equating the right hand sides, we get a singleequation for the mass.

mnvH+ mHvH=mnvN+ mNvN

mn=mNvNmHvH

vH vN .

We can now use this to find the initial velocityof the neutrons:

v=vH

2 1 +

mH(vH vN)mNvNmHvH

= vH2

mNvNmHvNmNvNmHvH

=

vHvN2

mNmHmNvNmHvH

.

We want to know the mass of the neutron inamu, so we plug in mH = 1 and mN = 14(greater accuracy is unnecessary, since the recoilvelocities are measured only to 10%). We alsolook at both boundaries of the nitrogen veloc-ity, calling these results m and v. Pluggingin numbers, the values ofmn are

mn= 1.159 amu

m+ = 1.415 amu

m= 0.911 amu.

Chadwicks experimental work is seen to be re-liable; todays accepted value for the neutronmass is 1.008665 amu, or 938.27231 0.00028

MeV/c2, well within his experimental range.The range of initial neutron velocity is given by

v= 3.07 107 m/secv+ = 2.82 107 m/sec

v= 4.13 107

m/sec.

7. K&K problem 4.13. The Lennard-Jonespotential is given by

U=

r0r

12 2

r0r

6

(a.) We find the minimum of this potential bydifferentiating it with respect to r and settingthe results equal to zero:

dUdr = 12r

r0r12 r0r

6= 0

This is easy to solve:r0r

12=r0

r

6 r= r0

The depth of the potential well is just U(r0) =. Thus the potential well has a depth .(b.) We find the frequency of small oscillationsby making a Taylor expansion of the potentialabout r = r0. Read section 4.10 in K&K for

more information on this. We can write thepotential as follows:

U(r) =U(r0) +

dU

dr

r=r0

(r r0)

+1

2

d2U

dr2

r=r0

(r r0)2 +

We know that dU/dr = 0 at r =r0, so we dropthe middle term.

U(r) +1

2d2U

dr2r=r0 (r r0)

2

This is exactly the form of the potential of a masson a spring. We only have to identify the springconstant. Remembering that Uspring = kx

2/2,we make the identification

k=

d2U

dr2

r=r0


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For the Lennard-Jones potential, we alreadyknow the first derivative, so we need to differen-tiate once more.

d2U

dr2 =

12

r2 13r0r

12

7r0r

6

Plugging in r = r0, we find the effective springconstant for this potential

k=72

r20

We now consider two identical masses m on theends of this spring. Their (coupled) equationsof motion are:

mr1= k(r

r0) mr2=

k(r

r0)

where r = r2 r1 is the distance between themasses. Subtracting these two equations, we get

mr= 2k(r r0)

The frequency of oscillation is then 2 = 2k/m.(Note that we could have obtained the sameresult by considering the two-mass system tobe a single mass of reduced mass mreduced =m1m2/(m1+ m2)). Plugging in the above value

for the effective spring constant k ,

= 12

r20m


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PROBLEM SET 6

1. K&K problem 6.1 Show that if the total

linear momentum....

2. K&K problem 6.3 A ring of mass M andradiusR lies....

3. Expansion of the previous problem:

(a.) Let the azimuth of the bug on the ring be that is, is zero when the bug startswalking, and 360 when the bug makes onerevolution. Assume (for part (a.) only) thatthe ring is fixed. Calculate the angular mo-mentum l of the bug about the pivot in the

previous problem, as a function of. Checkthat your result is consistent with what youused in the previous problem when was180.

(b.) Now assume that the bug is fixed at someazimuth on the ring, but that the ringitself is not fixed, having angular velocity about the pivot (opposite to the angularvelocity of the bug when the bug was mov-ing). Calculate the angular momentum l ofthe bug as a function of and .

(c.) Now assume that neither the bug nor thering are fixed. By requiring that the totalangular momentum l+l of the bug aboutthe pivot be balanced by the angular mo-mentum of the ring, obtain an expressionfor the angular velocity of the ring, as afunction of.

(d.) Get an integral expression for the angle through which the ring rotates, as a functionof time, assuming that d/dt = = con-stant. You need not evaluate the integral.

Note that the system is bootstrapping itsway around the pivot!

4. K&K problem 6.5 A 3,000-lb car is parkedon a....

5. A man begins to climb up a 12-ft ladder (seefigure). The man weighs 180 lb, and the ladder20 lb. The wall against which the ladder restsis very smooth, which means that the tangential

(vertical) component of force at the contact be-

tween ladder and wall is negligible. The foot ofthe ladder is placed 6 ft from the wall. The lad-der, with the mans weight on it, will slip if thetangential (horizontal) force at the contact be-tween the ladder and ground exceeds 80 lb. Howfar up the ladder can the man safely climb?

6. K&K problem 6.8 Find the moment ofinertia of a uniform sphere....

7. K&K problem 6.14 A uniform stick of massMand length L is....

8. K&K problem 6.18 Find the period of apendulum....


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1. K&K problem 6.1

(a.) We know that the total linear momentumof the system is zero. (This would occur, forexample, if we were in the center of mass frame.)

P=i

pi = 0

Examine the total angular momentum of thesystem. The vector from the origin to pointi isdenoted ri. The angular momentum in general

depends on where the origin is:

L=i

ri pi

We now want to find the angular momentumabout a new origin whose position vector is Rin the current coordinate system. In this newsystem, the position vector of point i becomesri R. Each point has its position changed bythe same amount. The new value of the angularmomentum is

Lnew=i

(ri R) pi

Expanding,

Lnew=i

ri pi i

R pi

BecauseR is the same for all points, we can pullit outside of the sum:

Lnew=i

ri pi Ri

pi

=i

ri pi RP

We know that P= 0, so we are done.

Lnew=L

(b.) The proof for this part is identical if angu-lar momentum is replaced by torque and linearmomentum is replaced by force.

2. K&K problem 6.3

This problem and the next concern the samesystem that of a bug walking along a hoopthat is free to pivot around a point on its edge.The hoop lies flat on a frictionless surface. Thering has mass Mand radius R, and the bug hasmassm and walks on the ring with speed v .

The key idea in this problem is conservationof angular momentum. About the pivot there isno net torque on the system, so the total angularmomentum about that point is conserved. Thering starts at rest with the bug on the pivot, andthe bug starts walking at speed v. Immediatelyafter the bug starts walking, the total angularmomentum measured about the pivot point con-tinues to be zero. The ring is not yet moving, soit has no angular momentum; the bug has begunto move, but it is at r= 0, so it has no angular

momentum yet.

