physics fundamentals.pdf

Fundamentals of Physics Halliday & Resnic

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CH11Rotational Motion II 5, 7, 9, 13, 15, 23, 25, 27, 31, 35, 39, 41, 45, 53, 55, 57, 58, 59, 61, 67

Problem 11-5 A 140 kg hoop rolls along a horizontal floor so that the hoops center of mass has a speed of 0.150 m/s. How much work must be done on the hoop to stop it?

140 0.150/

2 21 12 2

K I mv= + 2I mr= 140m kg=

0.15 /v m s= v

r =

2 21 12 2

K mvI= + 2 2 21 1( )( )2 2

vr

m vr m= + 2mv= 2(140)(0.15) 3.15J= =

Problem 11-7 In Fig. 11-31, a constant horizontal force appF

G of magnitude 10 N is applied to a wheel of mass 10

kg and radius 0.30 m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 20.6 /m s . (a) In unit-vector notation, what is the frictional force on the wheel? (b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?

11-31 appFG

10N 10 0.30m 20.6 /m s ab

11-31

10appF Ni=

G (a) app sF f ma = 10 (10)(0.6) 4s appf F ma N= = =


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(b) 0.3R m= 20.6 2 /

0.3coma rad sR

= = = I =

22(0.3 )(4 ) 0.62 /m NI kg m

rad s= = =

Problem 11-9 In Fig. 11-33, a solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance 6L m= down a roof that is inclined at the angle 030 = . (a) What is the angular speed of the cylinder about its center as it leaves the roof? (b) The roofs edge is at height

5H m= . How far horizontally from the roofs edge does the cylinder hit the level ground?

11-33 10cm 12 6L m= 030 = ab

5H m=

11-33

(a) We find its angular speed as it leaves the roof using conservation of energy. Its initial

kinetic energy is Ki = 0 and its initial potential energy is Ui = Mgh where 6.0sin 30 3.0 mh = = (we are using the edge of the roof as our reference level for

computing U). Its final kinetic energy (as it leaves the roof) is (Eq. 11-5) K Mv If = +12 2 12 2 .

Here we use v to denote the speed of its center of mass and is its angular speed at the moment it leaves the roof. Since (up to that moment) the ball rolls without sliding we can set v = R = v where R = 0.10 m. Using I MR= 12 2 (Table 10-2(c)), conservation of energy leads to

2 2 2 2 2 2 2 21 1 1 1 3 .2 2 2 4 4

Mgh Mv I MR MR MR = + = + = The mass M cancels from the equation, and we obtain

= = =1 43

1010

43

9 8 30 63R

gh.

. . .m

m s m rad s2c hb g (b) Now this becomes a projectile motion of the type examined in Chapter 4. We put the

origin at the position of the center of mass when the ball leaves the track (the initial


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position for this part of the problem) and take +x leftward and +y downward. The result of part (a) implies v0 = R = 6.3 m/s, and we see from the figure that (with these positive direction choices) its components are

0 0

0 0

cos30 5.4 m ssin 30 3.1 m s.

x

y

v vv v

= == =

The projectile motion equations become

x v t y v t gtx y= = +0 0 212and . We first find the time when y = H = 5.0 m from the second equation (using the quadratic formula, choosing the positive root):

20 0 2 0.74s.y y

v v gHt

g + += =

Then we substitute this into the x equation and obtain x = =54 0 74 4 0. . .m s s m.b gb g Problem 11-13 A bowler throws a bowling ball of radius 11R cm= along a lane. The ball (Fig. 11-37) slides on the lane with initial speed ,0 8.5 /comv m s= and initial angular speed 0 0 = . The coefficient of kinetic friction between the ball and the lane is 0.21. The kinetic frictional force kf

G acting on the

ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed comv has decreased enough and angular speed has increased enough, the ball stops sliding and then rolls smoothly. (a) What then is comv in terms of ? During the sliding, what are the balls (b) linear acceleration and (c) angular acceleration? (d) How long does the ball slide? (e) How far does the ball slide? (f) What is the linear speed of the ball when smooth rolling begins? 11R cm= 11-37 ,0 8.5 /comv m s= 0 0 = 0.21 comv a comv bcdef

11-37

(a) We choose clockwise as the negative rotational sense and rightwards as the positive

translational direction. Thus, since this is the moment when it begins to roll smoothly, Eq. 11-2 becomes v Rcom m= = 011. .b g This velocity is positive-valued (rightward) since is negative-valued (clockwise) as shown in Fig. 11-57.


