Two objects are attached to ropes that are attached to wheels on a common axle a
ID: 1351011 • Letter: T
Question
Two objects are attached to ropes that are attached to wheels on a common axle as shown below. The two wheels are glued together so that they form a single object. The total moment of inertia of the two wheels is 43 kg · m2. The radii of the wheels are R1 = 1.1 m and R2 = 0.20 m.
(a) If m1 = 24 kg, find m2 such that there is no angular acceleration of the wheels.
(b) If 12 kg is gently added to the top of m1, find the angular acceleration of the wheels.
rad/s2
Find the tension in the rope holding m1.
kN
Find the tension in the rope holding m2.
kN
Explanation / Answer
(a) asks for m2 which is found by considering the pulley wheel - system in static EQUILIBRIUM, so that the CW torque/moment of m2 is balanced by the CCW torque/moment of m1:
m2R2 = m1R1 {I write this abbreviated since "g" cancels}
m2 = m1R1/R2 <= (24*1.1)/0.20 = 132 kg
(b) If 12 kg is added to m1 making it = 36 kg and m1 was only large enough to balance 24 kg, there would be an unbalanced CCW torque/moment of:
12(1.1)(9.8) 129 m-N
This torque/moment = 129 = I = (43)
= 129/43 = 3 rad/s² ANS
To calculate tension consider that if the pully wheel - mass system was in staic equilibrium as in (a), the tensions in each rope would simply be the weights of the hanging masses. IF the wheel was accelerating as in (b) that acceleration must be added/subtracted from "g" to find the acceleration of the two masses (and hence indirectly by F = tension = ma) the answer/s.
We know from the first part of (b) that = 3 rad/s². Converting this angular acceleration into a linear one at each of the two "hubs" of the wheel from: a = R
for m1: a = (3)(1.1) = 3.3 m/s²
for m2: a = (3)(0.20) = 0.6 m/s²
Now imagine the wheel accelerating in the CCW direction this would tend to relieve tension in the rope to m1 but tighten or increase tension in the rope of m2. Therefore SUBTRACT 3 m/s² from "g" to get the acceleration of m1 and ADD 3 m/s² to "g" to get the acceleration of m2. Then multiply these two accelerations by their masses and U have the final two answers for rope tensions.
So, T in m1 = (9.8-3)*24 = 163.2 N = 0.16 kN
T in m2 = (9.8+3)*132 = 1689.6 N = 1.68 kN