We want to find the angular velocity ofthe ring when the bug is opposite to the pivot.The bug is moving at speed v on the ring, butthe ring is also moving. The bug is at a distance2R from the pivot, so the velocity of that por-tion of the ring which is under the feet of the


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2

bug is 2R. The total velocity of the bug is thusv+ 2R. Next we need to know the moment ofinertia of the hoop. A hoop has moment of in-ertiaI=MR2 about its center of mass. We usethe parallel axis theorem to find the moment ofinertia about a point on the edge.

I=ICM+ Md2

The distance d from the center of mass to thedesired axis in this case is just R, so the momentof inertia of the hoop about a point on the edgeisI= 2MR2. We can now find an expression forthe total angular momentum of the system. Forthe hoop we use L= I and for the bug we useL= mvr sin . The angle between the positionvector and the velocity vector of the bug in this

case is simply /2, so sin is just 1. We nowwrite the angular momenta of the two pieces

Lbug= 2mR(v+ 2R) Lhoop= 2MR2

Since angular momentum about the pivot is con-served throughout the motion, We know thatLbug+ Lhoop = 0. This gives the following ex-pression:

2mvR+ 4mR2 + 2MR2 = 0

We solve this equation for in terms ofv and get

= mvMR+ 2mR

Note that the minus sign means that the hooprotates in a direction opposite to that in whichthe bug moves. This makes sense because thetotal angular momentum about the pivot pointmust vanish.

3. We now study the bug and hoop system inmore detail. See the diagram in the previous

problem.

(a.) In this part we assume that the ring is fixed.We want to calculate the angular momentum ofthe bug about the pivot point. The first step isto find the distance r between the bug and thepivot. We can do this using the law of cosines.Consider the triangle made by the line betweenthe bug and the pivot, and the two radial lines

extending from the center of the hoop to the bugand pivot, respectively. This is an isosceles tri-angle, with two equal sides of length R havingan angle between them. If we define the az-imuth of the bug on the hoop to be zero at thepivot, the angle is simply the azimuth of thebug on the hoop. The length r of the third sideis found using the law of cosines:

r2 =R2+R22R2 cos r2 = 2R2(1cos)

Using the trigonometric identity 1 cos =2sin2

2, we can get a simple result for r:

r= 2R sin

2

Now we know v and r. The only thing left to

determine is the angle between the position andvelocity vectors. The first step is to find the an-gle between the position vector r of the bug andthe line from the center of the circle to the bug.The isosceles triangle (like any triangle) has atotal angle , and its central angle is . The re-maining two angles are equal, so they must be( )/2 each. Thus the three angles add up to radians. Because the velocity vector v is tan-gent to the circle, the angle between r and v is/2 minus this angle. Thus the angle between rand v is /2. We can now get the angular mo-

mentum of the bug. Assuming that the ring isfixed,l = mvr sin from the bug alone, so

l= 2mvR sin2

2

(b.) In this part we assume that the ring is ro-tating with angular velocity , but that the bugis fixed on the ring. The velocity of the bug isjust r, wherer is the same as was calculated inthe previous part. The velocity in this case is al-ways perpendicular to its position vector. This

can be seen by remembering that the bug isntmoving on the ring, so it must be in uniformcircular motion about the pivot, with a velocitythat is tangent to its present position. Thereforethe angular momentum l of the bug is simplymvr, yielding

l =mr2 = 4mR2 sin2

2


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3

(c.) We now allow both the bug and the ringto move. The total angular momentum of thebug isl + l from parts (a.) and (b.) respectively.To this we must add the angular momentum ofthe ring to get the total angular momentum ofthe system. From problem 2. we know that thetotal angular momentum must be zero. The an-gular momentum of the ring is I, so we get thefollowing equation.

4mR2 sin2

2+ 2mvR sin2

2+ 2MR2 = 0

We solve this for in terms of. The result is

= mv sin2 2

MR+ 2mR sin2 2

This agrees with the result of problem 2. whenthe bug is at = , opposite to the pivot.

(d.) Finally, we want to find an expression forthe angle through which the ring rotates. Weknow that is related by a simple differentialequation to the angular velocity of the hoop

=d

dt

but we want to express in terms of so thatwe can use the fact that Rd/dt, the speed v ofthe bug with respect to the rim of the hoop, isconstant. We apply the chain rule to get

=d

dt =

d

d

d

dt

Substituting d/dt = v/R, where v is constant,we can write an integral for :

= Rv

o

mv sin2(/2)

MR+ 2mR sin2(/2)d

We can simplify this a little, but doing the inte-gral is hard, which is why you werent asked toevaluate it. Setting the initial bug azimuth 0to zero and using the fact that d/dt= v/Ris aconstant so that = vt/R,

(t) = vt/R0

sin2(/2)

(M/m) + 2 sin2(/2)d

4. K&K problem 6.5

A car of mass m is parked on a slope of angle facing uphill. The center of mass is a distance dabove the ground, and it is centered between thewheels, which are a distance l apart. We want

to find the normal force exerted by the road onthe front and rear tires.

It is easiest to do this problem choosing theorigin as the point on the road directly below thecenter of mass. About this point there are threetorques. The normal force on the front (Nf) andrear (Nr) set of wheels provides a torque, andalso gravity provides a torque mgd sin becausethe car isnt horizontal. However, the forces offriction on the tires dont provide any torque be-cause they are in line with the direction to the

origin. The torque from the front wheels and thetorque due to gravity tend to want to flip thecar over backwards, while the torque on the rearwheels opposes this tendency. We want the sumof the torques to vanish, because the (static) caris not undergoing any acceleration, angular orlinear:

0 =Nfl

2+ mgd sin Nr

l

2

Nr Nf =2mgd

l sin

We can get one more condition from the factthat the car is not undergoing linear accelera-tion perpendicular to the road. This means thatthe normal forces exactly cancel gravity:

Nr+ Nf=mg cos

We can take the sum and difference of thesetwo equations to get expressions for Nf and Nr.These are

Nr=mg1

2cos +

d

l sin

Nf =mg

12

cos dl

sin

Plugging in = 30,mg = 3000 lb,d = 2 ft, andl= 8 ft, we get Nr =1674 lb and Nf=924 lb.


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4

5. We will solve this problem symbolically andwait until the end to plug in numbers. Thisis always good practice because it makes it alot easier to check the units of the result and toexplore whether the result is reasonable when theinputs have limiting values. We take Mto be themass of the man (Mg= 180 lb) and m to be themass of the ladder (mg = 20 lb). The lengthHof the ladder is 12 ft, and its point of contact withthe wall isd = 6 ft from the wall. The angle thatit makes with the wall is = arcsin (d/H) = 30.Finally, the force of friction on the ladder fromthe ground is Ff Fmaxf , whereFmaxf = 80 lb.

