Chapter 10, problem 6: The angular position of a point on the rim of a rotating wheel is given by, (t )
4 .0t 3.0t 2 t 3; 4.
(a) What is the angular velocity at t = 2.0 s? (b)What is the angular velocity velo city at t = 4.0 s. s. If we make the units explicit, the function is
g c
c
4.0 rad / s t 3.0 rad / s2 t 2 1.0 rad / s3 t 3
but generally we will proceed as shown in the problem — letting letting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. (a) Equation 10-6 leads to
Evaluating this at t = = 2 s yields
2 =
c4t 3t dt d
2
h
t 3 4 6t 3t 2 .
4.0 rad/s.
(b) Evaluating the expression in part (a) at t = = 4 s gives
4 =
28 rad/s.
(c) What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at a t t = 4.0 s? avg
2 rad / s2 . 12 rad 42
4
(d) What is the instantaneous angular acceleration at the beginning of this time interval? (e) What is the instantaneous angular acceleration at the end of this time interval? d d 4 6t 3t 2 6 6t . dt dt
c
h
2
Evaluating this at t = = 2 s produces 2 = 6.0 rad/s . Chapter 10, problem 22: An astronaut is being tested in a centrifuge. The centrifuge has a radius of 10 m and, 2 in starting, rotates according to θ = 0.30t , where t is in seconds and θ is in radians. (a) When t = 5.0 s, what is the magnitude of the astronaut's angular velocity? (b) When t = 5.0 s, what is the magnitude of the astronaut's linear speed?
(a) Using Eq. 10-6, the angular velocity velocit y at t = = 5.0s is d d 0.30t 2 dt t 5.0 dt
c h
2 (0.30)(5.0) 3.0 rad / s.
t 5.0
(b) Equation 10-18 gives the linear speed at t = = 5.0s: v r (3.0 rad/s)( rad/s)(10 10 m) 30 m/s. m/s. (c) When t = 5.0 s, what is the magnitude of the astronaut's tangential acceleration? (d) When t = 5.0 s, what is the magnitude of the astronaut's radial acceleration? (c) The angular acceleration is, from Eq. 10-8, d
dt
d dt
(0.60 60t ) 0.60 60 rad / s2 .
Then, the tangential acceleration at t = = 5.0s is, using Eq. 10-22,
c
at r (10 m) 0.60 rad / s2 6.0 m / s2 . (d) The radial (centripetal) acceleration is given by Eq. 10 -23:
g10 mg 90 m / s .
ar 2r 3.0 rad / s
2
2
Chapter 10, problem 29: An early method of measuring the speed of light makes use of a rotating slotted wheel. A beam of light passes through a slot at the outside edge of the wheel, as in the figure, travels to a distant mirror, and returns to the wheel just in time to pass through the next slot in the wheel. One such slotted wheel has a radius of 5.0 cm and 500 slots at its edge. Measurements taken when the mirror was L = 500 m from the wheel 8 indicated a speed of light of 3.00×10 m/s. (a) What was the (constant) angular speed of the wheel? (b) what is the linear speed of a point on the edge of the wheel?
(a) In the time light takes to go from the wheel to the mirror and back again, the wheel turns through an angle of – 2 = 2/500 = 1.26 10 rad. That time is 2 2(500 m) t 3.34 106 s 8 c 2.998 10 m / s so the angular velocity of the wheel is
t
126 . 102 rad 334 . 10
6
s
3.8 103 rad / s.
(b) If r is the radius of the wheel, the linear speed of a point on its rim is
v r 3.8 103 rad/s 0.050 m 1.9 10 2 m/s. 2
Chapter 10, problem 30: A gyroscope flywheel of radius 2.83 cm is accelerated from rest at 14.2 rad/s until its angular speed is 2760 rev/min. (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process?