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(b) The force of friction exerted on the ball of mass m is kmg (negative since it points left), and setting this equal to macom leads to

a gcom2 2m s m s= = = 0 21 9 8 21. . .b g c h

where the minus sign indicates that the center of mass acceleration points left, opposite to its velocity, so that the ball is decelerating.

(c) Measured about the center of mass, the torque exerted on the ball due to the frictional force is given by = mgR . Using Table 10-2(f) for the rotational inertia, the angular acceleration becomes (using Eq. 10-45)

( )( )( )

22

2

5 0.21 9.8 m/s5 47 rad s2 5 2 2 0.11 m

mgR gI m R R = = = = =

where the minus sign indicates that the angular acceleration is clockwise, the same direction as (so its angular motion is speeding up).

(d) The center-of-mass of the sliding ball decelerates from vcom,0 to vcom during time t according to Eq. 2-11: v v gtcom com,0= . During this time, the angular speed of the ball increases (in magnitude) from zero to according to Eq. 10-12:

= = =t gtR

vR

52

com

where we have made use of our part (a) result in the last equality. We have two equations involving vcom, so we eliminate that variable and find ( )

( )( )com,0 22 2 8.5 m/s

1.2 s.7 7 0.21 9.8 m/sv

tg= = =

(e) The skid length of the ball is (using Eq. 2-15)

( ) ( )( ) ( )( )( )22 2com,0 1 18.5 m/s 1.2 s 0.21 9.8 m/s 1.2 s 8.6 m.2 2x v t g t = = = (f) The center of mass velocity at the time found in part (d) is

( )( )( )2com com,0 8.5 m/s 0.21 9.8 m/s 1.2 s 6.1 m/s.v v gt= = = Problem 11-15 Nonuniform ball. In Fig. 11-39, a ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.48 m. The initial height of the ball is 0.36h m= . At the loop bottom, the magnitude of the normal force on the ball is 2.00Mg.The ball consists of an outer spherical shell (of a certain uniform density) that is glued to a central sphere (of a different uniform density). The rotational inertia of the ball can be expressed in the general form 2I MR= , but is not 0.4 as it is for a ball of uniform density. Determine . 11-39 M R 0.48m 0.36h m= 2Mg

2I MR= 0.4


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11-39

The physics of a rolling object usually requires a separate and very careful discussion (above and beyond the basics of rotation discussed in chapter 10); this is done in the first three sections of chapter 11. Also, the normal force on something (which is here the center of mass of the ball) following a circular trajectory is discussed in section 6-6 (see particularly sample problem 6-7). Adapting Eq. 6-19 to the consideration of forces at the bottom of an arc, we have

FN Mg = Mv2/r which tells us (since we are given FN = 2Mg) that the center of mass speed (squared) is v2 = gr, where r is the arc radius (0.48 m) Thus, the balls angular speed (squared) is

2 = v2/R2 = gr/R2, where R is the balls radius. Plugging this into Eq. 10-5 and solving for the rotational inertia (about the center of mass), we find

Icom = 2MhR2/r MR2 = MR2[2(0.36/0.48) 1] . Thus, using the notation suggested in the problem, we find = 2(0.36/0.48) 1 = 0.50.