There are five forces to consider in thisproblem. They are the two normal forces onthe ladder, Ng from the ground and Nw from

the wall; the force Ff of friction at the base ofthe ladder; and the two forces of gravity, Mgon the man and mg on the ladder. This isa torque balance problem, so choosing a goodorigin makes it a lot easier. With this choice ofthe point of contact with ground, two of the fiveforces contribute no torque about that point.Not bad! As a sanity check we evaluate , thecoefficient of friction between the ladder and theground. The normal forceNf from the floor isequal and opposite to (M+ m)g, the sum of theweights of the ladder and the man. We are given

the maximum frictional forceFmaxf , and we knowthat Fmaxf = N, so = F

maxf /((M+m)g) =

80/200 = 0.4, a reasonable value.

We now calculate the torques. To find themaximum height h to which the man can climbwithout the ladder slipping, we assume that theladder is about to slip. This means that the nor-mal forceNw from the wall is equal and oppositeto Fmaxf , exactly countering the maximum forceof friction: since these two forces are the onlyforces in the horizontal direction they must sumto zero. The torque from the wall is then w =Fmaxf Hcos , where the minus sign indicatesthat this torque pushes clockwise. The torquefrom the weight of the ladder is exerted at themidpoint of the ladder, its center of mass. Thevalue of this torque ism= mg

H2

sin . Similarly,the torque exerted by the weight of the man, whois a distance h up the ladder, is M=Mgh sin .

Requiring these three torques to sum to zero,

0 =M+ m+ w

=Mgh sin + mgH

2 sin Fmaxf Hcos

Solving for h,

h=Fmaxf Hcos mgH2 sin

Mg sin

h

H =

FmaxfMg

cot m2M

=(M+ m)

M cot m

2M

=(M+ m)cot (m/2)

M

=0.4(200)

3

10

180= 0.7142

h= 8.571 ft

6. K&K problem 6.8

Because of the spherical symmetry, we work inspherical polar coordinates. To find the momentof inertia we need to evaluate the integral

I= r2dvwhere in these coordinates r = r sin is theperpendicular distance to the axis and dv =r2dr d(cos ) d is the element of volume. Theintegral to evaluate is thus

I=M

R0 r4 dr

11

sin2 d(cos )20 dR

0 r2 dr

11

d(cos )20 d

where the denominator is the volume V of thesphere, needed to evaluate its density = M/V.

Substituting u r/R,

I

MR2 =

10u4 du

11

sin2 d(cos )20 d1

0u2 du

11

d(cos )20 d

In both the numerator and the denominator, allthree integrals have limits that do not dependon the other variables, so each integral can be


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5

evaluated independently. The integrals cancel,and the u integrals have the ratio

1/5

1/3=

3

5

The integrand in the cos integral in the numer-ator can be rewritten

sin2 d(cos ) = (1 cos2 )d(cos )

=d(cos ) d1

3cos3

Therefore the ratio of the integrals is

2 23

2 =

2

3

Putting it all together,

I=MR2 35 2

3=2

5MR2

7. K&K problem 6.14

When the stick is released, there are twoforces acting on it, gravity at the midpoint, andthe normal force at the point B. We use thepoint B as the origin, so the only torque aboutthis point is provided by gravity. At the mo-ment of release, the stick is still horizontal, sothe torque is

B = Mgl2

where the minus sign indicates that the torquepulls clockwise. We know that the momentof inertia of a thin stick about its endpoint isI = Ml2/3, so we can easily find the angularacceleration from =I.

Mgl2

=1

3Ml2 = 3g

2l

The vertical acceleration of the center of massis given by the simple formula a = r, where

r is the distance between the center of massand point b, about which the stick is (instan-taneously) executing circular motion. (This isanalogous to the expression v =r.) Here thisdistance is r = l/2. This gives the accelerationof the center of mass:

a= 34g

where the minus sign indicates that the accel-eration is downward. Finally we use Newtonssecond law to find the normal force at B. Weknow the acceleration and we know the force ofgravity, so this is a simple equation

NMg= 34Mg N=1

4Mg

where the positive direction is up, opposite tothe force of gravity.

8. K&K problem 6.18

We want to find the equation of motion ofthe pendulum to determine the frequency. Wewill use the torque equation = I. If we

choose the pivot point of the pendulum as theorigin, only one force provides torque, the forceof gravity. It acts on the center of mass of thependulum, a distance lcm from the pivot point.The magnitude of this force is just (M+m)g.Thus the total torque is

= (M+ m)glcmsin

where is the angular position of the pendulum.Writing the torque equation and approximating

sin , we get

I= (M+ m)glcm

We recognize that this is the equation for a sim-ple harmonic oscillator. The angular frequencyand period are thus

=

(M+ m)glcm

I T = 2

I

(M+ m)glcm

All that is left is to evaluate lcm and I. Theequation for the center of mass is easy to use.The center of mass of the rod is halfway alongits length, and the disk is a distance l from thepivot, so

lcm=ml/2 + Ml

M+ m


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This expression simplifies the formula for theperiod

T= 2

I

(M+ m/2)gl

In the first case, where the disk is tied to therod, the moment of inertia is determined usingthe parallel axis theorem. The disk is fixed tothe rod, so, as the rod pivots, the disk must ro-tate at the same angular velocity. The momentof inertia of a stick about its end is ml2/3. Themoment of inertia of a disk about its center isMR2/2. Because the center of mass is displaceda distance l from the origin, the parallel axistheorem tells us that the total moment of iner-tia of the disk is MR2/2 +Ml2. Thus the totalmoment of inertia of the pendulum is

I=1

2MR2 +

M+

1

3m

l2

This gives the period of oscillation

T =T = 2

MR2/2 + (M+ m/3)l2

(M+ m/2)gl

In the second case, where the disk is free to rotateon the rod, the moment of inertia is smaller.Because the disk is not fixed, it has no tendencyto rotate. Its effective moment of inertia aboutthe center of mass is thus zero. For our purposeit is the same as a point mass a distance l awayfrom the origin. The new moment of inertia is

I=

M+

1

3m

l2

The period of oscillation in this case is alsosmaller:

T = 2(M+ m/3)l(M+ m/2)g

This is the same as our answer for the first casein the limitR 0.


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PROBLEM SET 7

1. A trick cyclist rides his bike around a wall

of death in the form of a vertical cylinder (seefigure). The maximum frictional force parallelto the surface of the cylinder is equal to a frac-tion of the normal force exerted on the bike bythe wall. Assume that the cyclist and his bikeare small relative to the radius of the cylinder.

