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- Sourish Ghosh Sir, what will be the answer to this question?Upvote·0· Reply ·2014-05-09 21:27:14
Akash Anand I am getting 2g/3. I guess ..may be you have made some mistake in constraint relation part.
A cylindrical pulley of mass M and radius R is connected to another pulley of mass M/2 ,R/2 through a massless string as shown. If sufficient string is wrapped over larger pulley,find the accleration of mass M attached to the smaller pulley when whole system is released from rest .(assume there is no slipping in the smaller pulley)
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4 Answers
1133
213Sir,for the first constraint velocity of the string near the bigger pulley and near the smaller pulley must be same.
therefore,
α1R=α2R/2 + a
and the second one is α2R/2=a
solving these two equations
α1R=2a and α2R=2a.
where am i going wrong?
- Akash Anand Your are going wrong from the very first line.
- Akash Anand Distance between the two pulleys are changing with time then why the velocity should be same?
- Anonymous @Akash Sir will the string not break if it has tension t1 and t2 both.
1the correct answer is 2g/5. In AAKASH ANAND'S solution the first constraint relation is wrong. Instead of \alpha_1R=\frac{a}{2}, it must be \alpha_1R=2a. Rest of the equations are correct. This change will give the correct answer. that is \frac{2g}{5}.
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- Sourish Ghosh I gave this a lot of thought but still can't figure out how to derive the relation. It is clear that the smaller pulley will rotate quicker than the larger one. But what concept to use to find the relation? Do we have to use the "net work done = 0" concept? If so, can you show how?
- Akash Anand If you think a bit like..rotation of smaller pulley won't make any difference for length of string. If smaller pulley would be massless then also the constraint relation will be as in this case, because putting mass on the lower pulley is not going to contribute in any way. For the second constraint relation, I am sure you don't have any doubt.
- Anonymous Can you please explain how has the constraint been derived? And the answer given 2g/5
