# Inertia and Rotation

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1. Apr 10, 2017

### Faiq

1. The problem statement, all variables and given/known data

Value of g = $\small \rm 10~ms^{-2}$
Inertia of the individual square lamina = $\rm \small \frac{1}{12}mr^2 ~kgm^{2}$
3. The attempt at a solution
$$\ C=Ia$$
$$\ Moment~of~Weight~of~Particle=1.594a$$
$$\ 0.3*2g=1.594a$$
$$\ 6=1.594a$$
$$\ a=3.764$$
$$\ \omega^2 = 2a\theta$$
$$\ \omega^2 = 2*3.764*2*\pi$$
$$\ \omega^2 = 47.3$$
$$\ \omega = 6.88$$
However the answer given is $\small \rm 6.52 ~rads^{-1}$
Can somebody please check my working and show me the flaw in my method?
There is also a method to solve this using energy consideration. Can somebody show me that method?

Last edited by a moderator: Apr 10, 2017
2. Apr 10, 2017

### TSny

The force that creates the torque on the combined lamina is not the weight of the particle on the end of the string.

You might consider an alternate approach using energy concepts.

3. Apr 10, 2017

### kuruman

The flaw in your reasoning is that you assumed that the torque is generated by the hanging weight not by the tension in the string. The tension is less than mg otherwise the hanging mass would not accelerate down.

On edit: Apologies for repeating TSny's comment.

Last edited: Apr 10, 2017
4. Apr 10, 2017

### Faiq

So is there a way to calculate the tension?

5. Apr 10, 2017

### Faiq

This was given in response to this question.
" An alternative valid approach which was also seen is to relate the net force and the couple acting on the particle and the lamina to their linear and rotational acceleration respectively."

6. Apr 10, 2017

Nevermind
Got it.