(a.) At what minimum speed must the cyclistgo to avoid slipping down?

(b.) At what angle to the horizontal must hebe inclined at that minimum speed?

(c.) If =0.6 (typical of rubber tires on dryroads) and the radius of the cylinder is 5

m, at what minimum speed must the cyclistride, and what angle does he make with thehorizontal?

2. K&K problem 6.24 Drum A of mass M andradius R....

3. K&K problem 6.27 A yo-yo of mass M hasan axle....

4. Two men, each of mass 100 kg, stand at oppo-site ends of the diameter of a rotating turntableof mass 200 kg and radius 3 m. Initially the

turntable makes one revolution every 2 sec. Thetwo men make their way to the middle of theturntable at equal rates.

(a.) Calculate the final rate of revolution andthe factor by which the kinetic energy ofrotation has been increased.

(b.) Analyze, at least qualitatively, the meansby which the increase of rotational kinetic

energy occurs.

(c.) At what radial distance from the axis of ro-tation do the men experience the greatestcentrifugal force as they make their way tothe center?

5. K&K problem 7.4 In an old-fashioned rollingmill, grain....

6. K&K problem 7.5 When an automobilerounds a curve....

7. K&K problem 8.2 A truck....

8. K&K problem 8.4 The center of mass....


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1




1. We will need to use fictitious forces tosolve this problem easily. Fictitious forces arenever necessary, but they often simplify prob-lems greatly.

(a.) We want to know the minimum speed thecyclist needs not to slip down the side. Theforce of static friction must be mg to hold himup, so we require that N = mg. The force ofstatic friction can be less than N, but we aresetting it to the maximum to see what the limitis. Thus we need N > mg/. The only forceacting horizontally in the system is the normalforce, so it must entirely provide the centripetal

acceleration, which is v2/R. We thus obtain

mv2

R mg

v

gR

(b.) We now need to consider the fictitiouscentrifugal force. The cyclist is in an acceler-ating frame of reference because he is movingin a circle. To correctly apply Newtons secondlaw in the cyclists frame, we must introducethe centrifugal force, which points outward withmagnitude mv2/R. In the frame of the cy-clist there are four forces: the normal force andthe centrifugal force cancel each other, and thefriction and gravity cancel each other. This isrequired to insure that, by definition, the cyclistisnt accelerating in his own frame of reference.The problem is now reduced to a torque balanceto find the angle at which the cyclist is stable.

We choose the point of contact as the origin.There are two torques, g caused by gravity andc caused by the centrifugal force. Both act atthe center of mass. (This is important to note:fictitious forces always act at the center of mass!)We are assuming that the size of the cyclist l isvery small in comparison with the radius R, so

g = +mgl cos c = mlv2

R sin

We want to know the angle where the cyclist is

about to slip, so the normal force is mg/, equalto the centripetal force mv2/R. Substituting formv2/Rin the above equation,

mgl cos = mgl1

sin tan =

So when the cyclist is about to slip, he rides atan angle

= tan1

(c.) Taking = 0.6 and R= 5 meters, we find

that the cyclist must ride at a speed of least 9.0meters per second, or 29.5 ft/sec, or 20.1 mph.On a road bike this is a mellow cruising speed.At this minimum speed, the angle the cyclistmust make with the horizontal is 31 degrees. Wecaution you not to try this at home; its toughto get up to speed without crashing!

2. K&K problem 6.24

This problem is similar to many pulley problemsthat you have seen before. We need to applyboth Newtons second law and the torque equa-

tion to solve it. We denote the (positive down-ward) acceleration of the falling mass asa. Thereare two forces on it in the vertical direction, ten-sion and gravity. Newtons second law requires

Mg T=M a

We now need to apply the torque equation toboth drums. For each drum we choose its own


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2

center as the origin. Each drum feels only onetorque, the torque from the tension. This has amagnitude = T R in both cases. Notice thatboth of these torques have the same sign, thusthe drums will tend to angularly accelerate inthe same direction. Writing the torque equationfor each drum, with angular accelerations 1 forthe top drum and 2 for the bottom drum,

T R= I 1 T R= I 2

where the moments of inertia of both disks arethe same, I=M R2/2. From these equations itis easy to see that 1 =2 = T R/I, so theangular accelerations are both

= 2T

M R

We now need to find a relation betweena and .The linear acceleration due to each disk is givensimply by a = R. There are two disks, bothunwinding with the same angular acceleration,so the linear acceleration of the bottom one isjusta = 2R. The previous equation becomes

a=4T

M T = M a

4

Plugging this into the very first equation thatwe got from Newtons law, we find the initial ac-celeration of the drum, assuming that it movesstraight down, to be

a=4

5g

Will the drum in fact move straight down? Forthe moment assuming that the answer is yes,consider a (downward accelerating but nonrotat-ing) frame with its origin at the (instantaneous)point of tangency between the lower drum andthe tape. In this frame, the CM of the drum ex-periences a downward force mg and an upwardfictitious force 4

5mg which does not quite com-

pensate it. Therefore it feels a net downwardforce 1

5mg. About the chosen origin this force

causes a net (clockwise) torque, which causesthe lower drum to swing to the left like a pen-dulum bob in this frame. This contradicts our

assumption of a pure downward motion. There-fore the actual motion will be more complicatedthan this problem asks you to assume.

3. K&K problem 6.27

We need to apply both Newtons law and the

torque equation. The forces on the yo-yo hori-zontally are the force Fand the friction f. Thevertical forces are the normal force and grav-ity, which immediately tell us that N = M g.We want to find the maximum force we canapply with the yo-yo not slipping. It is impor-tant to note that the force of friction, whichstops the disk from slipping, is controlled by thecoefficient s of static friction because the sur-face of a rolling wheel is at rest with respectto the ground. Since we are concerned with

the maximum allowed force, we will considerthe maximum allowed friction, which is sN.Newtons law gives us

F sMg= M a

The moment of inertia of the yo-yo is I =M R2/2. Because we want the yo-yo to roll with-out slipping, we can use a = R. The torqueequation gives us

sMgR

F b= I =1

2

M Ra

We want to solve these two equations for F, themaximum allowed force. Eliminating a, we get

F sM g= 2sM g 2F b

R

Solving for F, we get

F =sMg 3R

R + 2b

Since R > b the applied force F is always largerthan the frictional force, so the yo-yo alwaysaccelerates to the right.

4. We will solve this problem symbolically andplug in numbers at the end. This is always agood practice because it makes it a lot easier togo back and check your work for correct dimen-sions and reasonable results for limiting cases.


