PHYS 1P21/1P91
Quiz 5 Solutions
4 June 2013
1. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. A motor spins up the flywheel
with a constant torque of 50 N·m. How long does it take the flywheel to reach top angular speed of
1200 rpm? [3 points]
Solution: First convert the units of angular velocity to rad/s:
∆ω = 1200
1200 × 2π rad
rev
=
= 40π rad/s
min
60
s
Now make use of the relations τ = Iα and I = (1/2)M R2 together with the relation ∆ω = α∆t to
determine the time:
∆ω
α
I∆ω
∆t =
τ
M R2 ∆ω
∆t =
2τ
(250)(0.75)2 (40π)
∆t =
2(50)
∆t = 180 s = 3 min
∆t =
2. The ropes in the figure are each wrapped around a cylinder, and the two cylinders are fastened
together and turn on a frictionless axle. The smaller cylinder has a diameter of 10 cm and a mass of
5.0 kg, and the larger cylinder has a diameter of 20 cm and a mass of 20 kg. Determine the angular
acceleration of the system. [3 points]
Solution: Draw a free-body diagram for each block. Choose down to be the positive direction for
the left block and choose up to be positive for the right block, so that these are compatible with
using counter-clockwise as the positive sense for rotation. Applying Newton’s second law to each
block, we obtain
m2 g − T2 = m2 a2
T1 − m1 g = m1 a1
which can be solved for the tensions to obtain
T2 = m2 g − m2 a2
T1 = m1 g + m1 a1
(1)
(2)
The “no-slipping” conditions result in the relations a2 = R2 α and a1 = R1 α.
The net torque on the system is
τ = R2 T2 − R1 T1
τ = R2 (m2 g − m2 a2 ) − R1 (m1 g + m1 a1 )
τ = R2 (m2 g − m2 R2 α) − R1 (m1 g + m1 R1 α)
The total moment of inertia of the two cylinders together is
1
1
I = M1 R12 + M2 R22
2
2
Thus, the angular acceleration of the system can be obtained from the relation τ = Iα:
1
1
2
2
R2 (m2 g − m2 R2 α) − R1 (m1 g + m1 R1 α) =
M1 R1 + M2 R2 α
2
2
1
1
R2 m2 g − m2 R22 α − R1 m1 g − m1 R12 α = M1 R12 α + M2 R22 α
2
2
1
1
R2 m2 g − R1 m1 g = m2 R22 α + m1 R12 α + M1 R12 α + M2 R22 α
2
2
R2 m2 g − R1 m1 g
α=
m2 R22 + m1 R12 + 21 M1 R12 + 12 M2 R22
(0.05)(4.0)(9.8) − (0.10)(2.5)(9.8)
α=
2
(4.0)(0.05) + (2.5)(0.10)2 + (0.5)(20)(0.10)2 + (0.5)(5.0)(0.05)2
α = −3.47 rad/s2
Thus, the pulley rotates clockwise with an angular acceleration of magnitude 3.5 m/s2 .
3. A 1.5 kg block is connected by a rope across a 50-cm-diameter, 2.0 kg, frictionless pulley, as shown
in the figure. A constant 10 N tension is applied to the other end of the rope. Starting from rest,
how long does it take the block to move 30 cm? [4 points]
Solution: Draw a free-body diagram for the 1.5 kg block, and also draw a free-body diagram for
the pulley. To make life easy, I’ll choose up to be the positive direction for the block, and I’ll choose
clockwise to be the positive direction for the rotation of the pulley (this is not the standard convention
for positive direction of rotations).
Applying Newton’s second law to the block, we obtain
T − 1.5g = 1.5a
(1)
where T is the tension in the left side of the rope. Applying the rotational analogue of Newton’s
second law to the pulley, we obtain
10R − T R = Iα
The no-slipping condition results in
α=
a
R
(2)
(3)
where R is the radius of the pulley.
Now solve equation (1) for T and then substitute this expression and the expression for α from
equation (3) into equation (2):
a
10R − (1.5g + 1.5a) R = I ·
R
Solving for a, we obtain
Ia
R
1 M R2 a
10 − (1.5g + 1.5a) = ·
2
R2
1
10 − 1.5g − 1.5a = M a
2
10 − 1.5g = (1.5 + 0.5M ) a
10 − 1.5(9.8)
a=
1.5 + 0.5(2.0)
a = −1.9 m/s2
10R − R (1.5g + 1.5a) =
Note that the acceleration is negative, so the block descends. Evidently the 10 N force is not strong
enough to lift the block.
The time needed for the block to move 30 cm is:
1
∆y = ay (∆t)2
2
s
2∆y
∆t =
ay
r
2(0.30)
∆t =
1.9
∆t = 0.56 s