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Unformatted text preview: ation of the bicycle wheel can be calculated from
Equation 8.4. Once the angular acceleration is known, Equation 9.7 can be used to find the
net torque caused by the brake pads. The normal force can be calculated from the torque
using Equation 9.1. SOLUTION The angular acceleration of the wheel is, according to Equation 8.4, α= ω − ω0
t = 3.7 rad/s − 13.1 rad/s
= −3.1 rad/s 2
3.0 s If we assume that all the mass of the wheel is concentrated in the rim, we may treat the
wheel as a hollow cylinder. From Table 9.1, we know that the moment of inertia of a
hollow cylinder of mass m and radius r about an axis through its center is I = mr 2 . The net
torque that acts on the wheel due to the brake pads is, therefore, ∑τ = Iα = (mr 2 )α (1) From Equation 9.1, the net torque that acts on the wheel due to the action of the two brake
pads is
(2)
∑τ = –2 f k l
where fk is the kinetic frictional force applied to the wheel by each brake pad, and
l = 0.33 m is the lever arm between the axle of the wheel and the brake pad (see the
drawing in the text). The factor of 2 accounts for the fact that there are two brake pads. The
minus sign arises because the net torque must have the same sign as the angular
acceleration. The kinetic frictional force can be written as (see Equation 4.8)
f k = µk FN (3) where µk is the coefficient of kinetic friction and FN is the magnitude of the normal force
applied to the wheel by each brake pad. 470 ROTATIONAL DYNAMICS Combining Equations (1), (2), and (3) gives –2( µk FN )l = (mr 2 )α
FN = – mr 2α –(1.3 kg)(0.33 m)2 (–3.1 rad/s 2 )
=
= 0.78 N
2µk l
2(0.85)(0.33 m) 46. REASONING AND SOLUTION The moment of inertia of a solid cylinder about an axis
2 coinciding with the cylinder axis which contains the center of mass is Icm = (1/2) MR . The
2 parallel axis theorem applied to an axis on the surface (h = R) gives I = Icm + MR , so that
I= 3
2 MR 2 47. REASONING AND SOLUTION Newton's law applied to the 11.0kg object gives
2 2 T2 − (11.0 kg)(9.80 m/s ) = (11.0 kg)(4.90 m/s ) or T2 = 162 N or T1 = 216 N A similar treatment for the 44.0kg object yields
2 2 T1 − (44.0 kg)(9.80 m/s ) = (44.0 kg)(−4.90 m/s )
For an axis about the center of the pulley
2 T2r − T1r = I(−α) = (1/2) Mr (−a/r)
Solving for the mass M we obtain
2 M = (−2/a)(T2 − T1) = [−2/(4.90 m/s )](162 N − 216 N) = 22.0 kg Chapter 9 Problems 471 48. REASONING
1
a. The kinetic energy is given by Equation 9.9 as KE R = 2 Iω 2 . Assuming the earth to be a
uniform solid sphere, we find from Table 9.1 that the moment of inertia is I = 2 MR 2 . The
5
mass and radius of the earth is M = 5.98 × 1024 kg and R = 6.38 × 106 m (see the inside of
the text’s front cover). The angular speed ω must be expressed in rad/s, and we note that the
earth turns once around its axis each day, which corresponds to 2π rad/day.
b. The kinetic energy for the earth’s motion around the sun can be obtained from
1
Equation 9.9 as KE R = 2 Iω 2 . Since the earth’s radius is small compared to the radius of
the earth’s orbit (Rorbit = 1.50 × 1011 m, see the inside of the text’s front cover), the moment
2
of inertia in this case is just I = MRorbit . The angular speed ω of the earth as it goes around
the sun can be obtained from the fact that it makes one revolution each year, which
corresponds to 2π rad/year. SOLUTION
a. According to Equation 9.9, we have
1
KE R = 2 Iω 2 = = 1
2 2
5 1
2 ω
c MR h
2 2
5 5
c.98 × 10 24 2 L2π rad IF day IF1 h IO
F J1 J J
mh M
GGGP
HHHQ
N1 day K24 h K3600 s K
2 h
c kg 6.38 × 10 6 2 = 2 .57 × 10 29 J
b. According to Equation 9.9, we have
1
KE R = 2 Iω 2 = = 1
2 1
2 MR
ω
ch 5
c.98 × 10 2
orbit 24 h
c 2 kg 1.50 × 10 L2π rad IF 1 yr IF day IF1 h IO
FJ
m hM
GGP
G yr K365 day J 124 h J 3600 s J
HHQ
HH
N1 G K K K
2 11 2 = 2 .67 × 10 33 J 49. SSM REASONING The kinetic energy of the flywheel is given by Equation 9.9. The
moment of inertia of the flywheel is the same as that of a solid disk, and, according to Table
1
9.1 in the text, is given by I = 2 MR 2 . Once the moment of inertia of the flywheel is
known, Equation 9.9 can be solved for the angular speed ω in rad/s. This quantity can then
be converted to rev/min. 472 ROTATIONAL DYNAMICS SOLUTION Solving Equation 9.9 for ω, we obtain, ω= 2 ( KE R )
I 2 ( KE R ) = MR 2 1
2 4(1.2 × 10 9 J)
= 6.4 × 10 4 rad / s
2
(13 kg)(0.30 m) = Converting this answer into rev/min, we find that c ω = 6.4 × 10 4 rad / s G rad JGmin J
hF1πrev I F60 s I =
H KH K
2
1 6.1 × 10 5 rev / min 1
50. REASONING The kinetic energy of a rotating object is expressed as KE R = 2 Iω 2
(Equation 9.9), where I is the object’s moment of inertia and ω is its angular speed.
According to Equation 9.6, the moment of inertia for rod A is just that of the attached
particle, since the rod itself is massless. For rod A with its attached particle, then, the
moment of inertia is I A = ML2 . According to Table 9.1, the moment of inertia for rod B is I B = 1 ML2 .
3 SOLUTION Using Equation 9.9 to calculate the kinetic energy, we find...
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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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