Problem 11-23 Force ( 8 ) (6 )F N i N j= +G acts on a particle with position vector (3 ) (4 )r m i m j= +G .What are (a) the torque on the particle above the origin, in unit-vector notation, and (b) the angle between the directions of rG and FG ? ( 8 ) (6 )F N i N j= +G (3 ) (4 )r m i m j= +G ab rG FG x y zF F i F j F k= + +

G r xi y j zk= + +G

x y z

i j kr F x y z

F F F =

GG

y z x yx z

y z x yx zi j k

F F F FF F= +

( ) ( ) ( )z y x z y xyF zF i zF xF j xF yF k= + + (a) r F = GG G (3)(6) (4)( 8) 50k N m= = (b) sinr F rF =GG 2 2 5r x y m= + = 2 2 10x yF F F N= + = r F = GG G 50 (5)(10)sin=

sin 1 = 1 0sin (1) 90 = =


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Problem 11-25 Force (2 ) (3 )F N i N k= G acts on a pebble with potion vector (0.5 ) (2 )r m j m k= G relative to the origin. In unit-vector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point (2 ,0, 3 )m m ? (2 ) (3 )F N i N k= G (0.5 ) (2 )r m j m k= G ab (2 ,0, 3 )m m x y zF F i F j F k= + +

G r xi y j zk= + +G

x y z

i j kr F x y z

F F F =

GG

y z x yx z

y z x yx zi j k

F F F FF F= +

( ) ( ) ( )z y x z y xyF zF i zF xF j xF yF k= + + (a) (0,0.5, 2)r = G (2,0, 3)F = G 1.5 4 1i j k N m = G (b) (0,0.5, 2) (2,0, 3) ( 2,0.5,1)r = = G

(2,0, 3)F = G 1.5 4 1i j k N m = G

Problem 11-27 In the instant of Fig. 11-41, two particles move in an xy plane. Particle 1P has mass 6.5 kg and speed 1 2.2 /v m s= , and it is at distance 1 1.5d m= from point O. Particle 2P has mass 3.1kg and speed 2 3.6 /v m s= , and it is at distance 2 2.8d m= from point O. What are the (a) magnitude and (b) direction of the net angular momentum of the two particles about O? 11-41 xy 1P 6.5 1 2.2 /v m s= O 1 1.5d m= 2P 3.1 2 3.6 /v m s= O 2 2.8d m= Oab

11-41


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r P= G GGA 1P ( )( )( ) 21 1 1 2.8 m 3.1 kg 3.6 m/s 31.2 kg m s.r mv= = = A +z

2P ( )( )( ) 22 2 2 1.5 m 6.5 kg 2.2 m/s 21.4 kg m s.r mv= = = A -z

(a) 21 2 31.2 21.4 9.8 /L kg m s= = = A A (b) +z

Problem 11-31 In Fig. 11-43, a 0.400 kg ball is shot directly upward at initial speed 40.0 m/s. What is its angular momentum about P, 2.00 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground? 11-43 0.4kg 40m/sP 2 P ab P cd

11-43

(a) 0 0

(b) With the convention (used in several places in the book) that clockwise sense is to be associated with the negative sign, we have L = r m v where r = 2.00 m, m = 0.400 kg, and v is given by free-fall considerations (as in chapter 2). Specifically, ymax is determined by Eq. 2-16 with the speed at max height set to zero; we find ymax = vo2/2g where vo = 40.0 m/s. Then with y = 12 ymax, Eq. 2-16 can be used to give v = vo / 2 . In this way we arrive at L = 22.6 kg.m2/s.

(c) (2)(0.4)(9.8) 7.84r F r mg N m = = = = (d) r F =

(2)(0.4)(9.8) 7.84r F r mg N m = = = = Problem 11-35 At time t, 2 24 (2 6 )r t i t t j= +G gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( rG is in meters and t is in seconds). (a)Find an expression for the torque acting on the particle relative to the origin. (b) Is the magnitude of the particles angular momentum relative to the origin increasing, decreasing, or unchanging?


