CHAPTER 10
SOLUTION FOR PROBLEM 16
(a) Use 1 rev = 2π rad and 1 min = 60 s to obtain
ω=
200 rev (200 rev)(2π rad/rev)
=
= 20.9 rad/s .
1 min
(1 min)(60 s/min)
(b) The speed of a point on the rim is given by v = ωr, where r is the radius of the flywheel
and ω must be in radians per second. Thus v = (20.9 rad/s)(0.60 m = 12.5 m/s.
(c) If ω is the angular velocity at time t, ω0 is the angular velcoty at t = 0, and α is the angular
acceleration, then since the anagular acceleration is constant ω = ω0 + αt and
α=
ω − ω0 (1000 rev/min) − (200 rev/min)
2
=
= 800 rev/min .
t
1.0 min
(d) The flywheel turns through the angle θ, which is
1
1
2
θ = ω0 t + αt2 = (200 rev/min)(1.0 min) + (800 rev/min )(1.0 min)2 = 600 rev .
2
2
CHAPTER 10
SOLUTION FOR PROBLEM 32
Use the parallel-axis theorem. According to Table 10–2, the rotational inertia of a uniform slab
about an axis through the center and perpendicular to the large faces is given by
M 2 2
(a + b ) .
12
0
A parallel axis through a corner is a distance h = (a/2)2 + (b/2)2 from the center, so
Icom =
M 2 2 M 2 2
M 2 2
(a + b ) +
(a + b ) =
(a + b )
12
4
3
o
0.172 kg J
(0.035 m)2 + (0.084 m)2 = 4.7 × 10−4 kg · m2 .
=
3
I = Icom + M h2 =
CHAPTER 10
SOLUTION FOR PROBLEM 41
(a) Use constant acceleration kinematics. If down is taken to be positive and a is the acceleration
of the heavier block, then its coordinate is given by y = 12 at2 , so
a=
2y 2(0.750 m)
2
=
= 6.00 × 10−2 m/s .
2
2
t
(5.00 s)
2
The lighter block has an acceleration of 6.00 × 10−2 m/s , upward.
(b) Newton’s second law for the heavier block is mh g − Th = mh a, where mh is its mass and
Th is the tension force on the block. Thus
2
2
Th = mh (g − a) = (0.500 kg)(9.8 m/s − 6.00 × 10−2 m/s ) = 4.87 N .
(c) Newton’s second law for the lighter block is ml g − Tl = −ml a, where Tl is the tension force
on the block. Thus
2
2
Tl = ml (g + a) = (0.460 kg)(9.8 m/s + 6.00 × 10−2 m/s ) = 4.54 N .
(d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the rim of
the pulley must be the same as the acceleration of the blocks, so
2
α=
a 6.00 × 10−2 m/s
2
=
= 1.20 rad/s .
R
5.00 × 10−2 m
(e) The net torque acting on the pulley is τ = (Th − Tl )R. Equate this to Iα and solve for I:
I=
(Th − Tl )R (4.87 N − 4.54 N)(5.00 × 10−2 m)
=
= 1.38 × 10−2 kg · m2 .
2
α
1.20 rad/s
CHAPTER 10
HINT FOR PROBLEM 9
Use the equation for constant-angular acceleration rotation: θ = ω0 t+ 12 αt2 , where θ is the angular
position at time t (measured from the angular position at t = 0), ω0 is the angular velocity at
t = 0, and α is the angular acceleration. Solve for α.
(b) the average angular velocity is the angular displacement divided by the time interval.
(c) Evaluate the constant-acceleration equation ω = ω0 + αt.
(d) Use θ = ω0 t + 12 αt2 . You might take the time to be zero at the end of the first 5 seconds and
take ω0 to be the answer to part (c).
J
2
ans: (a) 2.0 rad/s ; (b) 5.0 rad/s; (c) 10 rad/s; (d) 75 rad
o
CHAPTER 10
HINT FOR PROBLEM 19
The tangential component of the acceleration is given by at = αr and the radial component is
given by ar = ω 2 r, where α is the angular acceleration, ω is the angular velocity, and r is the
distance from the rotation axis to the point. Use ω = dθ/dt to find ω and α = dω/dt to calculate
α.
J
2
2
ans: (a) 6.4 cm/s ; (b) 2.6 cm/s
o
CHAPTER 10
HINT FOR PROBLEM 31
(a) The rotational inertia of a rod of length d and mass M , rotating about an axis through its
1
M d2 (see Table 10–2). Use the parallel-axis theorem to find
center and perpendicular to it, is 12
the rotational inertias of the rods in this problem. For one rod the rotation axis is d/2 from its
center and for the other it is 3d/2 from its center. The rotational inertia of either of the particles
is the product of its mass and the square of the its distance from the rotation axis. Sum the
rotational inertias to obtain the total rotational inertia.
(b) The rotational kinetic energy is given by K = 12 Iω 2 , where I is the total rotational inertia.
J
ans: (a) 0.023 kg · m2 ; (b) 11 mJ
o
CHAPTER 10
HINT FOR PROBLEM 39
Use τnet = Iα, where τnet is the net torque on the cylinder, I is its rotational inertia, and α is its
angular acceleration. The net torque is the sum of the individual torques. The magnitude of the
torque associated with F1 is F1 R; the magnitude of the torque associated with F2 is F2 R; the
magnitude of the torque associated with F3 is F3 r; the magnitude of the torque associated with F4
is 0. If the torque, acting alone, tends to produce an angular acceleration in the counterclockwise
direction it enters the sum as a positive value; if it tends to produce an angular acceleration in
the clockwise direction it enters the sum as a negative value. The rotational inertia is I = 12 M R2 ,
where M is the mass of the cylinder. Solve for α. If the result is positive the angular acceleration
is in the counterclockwise direction; if it is negative the angular acceleration is in the clockwise
direction.
J
o
2
ans: (a) 9.7 rad/s ; (b) counterclockwise Use τnet = Iα, where τnet is the net torque on the
cylinder, I is it rotational inertia, and α is its angular acceleration. The net torque is the sum of
the individual torques. The magnitude of the torque associated with F1 is F1 R; the magnitude
of the torque associated with F2 is F2 R; the magnitude of the torque associated with F3 is F3 r;
the magnitude of the torque associated with F4 is 0. If the torque, acting alone, tends to produce
an angular acceleration in the counterclockwise direction it enters the sum as a positive value; it
tends to produce an angular acceleration in the clockwise direction it enter the sum as a negative
value. The rotational inertia is I = 12 M R2 , where M is the mass of the cylinder. Solve for α. If
the result is positive it is in the counterclockwise direction; if it is negative the acceleration is in
the clockwise direction.
J
2
ans: (a) 9.7 rad/s ; (b) counterclockwise
o