(a) The tangential acceleration, using Eq. 10-22, is
c
at r 14.2 rad / s2 (2.83 cm) 40.2 cm / s2 . (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (b) In rad/s, the angular velocity is = (2760)(2/60) = 289 rad/s, so ar 2 r (289 rad / s) 2 (0.0283 m) 2.36 103 m / s2 . (c) Through what distance does a point on the rim move during the spin-up? (c) The angular displacement is, using Eq. 10-14, 2 (289 rad/s) 2 2.94 103 rad. 2 2 2(14.2 rad/s )
Then, using Eq. 10-1, the distance traveled is s r (0.0283 m) (2.94 103 rad) 83.2 m. Interlude (parallel axis theorem): consider an arbitrary body of total mass M. Define the coordinate system such that the center of mass has a radial coordinate of 0. Consider an axis of rotation which passes through the center of mass, and then consider an axis of rotation parallel to the aforementioned axis, a distance h away,
ICM R2 dm I R2 dm ( R h)2 dm R 2 dm 2h R dm h2 dm ICM 2h 0 h 2 M ICM h2 M
Interlude (perpendicular axis theorem): consider same setup as above, but let I x and I y be known. Consider
rotation about the z-axis. Then: I z rz 2 dm ( x 2 y 2 ) dm I x I y . Chapter 10, problem 37: Calculate the rotational inertia of a meter stick, with mass 0.56 kg, about an axis perpendicular to the stick and located at the 20 cm mark. (Treat the stick as a thin rod) 2
We use the parallel axis theorem: I = I com + Mh , where I com is the rotational inertia about the center of mass (see Table 10-2(d)), M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies h = 0.50 m – 0.20 m = 0.30 m, 1 1 2 I com ML2 0.56 kg 10 . m 4.67 10 2 kg m2 . 12 12
b
gb g
Consequently, the parallel axis theorem yields
g
g 9.7 10
I 4.67 102 kg m2 0.56 kg 0.30 m
Chapter 10, problem 38: Figure 10-32, shows three 0.0100 kg particles that have been glued to a rod of length L=6.0 cm and negligible mass. The assembly can rotate around a perpendicular axis through point O at the left end. If we remove one particle (that is 33% of the mass) by what percentage does the rotational inertia of the assembly around the rotation axis decrease when
2
2
kg m2 .
that removed particle is a.) the innermost one an d b.) the outermost one?
(a) Equation 10-33 gives 2
2
2
2
I total = md + m(2d ) + m(3d ) = 14 md . 2
2
2
If the innermost one is removed then we would only obtain m(2d ) + m(3d ) = 13 md . The percentage difference between these is (13 – 14)/14 = 0.0714 7.1%. 2
2
2
(b) If, instead, the outermost particle is removed, we would hav e md + m(2d ) = 5 md . The percentage difference in this case is 0.643 64%. Chapter 10, problem 42: The masses and coordinates of four particles are as follows:
m1 50 g; x1 2.0cm; y1 2.0 cm; m3 25 g ; x3 3.0cm; y3 3.0cm;
m2 25 g; x2 0 cm; y2 4.0 cm; m4 30 g; x2 2.0 cm; y2 4.0 cm;
What are the rotational inertias of this collection about the (a) x, (b) y, and (c) z axes?: (a) We apply Eq. 10-33: I x
4
m y i 1
i
2 i
2 2 2 2 50 2.0 25 4.0 25 3.0 30 4.0 g cm 2 1.3 103 g cm 2 .
(b) For rotation about the y axis we obtain I y
4
m x i
2 i
b g b gbg 2
2
bg bg 2
2
50 2.0 25 0 25 3.0 30 2.0 5.5 102 g cm2 .
i 1
(c) There are two ways to solve this problem (1) doing the same as parts (a) and (b), or (2) using the perpendicular axis theorem. Directly calculating (using the fact that the distance from the z axis is I z
4
m cx i
2 i
x 2 y 2 ),
h
yi2 I x I y 1.3 103 5.5 102 1.9 102 g cm2 .
i 1
(d) Clearly, the answer to part (c) is A + B. This verifies the perpendicular axis theorem. Chapter 10, problem 55: In Fig. 10-42a, an irregularly shaped plastic plate with uniform thickness and density (mass per unit volume) is to be rotated around an axle that is perpendicular to the plate face and through point O. The rotational inertia of the plate about that axle is measured with the following method. A circular disk of mass 0.500 kg and radius 2.0 cm is glued to the plate, with its center aligned with point O (Figure b). A string is wrapped around the edge of the disk (in the same the way a string is wrapped around a top). Then the string is pulled for 5.00 s. As a result, the disk and plate are rotated by a constant force of 0.400 N that is applied by the string tangentially to the edg e of the disk. The resulting angular speed is 114 rad/s.