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(a.) Let the disk have mass M and radius R,and the two men each have mass m. If the menare momentarily at a radius r from the center ofthe disk, the total moment of inertia is given by

I(r) =

1

2 M R

2

+ 2mr

2

The initial angular velocity of the disk is 0, so,when r = R, the initial angular momentum ofthe system is

L= I 0 =

1

2M+ 2m

R20

There are no net torques acting on the system,so L is conserved. We can use L conservation,I(r)(r) =I(R)0, to obtain the angular veloc-

ity of the system as a function of the radius ofthe men:

(r) =0MR2 + 4mR2

M R2 + 4mr2

The final angular velocity is just the an-gular velocity when the men reach the center,(r= 0).

=0M+ 4m

M =

1 +

4m

M

0

Plugging in values, we find that the final angularvelocity is 1.5 revolutions per second.

The factor by which the kinetic energy hasincreased is

KfKi

=1

2I(0)2

1

2I(R)2

0

Evaluating this, we find

KfKi

=1

4M R2

1 + 4mM

220

1

4M R22

0 mR22

0

Simplifying,

KfKi

= 1 +4m

M

K=4m

MKi

For the masses in the problem, Kf/Ki is equalto 3. The rotational kinetic energy is tripled.

(b.) The extra kinetic energy comes from thework that the men must do against the (ficitious)centrifugal force to make their way from the edgeof the turntable to the center. This is the qualita-tive statement which the problem requests. Op-tionally, one can perform a quantitative analysis:

Each man pushes against the centrifugalforce to get to the center. The work they dois converted to rotational kinetic energy. Thecentrifugal force on each man is given by

Fc =m((r))2r= m2

0r

1 + 4m/M

1 + 4mr2/(MR2)

2

The work done is just Fc integrated from zeroto R, doubled since each man does the same

amount of work.

W = 2m20

1 +

4m

M

2 R0

r dr

(1 + 4mr2

MR2 )2

You can look this up in a table, or notice thatthe top is proportional to the derivative of thebottom, so antidifferentiating is not too hard:

W =

2m2

0 1 +4mM

2

MR2/8m

(1 + 4mr2

MR2 )2

R

0

Evaluating this, we get

W =M R22

0

4

1 +

4m

M

2

1 +4m

M

Multiplying this out,

W =MR22

0

44m

M +

16m2

M2

Simplifying,

W =mR220

+4m2

M R22

0

=

1 +

4m

M

mR22

0


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4

This is 4m/Mtimes the initial kinetic energy, so,as expected, W is equal to the kinetic energygain Kthat we already calculated.

(c.) We want to find where the maximum cen-trifugal force is felt. This is just a maximization

problem. Differentiate Fc with respect to r andset it to zero, and also check the endpoints.

dFcdr

= d

dr

m2

0r

(1 + 4m/M)2

(1 + 4mr2/M R2)2

= 0

This gives

1 16mr2

M R21

1 + 4mr2/MR2 = 0

We can solve this for r. Set x = 4mr2/M R2.Then 1 (4x/(1 + x)) = 0 so x = 1/3, and

4mr2

M R2 =

1

3 r= R

M

12m

If we plug in the mass values for this problem, weobtain r =R/

6. Since the centrifugal force is

everywhere positive, and it is zero at the center,this extremum must in fact be the maximum.

5. K&K problem 7.4

Referring to the diagram, the stone orbits aroundthe vertical shaft with orbiting angular velocity. The velocity of the stones CM is thusv= R. The stone is rolling without slipping onthe flat surface, so its rolling angular velocity is

=v

b =

R

b

in magnitude. Since angular velocity is a vector,we can add these separate components to obtainthe full angular velocity vector. In cylindricalcoordinates, with z pointing along the axis ofthe orbit,

= Rbr + z

where the leading minus sign tells us that theradial component of is negative, i.e. the mill-stone is rotating clockwise about its horizontalaxle. To calculate the angular momentum, wechoose as an origin the intersection of the cen-terline of the vertical shaft and the centerlineof the horizontal axle. Both the shaft and theaxle are parallel to mirror symmetry axes of themillstone; thus we expect that the component

of angular momentum due to will be parallelto , and the component of angular momentumdue to will be parallel to . More quanti-tatively, the component Lz along z is equal to(I +M R2), where I = 1

4M b2 is the moment

of inertia of a disk about a diameterand M R2

is added to I by use of the parallel axis theo-rem. Since Lz is constant, no torque is requiredto maintain it and we dont need to consider itfurther. To calculate the radial component Lrof the angular momentum, we need I=M b2/2,the moment of inertia of a disk about its center:

Lr = 1

2Mb2

where the minus sign again reminds us that themillstone is rolling clockwise about its horizontalaxle. Remember that, in cylindrical coordinates,the only unit vector which is constant is z; theradial and azimuthal unit vectors depend on .Even though the magnitude of Lr is constant,its direction is changing. In a time increment dt,

the azimuth of the millstone axle with respectto the shaft changes by an angular incrementd= dt. This causes L to change by

dL=Lrd

=Lrdt

= 12

Mb2dt


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5

The torque is thus

=dL

dt = 1

2M b2

Finally we consider the forces on the sys-

tem. The vertical shaft exerts a force on thehorizontal axle, gravity pulls down on the mill-stones CM, and the normal force pushes up onthe millstone. However, with respect to the cho-sen origin, the first of these forces can exert notorque because it is applied at r = 0. The torquedue to gravity is in the + direction, and thetorque due to the normal force N of the flatsurface on the millstone is in the direction.Thus, in the + direction, we have

NR + M gR = =

1

2Mb2

M

g+1

2

b2

R

=N

Substituting= R/b,

N=M

g+

1

2b2

Of course, by Newtons third law, the contactforce exerted by the millstone upon the flat sur-face is equal and opposite to N. As advertised,the effective weight of the millstone for crushinggrain is greater than M g; this increment risesquadratically with the angular velocity.

What keeps the millstone from accelerat-ing upward, since the upward normal force on itis greater than the downward force of gravity?The force of the vertical shaft on the horizontalaxle, which we ignored in the torque equationbecause it is applied at the origin, must pushdownward, in alignment with gravity, with thevalueMb2/2.

Such millstones must have been in use be-fore the time of Newton, so the benefits of theirincreased effective weight when rolling in a cir-cle must have been discovered empirically ratherthan logically.

