1THATS.QUITE SIMPLE see this image u will understand
follow this link : http://imageshack.us/photo/my-images/577/ansd.jpg/
The figure shows mass m moves with velocity u. Ring is restricted to move on smooth fixed horizontal rod. Velocity of the ring at that moment?
I thought the answer to be u/2 but the correct answer is 2u. This can happen if we take the rod to be the hypotenuse ie v(ring) * cos60 = u. please explain why?
http://imageshack.us/photo/my-images/716/imagenax.jpg/
please check the pic in the link above if the link at the top is not working
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5 Answers
11Why can't it be ucos60=v?
Because the ring only has velocity in the x-direction.
So the x component of the ropes velocity is given to the ring giving(dropping a perpendicular from a point in the rope to the rod)
Ucos60=v?
1read this: velocity along the two ends of the string should be the same
1vatika u must have done pulley questions in this chapter when acceleration of a block going up=acceleration of another one coming down in a system.. this is also a system so velocity at both ends is same as mentioned by samagr@
1I do understand that the velocity/acceleration is same at both ends. I want to understand the fallacy in my solution when I take x-axis in the direction of v and decompose u on the x-axis as u.cos(60) and equate it with v.
I am missing something very basic.
Please point that out to me.
This image will be more illustrative.
http://imageshack.us/photo/my-images/189/photoqcfo.jpg/