# Slide L0902 pg3 Question

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Hi, I was wondering if you could explain this question. I was trying to figure this our using torque = r x F, but then noticed I have to use inertia since it's rotating. But I don't know what to do from there.

Here's the question:

+1 vote

You are thinking correctly. First find torque and then we have $\tau=I\alpha$ to get $\alpha$.

Here the mass is all on the rim of the wheel so $I=MR^2$.
answered Dec 6, 2013 by (21,750 points)
selected Dec 6, 2013 by Teemu
I got the answer but I would like to make sure I did it right.

So I found the sum of torques to be:

Fr = I a(angular)

then I changed inertia into MR^2

Fr = Mr^2 a

then

F = Mr a
(1) cos(60) = (1) (.5) a