Recent content by Amartya Sen

Yahoo...Finally I arrived the answer...It turns out that I had been replacing v2 by ω2l2 instead of ω2l2/4 in the final step ...?:)?:)?:) Thanks pals... And sorry for wasting your precious time..if I did.

Oh Yeah..Yeah...I'm sorry. It should be as BvU said. Thanks BvU.. Anyway, even after the correction, answer comes to be wrong...

Here's my solution anyway... Considering the body to be centred at the mid point of the middle leg of the 'H', we have, mgl/2 =mv2/2 + [(ml2/12 + ml2/4 + ml2/4)ω2]/2 which on simplifying and putting v=ωl/2 gives: ω=(12g/17l)^0.5...

Thanks BvU for your help..! :smile::smile::smile: In the first place, I do know the parallel+perpendicular axis theorems. :wink: But the formula says that we have to use the MoI about the COM of the object, not about the axis about which it is rotating. Saying this, I want to also add that it's...

Yes.! That's right! The correct answer is V/3R Pardon, but I didn't understand your solution. You conserved angular momentum about point of contact ? OK..., let me clear my basics first of all.., I suppose no matter about what point we are conserving angular momentum, we always use velocity...

Homework Statement Actually, my confusion originated from solving two different problems... 1) A point object of mass 'm' moving horizontally hits the lower end of the uniform thin rod of length 'l' and mass 'm' and sticks to it. The rod is resting on a horizontal, frictionless surface and...

I know that...But why it is so..? And what's wrong with the application of the original formula?

Homework Statement Respected Physics Gurus/experts...! I am confused in the application of Kinetic energy Expression, i.e, KE = (1/2)MVCM2+(1/2)ICMω2 I had been trying out this question actually(it's pretty simple though:-p)...--- "A rigid body is made of three identical thin rods each of...