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t 3kg 2 24 (2 6 )r t i t t j= +G xy rGtab

(a) 8 (2 12 )drv ti t jdt

= = +GG

2 2( ) (3) 4 2 6 08 2 12 0

i j kL r p m r v t t t

t t= = =

G G G G G 2(3)(8 )t k=

netdLdt

=GG (48 )tk N m=

(b) From our (intermediate) result in part (a), we see the angular momentum increases in proportion to t2.

Problem 11-39 In Fig. 11-45, three particles of mass 23m g= are fastened to three rods of length 12d cm= and negligible mass. The rigid assembly rotates around point O at angular speed 0.85 /rad s = . About O, what are (a) the rotational inertia of the assembly, (b) the magnitude of the angular momentum of the middle particle, and (c) the magnitude of the angular momentum of the assembly?

11-45 23m g= 12d cm= O 0.85 /rad s = Oabc

11-45

(a) 2 2 2(3 ) (2 )I m d m d md= + + 214md= 2 2 3 2(14)(2.3 10 )(0.12) 4.6 10 kg m = =

(b) 2 2(2 ) 4I m d md= = 24L I md = = 2 2 3 24(2.3 10 )(0.12) (0.85) 1.1 10 /kg m s = = (c) 214L I md = = 2 2 3 214(2.3 10 )(0.12) (0.85) 3.9 10 /kg m s = =

Problem 11-41 Figure 11-47 shows a rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 2.5 s. Assuming 0.5R m= and 2m kg= , calculate (a) the structures rotational inertia about the axis of rotation and (b) its angular momentum about that axis.


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11-47 R m R 2.5ab

11-47

(a) For the hoop, we use Table 10-2(h) and the parallel-axis theorem to obtain I I mh mR mR mR1

2 2 2 212

32

= + = + =com . Of the thin bars (in the form of a square), the member along the rotation axis has (approximately) no rotational inertia about that axis (since it is thin), and the member farthest from it is very much like it (by being parallel to it) except that it is displaced by a distance h; it has rotational inertia given by the parallel axis theorem:

I I mh mR mR22 2 20= + = + =com .

Now the two members of the square perpendicular to the axis have the same rotational inertia (that is I3 = I4). We find I3 using Table 10-2(e) and the parallel-axis theorem:

I I mh mR m R mR32 2

221

12 213

= + = + FHGIKJ =com .

Therefore, the total rotational inertia is

I I I I mR1 2 3 4219

616+ + + = = . .kg m2

(b) The angular speed is constant:

= = =t22 5

2 5.

. rad s.

Thus, L I= = total 2kg m s. 4 0. Problem 11-45 A wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels? (b) What fraction of the original rotational kinetic energy is lost? 800 /ab


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2 12I I= 1 800 / minrev =

(a) 1 1 2( )i fI I I = + 1 1

1 2 1 1

( )(800) 267 / min( ) (2 )

if

I I revI I I I

= = =+ +

(b) 2112i i

K I = 21 2

1 ( )2f f

K I I = + 211 11 1

1 ( 2 )( )2 2

iII II I

= + +2

11 1( )3 2 i

I = 13 i

K=

1

23 0.673

i ii f

i i

K KK KK K

= = = Problem 11-53 A horizontal vinyl record of mass 0.10 kg and radius 0.10 m rotates freely about a vertical axis through its center with an angular speed of 4.7 rad/s. The rotational inertia of the record about its axis of rotation is 4 25 10 kg m . A wad of wet putty of mass 0.020 kg drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately after the putty sticks to it?

0.1 0.10 4.7rad/s 4 25 10 kg m 0.02kg For simplicity, we assume the record is turning freely, without any work being done by its

motor (and without any friction at the bearings or at the stylus trying to slow it down). Before the collision, the angular momentum of the system (presumed positive) is Ii i where Ii = 50 10 4. kg m2 and i = 4 7. .rad s The rotational inertia afterwards is

I I mRf i= + 2 where m = 0.020 kg and R = 0.10 m. The mass of the record (0.10 kg), although given in the problem, is not used in the solution. Angular momentum conservation leads to

I I II mRi i f f f

i i

i

= = + =2 3 4. rad / s. Problem 11-55 A uniform thin rod of length 0.500 m and mass 4.00 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.00 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullets path makes angle 060 = with the rod (Fig. 11-52). If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullets speed just before impact?

