At the instant of release of an object from rest. Is the only force that can act its weight?

Question

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?

Actually there is movement because AB goes from horizontal initially to vertical when $\omega = 2.4$ — John Alexiou, Aug 26 '14 at 18:42
This question should not close as the OP has some confusion on how to apply the equations of motion. — John Alexiou, Aug 26 '14 at 18:50
See here on how to apply the equations of motion on a point away from the center of mass. — John Alexiou, Aug 26 '14 at 18:53
@ja72 I have checked your link carefully. But are you saying the mark-scheme you see above is correct or incorrect? As they have applied equations of motion at the point $A$ and NOT $G$ ($A$ is away from the COM). — BLAZE, Aug 26 '14 at 19:00

Floris · Accepted Answer · 2014-08-26T18:47:26.020

2

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...

edited Aug 26 '14 at 18:47

answered Aug 26 '14 at 18:20

Floris

118,905

okay thank you for your answer, i'm getting mixed views on this question as i also asked this under the stack-exchange mathematics site. Let me ask you this instead, I understand that maximum friction is reached when $F=\mu R$. So is there a frictional force acting on the object before you push it gently? – BLAZE Aug 26 '14 at 18:28
If you look at the microscopic mechanism behind friction it looks a bit like a car on a road full of pot holes. When the car is standing still in the pothole, there is no force in the direction it would like to travel: but in order to get out of the pot hole, it must climb out of the hole. Similarly a truly stationary object "feels no friction" - but it cannot move until it overcomes the frictional force. The question clearly stated "a frictional couple of constant moment opposes the motion". Before you move you must overcome it. – Floris Aug 26 '14 at 18:34

At the instant of release of an object from rest. Is the only force that can act its weight?

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