6. K&K problem 7.5

(a.) If the flywheel were horizontal with its spinaxis pointing up, it would have little effect, since

the direction of its angular momentum wouldnot change as the car turns left or right. Sothe flywheel should be vertical with its spinaxis pointing either sideways or forward. Decid-ing between these alternatives requires a morequantitative analysis.

Lets look at the car from the rear while itis in motion with speed v. First well considerthe car without any flywheel. Suppose that thecar is in the process of turning to the left, tak-ing a turn of radiusR0. Adopt a reference frameattached to the car, with an origin halfway be-tween the tires at the level of the road. It is easyto see why the act of turning causes the nor-mal forces on the tires to become unbalanced.The sum of the torques on the car must remainzero if the car (assumed to have no suspension

system, so it doesnt lean) keeps all four tireson the road. With respect to the origin chosen,the forces of friction on both tires can exert notorque, because these forces act directly towardor away from the origin. Neither can the force ofgravity exert a torque about this origin, for thesame reason. That leaves Nl and Nr, the nor-mal forces on the left and right sets of tires, andM v2/R0, the fictitious centrifugal force whichpulls the CM to the right in this acceleratingframe. Let the CM be a distance d above theroad; let the right-left separation of the wheelsbe 2D. Along v, the sum of the torques is then

NlD+ NrD M v2

R0d= 0

Clearly Nr must exceed Nl if this equation is tobe satisfied. This is the problem we are tryingto solve with the flywheel.


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6

The torques that we just considered werealong v. If the flywheel is to help, its an-gular momentum L should be directed so that,when the car turns left, the flywheel producesa torque on the car equal to +M v2d/R0 along

v. By Newtons third law, the torque of the

car on the flywheel should correspondingly beequal to +vM v2d/R0. So, as the car turns left,the change in L of the flywheel should be di-rected along +v. This will happen if the angularmomentum vector of the (vertical) flywheel ispointing to the right. This means that the fly-wheel should rotate in the opposite direction asthe tires. For simplicity, well install it at heightd from the road so as not to perturb the CM.

Its not necessary to reverse the flywheel di-rection for right as opposed to left turns, because

both the centrifugal force and the change in Lwill correspondingly reverse direction.

(b.) Now that we have determined the flywheeldirection, we can calculate the desired magnitudeLof the flywheels angular momentum. We have

dL

dt =L =M v2d/R0

where = v/R0 is the angular velocity of thecar around the turn. Solving,

L= M vd

For a disk-shaped flywheel of mass m and ra-dius r, I = mr2/2, and the flywheels angularvelocity should be

=2Mvd

mr2

This is independent of the turn radiusR0, whichis very nice. Weve achieved perfectly flat cor-

nering for a turn of any radius! Unfortunately, depends linearly on the velocity v of the car.So, unless we can come up with a quick easy wayof varying the kinetic energy of a big flywheel inconcert with the square of the cars speed, werenot going to get rich installing these devices ashigh-performance vehicle options.

7. K&K problem 8.2

(a.) The acceleration of the truck is A, andthe mass and width of its rear door are M andw. The door starts fully open. The door canbe thought of as a series of thin sticks, pivotedabout their ends. The moment of inertia of thedoor is thus

I=13

M w2

The easiest way to find the angular velocity ofthe door is to use work and energy. The rota-tional kinetic energy is given by

K=1

2I2

The work done by a torque on a system is givenby

W= d

In this system there is one torque of interest.We use the hinge of the door as the origin, sothe only torque comes from the fictitious force ofacceleration. When the door has swung throughan angle , this torque is given by

=1

2M Aw cos

Note that w/2 is just the distance to the centerof mass. From this we can easily calculate thework done from 0 to 90 degrees.

W=

/20

1

2M Aw cos d=

1

2M Aw

Substituting the expression for rotational kineticenergy, we find the angular velocity of the doorafter it has swung through 90:

1

2I2 =

1

2M Aw =

3A

w

(b.) The force on the door needs to do two

things. It needs to accelerate the door at a rateA, and it needs to provide the centripetal ac-celeration to make the door rotate. At =90degrees, the torque is zero, so the angular veloc-ity is not changing. Instantaneously, the door isin uniform circular motion. The force requiredto accelerate the door is just

FA= M A


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7

The force required to provide the centripetalacceleration is

Fc = M 2

w

2

The total force is the sum of these, and they actin the same direction. We substitute the valuefor from part (a.) to get

F =FA+ Fc = M A +3

2MA=

5

2M A

8. K&K problem 8.4

(a.) This is a torque balance problem. A car ofmassm has front and rear wheels separated by adistance l, and its center of mass is midway be-tween the wheels a distance d off the ground. Ifthe car accelerates at a rateA, it feels a fictitiousforce acting on the center of mass. This tends to

lift the front wheels. When the front wheels areabout to lift off the ground, the normal force onthe front wheels,Nfis zero. This means that thenormal force on the back wheels Nb must equalthe weight of the car,Nb= mg. The simplest ori-gin to use in this problem is the point on the roaddirectly under the center of mass. Here there areonly three torques, due to the two normal forcesand fictitious force. The torque from Nb exactlybalances the torque from the fictitious force whenthe wheels are about to lift, so we have

12

Nbl= mAd = 12

mgl A= l2d

g

For the numbers given,A= 2g= 19.6 m/sec2.

(b.) For deceleration at a rate g, again we sim-ply apply torque balance about the same origin.We also need the fact that

Nf+ Nb= mg

Torque balance gives

1

2Nfl

1

2Nbl mgd = 0

Substituting from the previous equation, we get

Nf=

1

2+

d

l

mg

Nb=

1

2 d

l

mg

Plugging in the numbers, we get Nf= 3mg/4 =2400 lb, and Nb= mg/4 = 800 lb.


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PROBLEM SET 8

1. K&K problem 8.5 Many applications....

2. K&K problem 8.11 A high speed hydro-foil....

3. K&K problem 9.3 A particle moves....

4. K&K problem 9.4 For what values ofn....

5. K&K problem 9.6 A particle of mass m....

6. K&K problem 9.12 A space vehicle is incircular orbit....

7. A satellite of mass m is travelling at speedV in a circular orbit of radius R under the

gravitational force of a fixed mass at the origin.

a. Taking the potential energy to be zero atinfinite radius, show that the total mechanicalenergy of the satellite is mV2/2.

b. At a certain pointB in the orbit (see fig-ure), the direction of motion of the satellite issuddenly changed without any change in themagnitude of the velocity. As a result, thesatellite goes into an elliptical orbit. Its closestapproach to the origin is now R/5. What is thespeed of the satellite at this distance, expressedas a multiple ofV?

c. Through what angle (see figure) was thevelocity of the satellite turned at point B?

