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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Angular 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

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Does 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 \( \alpha \).

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Ah, that makes sense. Thanks.
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