What is the rotational inertia of the plate about the axle?
Combining Eq. 10-34 and Eq. 10-45, we have RF = I , where is given by o = 0 in this case). Computing,
/t (according
to Eq. 10-12, since
Itot Rplate 2 dm Rdisk 2 dm I plate Idisk Iplate 12 MR2 ; 1
2
where I disk = 2 MR (item (c) in Table 10-2). Therefore, I plate =
RFt
1
2
4
– 2 MR = 2.51 10
kg m 2 .
Chapter 10, problem 65: A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 55.0 m. At the instant it makes an angle of 35.0° with the vertical as it falls, what is: a) The radial acceleration of the top? 2
(a) We use conservation of mechanical energy to find an expression for as a function of the angle that the chimney makes with the vertical. The potential energy of the chimney is given by U = Mgh, where M is its
mass and h is the altitude of its center of mass above the ground. When the chimney makes the angle with the vertical, h = ( H /2) cos . Initially the potential energy is U i = Mg ( H /2) and the kinetic energy is zero. The kinetic energy is 21 I 2 when the chimney makes the angle with the vertical, where I is its rotational inertia about its bottom edge. Conservation of energy then leads to 1 MgH / 2 Mg( H / 2) cos I 2 2 ( MgH / I ) (1 cos ). 2
2
The rotational inertia of the chimney about its bas e is I = MH /3 (found using Table 10-2(e) with the parallel axis theorem). Thus
3 g H
(1 cos )
3(9.80 m/s2 ) 55.0 m
(1 cos 35.0) 0.311 rad/s. 2
The radial component of the acceleration of the chimney top is given by ar = H , so 2 2 ar = 3 g (1 – cos ) = 3 (9.80 m/s )(1 – cos 35.0 ) = 5.32 m/s . b) The tangential acceleration of the top? The tangential component of the acceleration of the chimney top is given by at = H , where is the angular acceleration. We are unable to use Table 10-1 since the acceleration is not uniform. Hence, we differentiate 2 = (3 g / H )(1 – cos ) with respect to time, replacing d / dt with , and d / dt with , and obtain d 2 dt
2 (3 g / H ) sin (3 g / 2 H )sin .
Consequently, 3(9.80 m/s2 ) 3 g a H sin sin 35.0 8.43 m/s 2. t 2 2
c) At what angle is the tangential acceleration equal to g? The angle at which at = g is the solution to 3 g g . Thus, sin = 2/3 and we obtain = 41.8°. 2 sin
Chapter 11, Problem 1: A car travels at 80 km/h on a level road in the positive direction of an x axis. Each tire has a diameter of 66 cm. Relative to a woman riding in the car, what are the following values? (a) the velocity v at the center of each tire (b) the velocity v at the top of each tire (c) the velocity v at the bottom of each tire?