0.5 4


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3g 060 = 11-52 10 rad/s

11-52

0.5m 0.25r m= 0sin (0.25 )(0.003 )( )sin 60rmv m kg v = 2rodI I mr= +

22

12ML mr= +

2

2( )12

ML mr +2

2(4)(0.5)[ (0.003)(0.25) ](10)12

= +

2

2sin ( )12

MLrmv mr = +

22( )

12sin

ML mrv

rm

+=

22

30

(4)(0.5)[ (0.003)(0.25) ](10)12 1.3 10 /

(0.25)(0.003)sin 60m s

+= =

Problem 11-57 Figure 11-53 is an overhead view of a thin uniform rod of length 0.800 m and mass M rotating horizontally at angular speed 20.0 rad/s about an axis through its center. A particle of mass M/3.00 initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particles speed pv is 6.00 m/s greater than the speed of the rod end just after ejection, what is the value of pv ?

11-53 0.8M 20/ M/3 pv 6/ pv

11-53

By angular momentum conservation (Eq. 11-33), the total angular momentum after the explosion must be equal to before the explosion:

p r p rL L L L + = +


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L2 mvp + 112 ML2 = Ip + 112 ML2 where one must be careful to avoid confusing the length of the rod (L = 0.800 m) with the angular momentum symbol. Note that Ip = m(L/2)2 by Eq.10-33, and

= vend/r = (vp 6)/(L/2), where the latter relation follows from the penultimate sentence in the problem (and 6 stands for 6.00 m/s here). Since M = 3m and = 20 rad/s, we end up with enough information to solve for the particle speed: vp = 11.0 m/s.

Problem 11-58 In Fig. 11-54, a 1.0 g bullet is fired into a 0.50 kg block attached to the end of a 0.60 m nonuniform rod of mass 0.50 kg. The block rod bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertia of the rod alone about that axis at A is 20.06kg m . Treat the block as a particle (a) What then is the rotational inertia of the block rod bullet system about point A? (b) If the angular speed of the system about A just after impact is 4.5 rad/s, what is the bullets speed just before impact?

11-54 1g 0.5gk 0.6 0.5kg-- A A 20.06kg m a A--b 4.5 rad/s

11-54

(a) With r = 0.60 m, we obtain I = 0.060 + (0.501)r2 = 0.24 kg m2. (b) Invoking angular momentum conservation, with SI units understood,

( ) ( ) ( )( )0 0 00.001 0.60 0.24 4.5fL mv r I v= = =A which leads to v0 = 1.8 103 m/s.

Problem 11-59 A uniform disk of mass 10m and radius 3.0r can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass m and radius r lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of 20 rad/s. Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their angular velocity about the center of


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the larger disk? (b) What is the ratio 0/K K of the new kinetic energy of the two-disk system to the systems initial kinetic energy?

10m 3r m r 20rad/sa

b 0/K K

10M m= 3R r= 2 2 21 1 90(10 )(3 )2 2 2big

I MR m r mr= = =

m m= r r= 212small

I mr=

2 2 2090 1 912 2 2big small

I I I mr mr mr= + = + = 3 2h R r r r r= = = 20 ( )big small big smallI I I I I mh= + = + + 2 2 2 290 1 99(2 )2 2 2mr mr m r mr= + + =

(a) 0 0I I =

0 0II

=2

2

912 (20) 18 /992

mrrad s

mr= =

(b) 0

9991

II

= 0

9199

=

2

200 0

1212

IKK I

= 2 20 0

99 91( ) ( ) 0.9291 99

II

= = =

Problem 11-61 The uniform rod (length 0.60 m, mass 1.0 kg) in Fig. 11-55 rotates in the plane of the figure about an axis through one end, with a rotational inertia of 20.12kg m . As the rod swings through its lowest position, it collides with a 0.20 kg putty wad that sticks to the end of the rod. If the rods angular speed just before collision is 2.4 rad/s, what is the angular speed of the rodputty system immediately after collision?

