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Q3 from a mechanics exam past paper:enter image description here

I can do parts i) and ii) but for iii) in finding the angular acceleration, i used $C=I\alpha$, where $C$ is the applied couple or torque, $I$ is the moment of inertia for the lamina about A and $\alpha$ is the angular acceleration.

At the instant the object is released the only force acting on it is its own weight. Hence, $6g*0.8=9\alpha$ which yields $\alpha=5.227$.

However the mark-scheme says the answer is $\alpha=1.65$, as they have taken into account the frictional couple. Here is the mark-scheme:enter image description here

It was my understanding that at the moment of release no friction (and hence no frictional couple) can act as there is no movement (yet).

So could someone please kindly explain what the mark-scheme is talking about?

ACuriousMind
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1 Answers1

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Friction does not depend on velocity (unlike viscous drag). An object that is stationary on a table will continue to be stationary when you push it gently - because there is an opposing force of friction.

So no, your understanding is wrong: friction is present even when the object is just starting to move.

Let me draw a diagram:

enter image description here

That ought to clear it up...

Floris
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