The velocity of the car is a constant
v 80 km/h (1000 m/km)(1 h/3600 s) i (22m s)i, ˆ
ˆ
and the radius of the wheel is r = 0.66/2 = 0.33 m. (a) In the car’s reference frame (where the lady p erceives herself to be at rest) the road is moving toward the rear at vroad v 22 m s , and the motion of the tire is purely rotational. In this frame, the center of the tire is ―fixed‖ so vcenter = 0. (b) Since the tire’s motion is only rotational (not translational) in this frame, Eq. 10-18 gives vtop (22m/s)i. ˆ
(c) The bottom-most point of the tire is (momentarily) in firm contact with the road (not skidding) and has the same velocity as the road: v bot tom (22m s)i. This also follows from Eq. 10-18. ˆ
Relative to a woman riding in the car, what are (d) the magnitude a of the acceleration at the center of each tire (e) the magnitude a of the acceleration at the top of each tire (f) the magnitude a of the acceleration at the bottom of each tire? (d) This frame of reference is not accelerating, so ―fixed‖ points within it have zero acceleration; thus, acenter = 0. (e) Not only is the motion purely rotational in this frame, but we also have = constant, which means the only acceleration for points on the rim is radial (centripetal). Therefore, the ma gnitude of the acceleration is v 2 (22 m/s) 2 2 atop 1.5 103 m s . r 0.33 m 3
2
(f) The magnitude of the acceleration is the same as in part (d): a bottom = 1.5 10 m/s . Relative to a hitchhiker sitting next to the road, what are the following values? (g) the velocity v at the center of each tire (h) the velocity v at the top of each tire (i) the velocity v at the bottom of each tire? (g) Now we examine the situation in the road’s frame of reference (where the road is ―fixed‖ and it is the car that appears to be moving). The center of the tire undergoes purely translational motion while points at the rim undergo a combination of translational and rotational motions. The velocity of the center of the tire is v ( 22 m s)i. ˆ
(h) In part (b), we found vtop,car v and we use Eq. 4-39:
vt op , gr ou nd vt op , car vca r, g ro un d v i v i 2v i ˆ
ˆ
ˆ
which yields 2v = +44 m/s. (i) We can proceed as in part (h) or simply recall that the bottom-most point is in firm contact with the (zerovelocity) road. Either way, the answer is zero. Relative to a hitchhiker sitting next to the road, what are the following values? (j) the magnitude a of the acceleration at the center of each tire (k) the magnitude a of the acceleration at the top of each tire (l) the magnitude a of the acceleration at the bottom of each tire (j) The translational motion of the center is constant; it do es not accelerate. (k) Since we are transforming between constant-velocity frames of reference, the accelerations are unaffected. 3 2 The answer is as it was in part (e): 1.5 10 m/s . 3
2
(1) As explained in part (k), a = 1.5 10 m/s . Chapter 11, Problem 3: A 140 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.150 m/s. How much work must be done on the hoop to stop it?
By Eq. 10-52, the work required to stop the hoop is the negative of the initial kinetic energy of the hoop. The 2 initial kinetic energy is K 21 I 2 21 mv 2 (Eq. 11-5), where I = mR is its rotational inertia about the center of mass, m = 140 kg, and v = 0.150 m/s is the speed of its center of mass. Equation 11-2 relates the angular speed to the speed of the center of mass: = v/ R. Thus,
v2 1 2 2 K mR 2 mv mv2 140 kg 0.150 m/s 3.15 J 2 R 2 1
2
which implies that the work required is W K 0 3.15 J 3.15 J . 2
Chapter 11, Problem 17: A yo-yo has a rotational inertia of 800 g·cm and a mass of 120 g. Its axle radius is 3.2 mm and its string is 120 cm long. The yo-yo rolls from rest down to the end of the string. As the yo-yo reaches the end of the string, what are the following values?
(a) magnitude of its linear acceleration? Static friction keeps the string on the yo-yo, so Fr I CM RF Fr , in which
aCM / R aCM , x / R . This lets us compute the friction force as,
F Fr ICM
aCM ,x R2
aCM ,x
R 2 FFr ICM
MR 2 g sin MR2 I CM
The derivation of the acceleration is found in §11-4; Eq. 11-13 gives g acom 1 I com MR02 2 2 where the positive direction is upward. We use I com 950 g cm , M =120g, R0 = 0.320 cm, and g = 980 cm/s and obtain
| acom |
980 cm/s2 1 950 g cm
2
120 g 0.32 cm
2
12.5 cm/s 2 13 cm/s 2.
(b) length of time for yo-yo to reach end of string? Taking the coordinate origin at the initial position, Eq. 2-15 2 leads to ycom 21 acom t . Thus, we set ycom = – 120 cm, and find t
2 ycom acom
2 120cm
12.5 cm s 2
4.38 s 4.4 s.