0.6 1 11-55 20.12kg m 0.2

2.4/-


25 26 !

11-55

We make the unconventional choice of clockwise sense as positive, so that the angular

velocities in this problem are positive. With 0.6r m= and 20 0.12I kg m= , the rotational inertia of the putty-rod system (after the collision) is

2 20 (0.2) 0.19I I r kg m= + =

Invoking angular momentum conservation 0 fL L= or 0 0I I = , we have

( )20 0 20.12 kg m 2.4 rad/s 1.5rad/s.0.19 kg mII

= = = Problem 11-67 Two 2.00 kg balls are attached to the ends of a thin rod of length 50.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (Fig. 11-60), a 50.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 3.00 m/s and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle will the system rotate before it momentarily stops?

2 50 11-60 50 3.00/abc

11-60

(a) If we consider a short time interval from just before the wad hits to just after it hits and sticks, we may use the principle of conservation of angular momentum. The initial


26 26

angular momentum is the angular momentum of the falling putty wad. The wad initially moves along a line that is d/2 distant from the axis of rotation, where d = 0.500 m is the length of the rod. The angular momentum of the wad is mvd/2 where m = 0.0500 kg and v = 3.00 m/s are the mass and initial speed of the wad. After the wad sticks, the rod has angular velocity and angular momentum I, where I is the rotational inertia of the system consisting of the rod with the two balls and the wad at its end. Conservation of angular momentum yields mvd/2 = I where

I = (2M + m)(d/2)2 and M = 2.00 kg is the mass of each of the balls. We solve

mvd M m d2 2 2 2= +b gb g for the angular speed:

( )( )( )

( )( )( )2 0.0500 kg 3.00 m/s2 0.148 rad s.

2 2 2.00 kg 0.0500 kg 0.500 mmv

M m d = = =+ +

(b) The initial kinetic energy is K mvi = 12 2 , the final kinetic energy is K If = 12 2 , and their ratio is K K I mvf i = 2 2 . When I M m d= +2 42b g and = +2 2mv M m db g are substituted, this becomes

( )0.0500 kg 0.0123.

2 2 2.00 kg 0.0500 kgf

i

K mK M m

= = =+ + (c) As the rod rotates, the sum of its kinetic and potential energies is conserved. If one of the

balls is lowered a distance h, the other is raised the same distance and the sum of the potential energies of the balls does not change. We need consider only the potential energy of the putty wad. It moves through a 90 arc to reach the lowest point on its path, gaining kinetic energy and losing gravitational potential energy as it goes. It then swings up through an angle , losing kinetic energy and gaining potential energy, until it momentarily comes to rest. Take the lowest point on the path to be the zero of potential energy. It starts a distance d/2 above this point, so its initial potential energy is Ui = mgd/2. If it swings up to the angular position , as measured from its lowest point, then its final height is (d/2)(1 cos ) above the lowest point and its final potential energy is

U mg df = 2 1b gb gcos . The initial kinetic energy is the sum of that of the balls and wad:

( )( )22 21 1 2 2 .2 2i

K I M m d = = + At its final position, we have Kf = 0. Conservation of energy provides the relation:

mg d M m d mg d2

12

22 2

12

2+ + FHGIKJ = b g b g cos .

When this equation is solved for cos , the result is

( )

( )( ) ( )22 22 2.00 kg 0.0500 kg1 2 1 0.500 mcos 0.148 rad s

2 2 2 20.0500 kg 9.8 m s

0.0226.

M m dmg

+ + = =

=

Consequently, the result for is 91.3. The total angle through which it has swung is 90 + 91.3 = 181.

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