8. The commander of a spaceship that has shutdown its engines and is coasting near a strange-appearing gas cloud notes that the ship is follow-ing a path that will take it directly intothe cloud(see the figure). She also deduces from the ships

motion that its angular momentum with respect

to the cloud is not changing. What attractive(central) force could account for such an orbit?


57/289

1




1. K&K problem 8.5

A gyroscope with mass M has angular veloc-ity s and moment of inertia Is. It pivots atone end, and the center of mass is a distance lfrom the pivot. The angular momentum of thegyroscope is thus

L= Iss

The gyroscope undergoes an acceleration a per-pendicular to the spin axis. The fictitious force

will create a torque of magnitude

=M al

The direction of this torque is perpendicular toboth the acceleration and the gyroscope axis(down in the figure), causing the gyroscope axisto precess in the direction indicated by the an-gle . The magnitude of the angular momentumwill not change, but the direction will. Thus thegyroscope axis will rotate around the direction

of acceleration. The rate at which this happensis , anddL

dt =L =

This gives the following relation

Mla(t) =Iss(t)

Both the acceleration and the angular velocitycan depend on time. If we integrate both sidesof this equation, we can get a relation betweenthe final velocity and the total angle of rotation.

The integral of the acceleration is just the ve-locity and the integral of the angular velocity isjust the angle:

M lv = Iss v= IssMl

2. K&K problem 8.11

A hydrofoil moves with respect to the earthssurface at the equator with velocity v = 200mi/hr directed along each of the four points ofthe compass. At rest with respect to the surfaceof the earth, the acceleration of gravity is g. Weare asked to find the effective gravitational ac-celeration g that is felt by a passenger (of massm) who is at rest with respect to the hydrofoil.

The (fictitious) Coriolus force on the pas-senger, which is proportional to the passengers(vanishing) velocity in the hydrofoils frame,

must be zero if evaulated in this frame. The(fictitious) centrifugal force on the passenger is

Fcent = m ( R)= m(( R) R( ))=m(( R) + R 2)

where (v) is the totalangular velocity of thepassenger, due both to the rotation of the earthand to the motion of the hydrofoil; R = Rx isa vector pointing from the earths center to thehydrofoil at the equator; and the bac cab rule

is applied to the first line. The hydrofoils ve-locity has one of the four directions (E,W,N,S)= (y,y, z,z), yielding an angular velocity due to hydrofoil motion relative to the earthssurface:

= v

R(z,z,y, y)

To this one must add the earths angular velocity

= z

in order to get the total angular velocity = + of the hydrofoil. Evidently is perpen-

dicular to R, so

Fcent(v) =xmR((v))2

Therefore g points alongx, i.e. toward theearths center, for all four directions (E,W,N,S)ofv, as does g. Thus

g gg g

g =

R(( 2 2)g


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2

For these four directions,

2 2 = 2+ 2 (E)= 2+ 2 (W)=2 (N and S)

Therefore

g

g =

R

g(2 2) (E)

=R

g(2 2 (W)

=R

g(2) (N and S)

Using / = 0.1931 and R2/g = 0.003432, wecalculate

|g/g

|= 0.001325 from the 2 term

and|g/g| = 0.000128 from the 2 term. Thus

g/g = 0.001453 (E)= +0.001197 (W)

= 0.000128 (N and S)

3. K&K problem 9.3

A particle moves in a circle under the influenceof an inverse cube law force. This means that thepotential is inverse squared and it is attractive.

The effective potential is given by

Ueff(r) = L2

2mr2 A

r2

The radial force is zero for a circular orbit, sowe can find the radius.

0 =dUeff

dr = L

2

mr3+

2A

r3

This shows that a circular orbit can have any ra-dius, but there is only one possible magnitude ofangular momentum, given by

L2 = 2Am

Plugging this value of the angular momentuminto the effective potential, we find the peculiarresult that

Ueff= 0

SinceUeff is constant,d2r/dt2 = 0, so if the par-

ticle acquires a nonzero radial velocity it willcontinue with the same radial velocity. If theparticle moves with uniform radial velocity vr,the following equations are satisfied

drdt

=vrddt

= Lmr2(t)

Solving the first is easy: r(t) =r0+vrt. Pluggingthis result into the second equation, we find

d

dt =

L

m(r0+ vrt)2

We can solve this equation by direct integration,assuming that (0) = 0:

(t) = t0

L dtm(r0+ vrt)2

= Lmvr

1r0 1

r(t)

We replaceL with

2mAto get the final answer

(t) = 1

vr

2A

m

1

r0 1

r(t)

4. K&K problem 9.4

A particle moves in a circular orbit in the po-tentialU= A/rn. We want to know for whichvalues of n the orbit is stable. The effectivepotential is given by

Ueff= L2

2mr2 A

rn

To find the circular orbit radius we evaluatedUeff/dr= 0:

dUeffdr

= L2

mr3+

nA

rn+1

This gives the radius of the circular orbit r0when we set it to zero.

rn20 =nAm

L2

Since rn20 must be a positive quantity for anyvalue ofn, and A >0, this equation requires

n >0


59/289

3

We now look at the second derivative of theeffective potential atr = r0. If it is positive, thenit is a potential minimum and the orbit is stable.

d2Ueffdr2

= 3L2

mr4 n(n + 1)A

rn+2

= 1

r4

3L2m n(n + 1)A

rn2

>0

Sincer is always greater than zero we can divideit away. Substituting for rn2 at r = r0,

3L2

m (n + 1)L

2

m >0 n


60/289

4

for the second. The energy input needed to gofrom one orbit to the other is at least

E= GMmRe

1

8 1

4

=

GMm

8Re

Plugging in the values given,

E= 2.34 1010 joules

(b.) At point A, the rocket is fired, puttingthe spacecraft in an elliptical orbit. The majoraxis of this orbit is A= 6Re. To find the initialspeed, we use the energy equation. The energy ispartly gravitational potential energy and partlykinetic energy:

E=1

2 mv20

GMm

2Re = GMm

4Re

Solving this equation for v0, we find the orbitalspeed

v0=

GM

2Re

The energy of the elliptical orbit is given by

E= GMm6Re

E= GMm12Re

The change in energy at point A is entirely dueto a change in speed.

E=1

2mv21

1

2mv20 v1=

2GM

3Re

The change in speed required at point A is thus

vA =

2

3

1

2

GM

Re= 865 m/sec

We repeat this analysis at pointB. Conservation

of angular momentum gives

2Rev1= 4Rev2 v2=

GM

6Re

The energy of the new circular orbit is given by

E= GMm8Re

E= GMm24Re

Again this is due to the change in speed.

