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Joined:	16 Mar 2012
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16 Mar 2012 19:04:03 IST

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RIGID BODY DYNAMICS

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in any general problem of rigid body dynamics involving a pulley(when not considered light and has some inertia),we consider the tension on both the sides of ideal string to be different.....but when the string is ideal,it is considered to be massless,meaning thereby no gravitational force on it and no normal rkn force to the pulley....When the normal rkn force is zero,no frictional force should exist ....then how can tension be different????

Comments (3)

boywholived

Scorching goIITian

Joined: 10 Feb 2012

Posts: 262

17 Mar 2012 14:55:24 IST

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It's not that way. According to the problem we know that for the blocks to move the pulley should rotate. Since the pulley is mass less there should be some external torque it should have to rotate. First just assume the tension to be equal and calculate the net torque you find that the pulley experiences zero torque i.e it won't rotate. now you are clear that the tension should be different. hoping you understand.If yes please like..

Viraat Gupta

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Joined: 21 Jun 2011

Posts: 743

29 Mar 2012 21:14:13 IST

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just as said above , if the difference in tension occurs due to a nodal point , that is the pulley ... if there is no tension difference that means the mass at either side of the pulley is equal , hence no rotation of the pulley is possible

Viraat Gupta

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Joined: 21 Jun 2011

Posts: 743

29 Mar 2012 21:14:47 IST

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