Question

Help Needed... (Rotation)

Answer 1

Angular acc = torque / Moment of inertia about the axis of rotation

I = ML^2/3 since axis of rotation passes through the end point.
Now since the rod is of uniform density , the centre of gravity is at the mid point of rod.
torque equals : force * (perpendicular distance between point of suspension and the line of force ) which is = mg * (Lcos68)/2

Answer 2

Does what Diwakar wrote make sense? He has laid it out for you.

In notation, \( \tau = I\alpha \) and hence \( \alpha = \tau/I \).

Here, \( I = \frac{1}{4}mL^2 \).

Now \( \tau = \frac{1}{2} mgL\cos(68) \) and you can therefore calculate \( \alpha \).

Answer 3

Ah, that makes sense. Thanks.