E=1

2mv23

1

2mv22 v3=

GM

4Re

Finally we obtain the change in speed at point B

vB =

1

2

1

6

GM

Re= 726 m/sec

Since the two velocities at point A are both tan-gent to each other, and similarly for point B,the only changes in the velocities at either pointare the changes in their magnitudes.

7. A satellite of mass m moves in a circular or-bit of radius R at speed v. It is influenced bythe gravity of a fixed mass at the origin.

(a.) The mechanical energy of the satellite isgiven by

E=1

2mV2 GMm

R

We know that gravity exactly provides the cen-tripetal acceleration.

GMm

R2 =

mV2

R GM

R =V2

The total energy is thus

E= 12 mV2 mV2 = 12 mV2

(b.) At a certain point on the orbit, the directionof travel of the satellite changes. The magnitudeof the velocity does not change, so the total en-ergy of the orbit doesnt change. We can nowfind the kinetic energy at closest approach:

E= 12

mV2 =1

2mv2 5GMm

R

=1

2

mv2

5mV2

v= 3V

(c.) The circular orbit has angular momentumL1 = mRV, while the elliptical orbits angularmomentum, evaluated at the perigee, is

L2= mR

53V =

3

5L1


61/289


62/289


PROBLEM SET 9

1.

a. Expand in Taylor series: f(x) = ln(1 x);f(x) = 1/(1 +x).

b. Given two functions s(x) and c(x) suchthat ds/dx = c and dc/dx = s, prove thats(x) +c(x) = [s(0) +c(0)]ex.

2.

a. French problem 1-4(b).

b. French problem 1-9.

c. Prove DeMoivres theorem,

(cos +i sin )n = cosn+i sinn.

3. French problem 3-15.

4. At t = 0 a bullet of mass m and velocityv0 strikes a motionless block of mass M whichis connected to a wall by a spring of constantk. The block moves with coordinatex(t) (alongthe direction of the bullet) on a frictionless ta-ble next to the wall. The bullet embeds itself inthe block. Ifx(t) = [A exp(i(t+))], with A

real, evaluate , A, and .



7. A piano has middle C = 256 Hz, and C-above-middle-C = 512 Hz. The white keys ofits middle octave consist of middle C; D (1 stepabove middle C); E (2 steps); F (2.5); G (3.5); A(4.5); B (5.5); and C-above-middle-C (6 steps).Each step causes the frequency to be multipliedby a fixed factor.

a. Find the frequencies of D, E, F, G, A, andB.

b. If G were tuned to a perfect fifth, its sec-ond harmonic (2 the fundamental frequency)would be the same as middle Cs third harmonic.Find thebeat frequencybetween the second har-monic of G and the third harmonic of middle C.(Pianos are tuned by listening for such beats.)

Stringed instruments are tuned (for compositions

in the key of C) so that this beat frequency iszero, producing a smoother tone. When a com-position in a different key is played, the stringedinstrument can be retuned for that key, whichwould be impossible for the piano.

8. For an underdamped undriven harmonic os-cillator with 0/ Q = 100, find the numberof oscillations required to reduce the amplitudeof oscillation by the factor e 23.1.


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64/289

2

to E = E0et . We define the Q value as

Q 0/.(a.) Middle C on a piano is played, and the en-ergy decreases to half of its initial value in onesecond. The frequency is 256 Hz. The angular

frequency is this times 2, so 0 =1608.5/sec.We find from

1

2= (e/2)2 = 0.693

Lastly , the Q of the oscillator is

Q= 1608.5/0.693 = 2321

(b.) The note one octave above is struck (512Hz). The decay time is the same, so the Q value

is simply doubled: Q=4642.(c.) A damped harmonic oscillator has massm= 0.1 kg, spring constant k = 0.9 N/m, and adamping constant b. The energy decays to 1/ein 4 seconds. This means that

1

e=e4 = 0.25 sec1

b= m= 0.025 kg sec1

The natural frequency 0 =

k/m = 3 Hz. Fi-

nally, the Q of the oscillator is Q= 0/= 12.

4. Att= 0, the bullet collides inelastically withthe block, so only the momentum is conserved.The final velocity of the block and bullet is givenby

mv0= (M+ m)v v= mv0M+ m

We now have the initial conditions for the os-cillation. The initial position is x(0) = 0 andthe initial velocity is v(0) =v . The frequency

is given as usual by

k/mass, but the mass inquestion is the total mass of the system:

=

k

M+m

The solution is given by x(t) =[A exp(i(t+))]. The initial position tells us that cos= 0.This is ambiguous because the cosine is zero in

two places in one oscillation. We want a placewhere it is zero and rising, because we know thatx is increasing at the instant of contact. This is

= /2

The solution is now

x(t) = [A exp(i(t /2)])

We differentiate to get the velocity, and evaluatethis att= 0.

v(0) =v = A sin(/2) =A A= v

This gives the final result for the amplitude

A= mv0k(M+ m)

5. French 4-5.

(a.) A pendulum is forced by moving the pointof support. The coordinatex gives the locationof the pendulum bob, and gives the locationof the point of support. The forces on the pen-dulum are the damping force, which we mustassume to be proportional to the absolute veloc-ity of the pendulum, and the force of gravity.The force of gravity depends on the angle by

which the pendulum is raised. This is propor-tional to the distance that the pendulum bob isdisplaced from the point of support, x . Thisgives the equation of motion

md2x

dt2 = bdx

dt mg

l (x )

Using20 =g/l and =b/m, we put this in thestandard form with as a forcing term.

d2x

dt2 +

dx

dt

+20x= 20

(b.) The motion of the point of support is givenby (t) = 0cos t. We use the formula for theamplitude of forced oscillation, but we note thatin this case, the equation is

d2x

dt2 +

dx

dt +20x=

200cos t


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3

The constant 200 takes the place of the F0/mwe normally see in this type of equation. Theamplitude of the oscillation is thus given by

A() = 200

(

2

0 2

)2

+2

2

At exact resonance, = 0 and the amplitude is

A(0) =00

=Q0

Now we want to find Q. We are given that theforcing amplitude is 0 = 1 mm. The length ofthe pendulum is l = 1 m, so this gives 0= 3.13sec1. We know that the amplitude falls off bya factor ofe after 50 swings, or 50 periods. We

know that A= A0exp(t/2) so t = 2. t is 50periods, or 100/, where is the frequency offree oscillation

=

20+ 2/4

Thent = 2

100

= 2

50= 20+

2/4

(50)22 =20+ 2/4

Plugging in the numbers, we get = 0.

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