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        a rod rotates about point A .The rod is released from rest in the horizontal position so that it rotates about point A.[moment of inertia about point A =ml^2/3].find initial angular acceleration. [in the question how can we equate torque at A and torque due to mg. torque at  A is 0.]
9 months ago

chetan jangir
65 Points
							Hey Liza I hope you know how to calculate torque ,Torque due to force about a point is Forcein perpendicular direction *(into)its perpenidular distance about that point (here A) SO we know equation of rotation motion torque net= I(moment of inertia)*angular acceleration .$Mg*L/2=Ml^{^{2}}/3\alpha$ (here $\alpha$= angular acceleration) Calculating $\alpha$=3g/2lHere at A a tension acts but it’s torque is zero as it’s perdicular distance is zero and also it passes through A

8 months ago
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### Course Features

• 728 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions