I just need some clarification on part (c). Here's the whole problem for context:

Question

A uniform solid cylinder of mass 10 kg can rotate about a frictionless axle through its center O, as shown in the cross-sectional view in the figure. A rope wrapped around the outer radius R1 = 1.0 m exerts a force of magnitude F1 = 5.0 N to the right. A second rope wrapped around another section of radius R2 = 0.50 m exerts a force of magnitude F2 = 6.0 N downward. (The moment of inertia of a uniform solid cylinder is I = ½ m r^2)

a) What is the angular acceleration of the cylinder?
T = (F1*R1)-(F2*R2) = (1/2)mr^2(alpha)
(5*1)-(6*.50)-/(.5)(10)(1) = 0.4
angular acceleration = 0.4 rad/s^2

b) How many revolutions does the cylinder rotate through the first 5.0 seconds if it starts from rest?
theta = w_0 + .5(angular acceleration)(t)^2
0 + .5(.4)(5)^2 = 5.0 rad

c) Which way does it rotate, clockwise or counterclockwise?

I want to say counterclockwise, but the key I have says it should be clockwise. I thought it should be counterclockwise because the displacement is positive. But I'm not sure if that reasoning is right. Can you explain why it's clockwise? It's the only part of this problem I don't understand.

Answer 1

jrt

Answer 2

(cw = clockwise, ccw = counterclockwise). It should be clockwise since it is negative rotation. The force has Torque(1) = -5 meaning it rotates cw and Torque(2) = 3 meaning it rotates ccw. T(1) + T(2) = -2 so overall it is clockwise.