(c) linear speed at the end of the string? As it reaches the end of the string, its center of mass velocity is given by Eq. 2-11:
vcom acomt 12.5 cm s 2 4.38s 54.8 cm s , so its linear speed then is approximately | vcom | 55 cm/s. (d) translational kinetic energy at the end of the string? The translational kinetic energy is 1 1 2 2 0.120 kg 0.548 m s 1.8 1 0 2 J . K trans mvcom 2 2
(e) rotational kinetic energy at the end of the string? The angular velocity is given by = – vcom/ R0 and the rotational kinetic energy is K rot
1 2
I com 2
2
2
v 1 0.548 m s I com co m (9.50 10 5 kg m 2 ) 1.4 J . 3 2 R 2 3.2 10 m 0 1
(f) angular speed at the end of the string? The angular speed is
vcom R0
0.548 m/s 3
3.2 10 m
1.7 102 rad/s 27rev s .
Note: As the yo-yo rolls down, its gravitational potential energy gets converted into both translational kinetic energy as well as rotational kinetic energy o f the wheel. To show that the total energy remains conserved, we note that the initial energy is U i Mgyi (0.120 kg)(9.80 m/s 2 )(1.20 m) 1.411 J
which is equal to the sum of K trans (= 0.018 J) and K ro t (= 1.393 J). Chapter 11, problem 11: In the figure below, a constant horizontal force Fapp 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 2 acceleration of its center of mass has magnitude 0.60 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? With F app (10 N)i , we solve the problem by applying Eq. 9-14 and Eq. 11-37. ˆ
(a) Newton’s second law in the x direction leads to Fapp f s ma
f s 10N 10kg 0.60 m s 2 4.0 N.
In unit vector notation, we have f s (4.0 N)i , which points leftward. ˆ
(b) With R = 0.30 m, we find the magnitude of the angular acceleration to be 2 | | = |acom| / R = 2.0 rad/s ,
from Eq. 11-6. The only force not directed toward (or away from) the center of mass is f s , and the torque it produces is clockwise:
I
0.30 m 4.0 N I 2.0 rad s 2
2 which yields the wheel’s rotational inertia about its center of mass: I 0.60 kg m .
chapter 11, problem 13: In Figure 11-36, a ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.45 m. The initial height of the ball is h = 0.37 m. At the loop bottom, the magnitude of the normal force on the ball is 1.94Mg. 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
2
expres sed in the general form I = βMR , but β is not 0.4 as it is for a ball of uniform density. Determine β.
Adapting Eq. 6-19 to the consideration of forces at the bottom of an arc, we have 2 F N – Mg = Mv /r 2
which tells us (since we are given F N = 2 Mg ) that the center of mass speed (squared) is v = gr , where r is the arc radius (0.48 m) Thus, the ball’s angular speed (squared) is 2 2 2 2 = v /R = gr/R , where R is the ball’s radius. Plugging this into Eq. 10-5 and solving for the rotational inertia (about t he center of mass), we find 2 2 2 I com = 2 MhR /r – MR = MR [2(0.36/0.48) – 1] . Thus, using the notation suggested in the problem, we find = 2(0.36/0.48) – 1 = 0.50.
Chapter 11, problem 14: In Fig. 11-37, a small, solid, uniform ball is to be shot from point P so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau. Then it leaves the plateau horizontally to land on a game board, at a horizontal distance d from the right edge of the plateau. The vertical heights are h1 = 5.00 cm and
h2 = 1.60 cm. With what speed must the ball be shot at point P for it to land at d = 6.00 cm?
To find the center of mass speed v on the plateau, we use the projectile motion equations of Chapter 4. With vo y = 0 (and using ―h‖ for h2) Eq. 4-22 gives the time-of-flight as t = 2h/g . Then Eq. 4-21 (squared, and using d 2 2 for the horizontal displacement) gives v = gd /2h. Now, to find the speed v p at point P , we apply energy conservation, that is, mechanical energy on the plateau is equal to the mechanical energy at P . With Eq. 11-5, we obtain 1 1 2 2 mv + 2 I com + 2
mgh1 =
1 1 2 2 mv p + 2 I com p . 2 2
2
Using item ( f ) of Table 10-2, Eq. 11-2, and our expression (above) v = gd /2h, we obtain 2
gd /2h + 10 gh1/7 = v p
2
which yields (using the values stated in the problem) v p = 1.34 m/s.
