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0 votes
1.4k views
in Physics by (15 points)

A bobbin rolls without slipping on a horizontal surface. Moment of inertia of the bobbin I= MR2

Find the angular velocity of the bobbin. The answer is 2g divided by 25R but i didn't get it.

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1 Answer

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by (10.9k points)
I think that angular acceleration of the bobbin created is to be found out here. The dimension of angular velocity does not match with that of the quantity (depending on g/R) given in the answer..

Here the gravitational pull will produce a torque and that torque will generate angular acceleration of the bobbin.

If a represents the angular acceleration then

I×a = mg×R/2= the torque

=>3 mR^2a=R/2×mg

=>a=g/(2×3R) rad/sec^2

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