Unformatted text preview: 7.0 m/s 55. SSM REASONING AND SOLUTION The only force that does work on the cylinders as
they move up the incline is the conservative force of gravity; hence, the total mechanical
energy is conserved as the cylinders ascend the incline. We will let h = 0 m on the
horizontal plane at the bottom of the incline. Applying the principle of conservation of
mechanical energy to the solid cylinder, we have
2
2
mghs = 1 mv0 + 1 Isω0
2
2 (1) where, from Table 9.1, Is = 1 mr 2 . In this expression, v0 and ω0 are the initial translational
2
and rotational speeds, and hs is the final height attained by the solid cylinder. Since the
cylinder rolls without slipping, the rotational speed ω0 and the translational speed v0 are
related according to Equation 8.12, ω0 = v0 / r . Then, solving Equation (1) for hs , we obtain
hs = 2
3v0 4g Repeating the above for the hollow cylinder and using I h = mr 2 we have 476 ROTATIONAL DYNAMICS hh = 2
v0 g The height h attained by each cylinder is related to the distance s traveled along the incline
and the angle θ of the incline by
h
h
ss = s
and
sh = h
sin θ
sin θ
Dividing these gives
ss
= 3/4
sh 56. REASONING
a. The kinetic energy of the rolling wheel is the sum of its translational
rotational ( 12 mv2 ) and ( 12 Iω 2 ) kinetic energies. In these expressions m and I are, respectively, the mass and moment of inertia of the wheel, and v and ω are, respectively, its linear and angular
speeds. The sliding wheel only has translational kinetic energy, since it does not rotate. b. As the wheels move up the incline plane, the total mechanical energy is conserved, since
only the conservative force of gravity does work on each wheel. Thus, the initial kinetic
energy at the bottom of the incline is converted entirely into potential energy when the
wheels come to a momentary halt. The potential energy PE is given by PE = mgh
(Equation 6.5), where h is the height of the wheel above an arbitrary zero level.
SOLUTION
a. Since the rolling wheel is a disk, its moment of inertia is I = 1 mR 2
2 (see Table 9.1), where R is the radius of the disk. Furthermore, its angular speed ω is related to the linear
speed v of its center of mass by Equation 8.12 as ω = v/R. Thus, the total kinetic energy of
the rolling wheel is Rolling
Wheel KE = 1
2 2 1
2 mv + = 3 mv 2 =
4 3
4 I ω = mv +
2 1
2 2 ( 11
22 ( 2.0 kg )( 6.0 m/s )2 = mR 2 54 J The kinetic energy of the sliding wheel is
Sliding
Wheel KE = 1 mv 2
2 = 1
2 ( 2.0 kg )( 6.0 m/s )2 = ) 36 J v R 2 Chapter 9 Problems 477 b. As each wheel rolls up the incline, its total mechanical energy is conserved. The initial
kinetic energy KE at the bottom of the incline is converted entirely into potential energy PE
when the wheels come to a momentary halt. Thus, the potential energies of the wheels have
the values calculated in part a for the total kinetic energies.
The potential energy of a wheel is given by Equation 6.5 as PE = mgh, where g is the
acceleration due to gravity and h is the height relative to an arbitrary zero level. Therefore,
the height reached by each wheel is as follows:
Rolling
Wheel h= PE
54 J
=
= 2.8 m
mg
( 2.0 kg ) 9.80 m/s2 Sliding
Wheel h= PE
36 J
=
= 1.8 m
mg
2.0 kg ) 9.80 m/s 2
( ( ) ( ) 57. REASONING AND SOLUTION The conservation of energy gives
2 2 2 mgh + (1/2) mv + (1/2) Iω = (1/2) mv0 + (1/2) Iω0 2 If the ball rolls without slipping, ω = v/R and ω0 = v0/R. We also know that I = (2/5) mR2.
Substitution of the last two equations into the first and rearrangement gives
2
v = v0 − 10 gh =
7 ( 3.50 m/s )2 − 10 ( 9.80 m/s2 ) ( 0.760 m ) =
7 1.3 m/s 58. REASONING We first find the speed v0 of the ball when it becomes airborne using the
conservation of mechanical energy. Once v0 is known, we can use the equations of
kinematics to find its range x.
SOLUTION When the tennis ball starts from rest, its total mechanical energy is in the form
of gravitational potential energy. The gravitational potential energy is equal to mgh if we
take h = 0 m at the height where the ball becomes airborne. Just before the ball becomes
airborne, its mechanical energy is in the form of rotational kinetic energy and translational
2
kinetic energy. At this instant its total energy is 1 mv0 + 1 I ω 2 . If we treat the tennis ball as
2
2
a thinwalled spherical shell of mass m and radius r, and take into account that the ball rolls
down the hill without slipping, its total kinetic energy can be written as
1
2
mv0
2 + Iω =
1
2 2 1
2
mv0
2 + v
12
( mr 2 ) 0
23
r 2 5
2 = 6 mv0 478 ROTATIONAL DYNAMICS Therefore, from conservation of mechanical energy, he have
5
6 2
mgh = mv0 or v0 = 6 gh
5 The range of the tennis ball is given by x = v0 xt = v0 (cos θ ) t , where t is the flight time of
the ball. From Equation 3.3b, we find that the flight time t is given by t= v − v0 y
ay = (−v0 y ) − v0 y
ay =– 2v0 sin θ
ay Therefore, the range of the tennis ball is 2v sin θ
x = v0 xt = v0 (cos θ ) – 0 ay If we take upward as the positive direction, then using the fact that a y = – g and the
expression for v0 given above, we find
2 2cos θ sin θ 2 2cos θ sin θ 6 gh 12
x= = h cos θ sin θ v0 = g
g
5 5 12
(1.8 m) (cos 35°)(sin 35°)= 2.0 m
5 = 59. SSM WWW REASONI...
View
Full
Document
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

Click to edit the document details