Unformatted text preview: rizontally, the effects of gravity are negligible.
Momentum is conserved. Since the initial momentum of the system is zero, it must remain
zero as the ball is thrown and caught. While the ball is in motion, the platform will recoil in
such a way that the total momentum of the system remains zero. As the ball is caught, the
system must come to rest so that the total momentum remains zero. The distance that the
platform moves before coming to rest again can be determined by using the expressions for
momentum conservation and the kinematic description for this situation. SOLUTION While the ball is in motion, we have
MV + mv = 0 (1) where M is the combined mass of the platform and the two people, V is the recoil velocity of
the platform, m is the mass of the ball, and v is the velocity of the ball.
The distance that the platform moves is given by
x = Vt (2) where t is the time that the ball is in the air. The time that the ball is in the air is given by
t= L
v −V (3) where L is the length of the platform, and the quantity (v – V) is the velocity of the ball
relative to the platform. Remember, both the ball and the platform are moving while the ball
is in the air. Combining equations (2) and (3) gives V
x= v −V L From equation (1) the ratio of the velocities is V / v = −m / M . Equation (4) then gives
x= (V / v ) L = ( −m / M ) L = − mL = − ( 6.0 kg ) ( 2.0 m ) = −0.097 m
1 − (V / v ) 1 + ( m / M )
M +m
118 kg + 6.0 kg (4) 390 IMPULSE AND MOMENTUM The minus sign indicates that displacement of the platform is in the opposite direction to the
displacement of the ball. The distance moved by the platform is the magnitude of this
displacement, or 0.097 m . CHAPTER 8 ROTATIONAL KINEMATICS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) Using Equation 8.1 (θ = Arc length / Radius) to calculate the angle (in radians) that each
object subtends at your eye shows that θMoon = 9.0 × 10−3 rad, θPea = 7.0 × 10−3 rad, and
θDime = 25 × 10−3 rad. Since θPea is less than θMoon, the pea does not completely cover your
view of the moon. However, since θDime is greater than θMoon, the dime does completely
cover your view of the moon.
2. 2.20 cm
3. 38.2 s
4. (a) An angular acceleration of zero means that the angular velocity has the same value at all
times, as in statements A or B. However, statement C is also consistent with a zero angular
acceleration, because if the angular displacement does not change as time passes, then the
angular velocity remains constant at a value of 0 rad/s.
5. (c) A nonzero angular acceleration means that the angular velocity is either increasing or
decreasing. The angular velocity is not constant.
6. (b) Since values are given for the initial angular velocity ω0, the final angular velocity ω, and
the time t, Equation 8.6 [ θ = 1 (ω0 + ω ) t ] can be used to calculate the angular displacement
2 θ.
7. 32 rad/s
8. 88 rad
9. (c) According to Equation 8.9 (vT = rω), the tangential speed is proportional to the radius r
when the angular speed ω is constant, as it is for the earth. As the elevator rises, the radius,
which is your distance from the center of the earth, increases, and so does your tangential
speed.
10. (b) According to Equation 8.9 (vT = rω), the tangential speed is proportional to the radius r
when the angular speed ω is constant, as it is for the merrygoround. Thus, the angular 2.1 m speed of the second child is vT = ( 2.2 m/s ) . 1.4 m 392 ROTATIONAL KINEMATICS 11. 367 rad/s2
12. (e) According to Newton’s second law, the centripetal force is given by Fc = mac , where m
is the mass of the ball and ac is the centripetal acceleration. The centripetal acceleration is
given by Equation 8.11 as ac = rω2, where r is the radius and ω is the angular speed.
Therefore, Fc = mrω 2 , and the centripetal force is proportional to the radius when the mass
and the angular speed are fixed, as they are in this problem. As a result, 33 cm Fc = (1.7 N ) . 12 cm 13. (d) Since the angular speed ω is constant, the angular acceleration α is zero, according to
Equation 8.4. Since α = 0 rad/s2, the tangential acceleration aT is zero, according to
Equation 8.10. The centripetal acceleration ac, however, is not zero, since it is proportional
to the square of the angular speed, according to Equation 8.11, and the angular speed is not
zero.
14. 17.8 m/s2
15. (a) The number N of revolutions is the distance s traveled divided by the circumference 2πr
of a wheel: N = s/(2πr).
16. 27.0 m/s Chapter 8 Problems 393 CHAPTER 8 ROTATIONAL KINEMATICS
PROBLEMS 1. SSM REASONING AND SOLUTION Since there are 2π radians per revolution and it
is stated in the problem that there are 100 grads in onequarter of a circle, we find that the
number of grads in one radian is 100
(1.00 rad ) 21 rev 0.250grad = 63.7 grad rev π rad 2. REASONING The average angular velocity ω has the same direction as θ − θ 0 , because
θ − θ0
ω=
according to Equation 8.2. If θ is greater than θ0, then ω is positive. If θ is
t − t0
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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