## geerky42 3 years ago Help Needed... (Rotation)

1. geerky42

2. anonymous

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

3. JamesJ

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 $$\alpha$$.

4. geerky42

Ah, that makes sense. Thanks.