PHY 2053, Spring 2009, Quiz 9 — Whiting
1. A uniform solid cylinder of mass M = 50.0 kg and radius R = 1.00
m rotates of a frictionless horizontal axle. Two objects with equal mass
m = 12.5 kg hang from light cords wrapped around the cylinder. If the
system is released from rest, find:
a) the tension in each cord, and
Vertically, for each object, we have T − m × g = m × a. For the cylinder we have
τ = −2R × T = I × α = a × I/R.
Substitute for a from the first equation to find T :
(T − m × g) × I
m×g×I
⇒T =
.
m×R
2m × R2 + I
m×M ×g
12.5 × 50.0 × 9.80
Then use I = M × R2 /2 to find : T =
=
= 61.3 N.
4m + M
4 × 12.5 + 50.0
−2R × T =
b) the acceleration of each hanging object.
a=
T −m×g
−4 × m × g
−4 × 12.5 × 9.80
=
=
= −g/2.00 = −4.90 m/s2 .
m
4m + M
4 × 12.5 + 50.0
2. A 4.00 kg mass is connected by a light cord to a 3.00 kg mass on a
smooth surface. The pulley rotates about a frictionless axle and has a
moment of inertia of 0.250 kg.m2 and a radius of 0.500 m. Assuming that
the cord does not slip on the pulley, find:
a) the acceleration of the two masses, and
Horizontally, T2 = m2 × a. Vertically, T1 − m1 × g = −m1 × a. At the pulley, we have:
τ = r × (T2 − T1 ) = I × α = −a × I/r. Substituting for T1 and T2 we can solve for a:
a × (m1 × r2 + m2 × r2 + I) = m1 × g × r2 , which gives :
a=
m1 × r2 × g
4.00 × 0.5002 × 9.80
=
= g/2.00 = 4.90 m/s2 .
2
2
2
2
m1 × r + m2 × r + I
4.00 × 0.500 + 3.00 × 0.500 + .250
b) the tensions T1 and T2 .
T1 = m1 × (g − a) = 4.00 × (9.80 − 4.90) = 19.6 N, and T2 = m2 × a = 3.00 × 3.00 = 14.7 N.