Work Energy And Power Question 311
Question: A chain of mass m and length $ l $ lies on the surface of a rough sphere of radius $ R(>1) $ such that one end of chain is at the top most point of sphere. The chain is held at rest because of friction. The gravitational potential energy of the chain in this position (considering the horizontal diameter of sphere as reference level for gravitational potential energy), is
Options:
A) $ \frac{mgR^{2}}{l} $
B) $ \frac{mgR^{2}}{l}\sin ( \frac{l}{R} ) $
C) $ \frac{mgR^{2}}{l}\cos ( \frac{l}{R} ) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \sin \alpha =\frac{l}{R} $ Consider the element of the chain as shown in the figure below. Its mass is
$ dm=\frac{m}{l}\times Rd\theta $
Its $ PEwrt $ horizontal diagram as the reference position is
$ dU=dmg\times R\sin \theta $
$ dU=\frac{mR^{2}2}{l}\sin \theta d\theta $
$ U=\int _{{}}^{{}}{dU=\int _{\pi /2-\alpha }^{\pi /2}{\frac{mR^{2}g}{l}}\sin \theta d\theta } $
$ =\frac{mR^{2}g}{l}\sin \alpha =\frac{mR^{2}g}{l}\sin ( \frac{l}{R} ) $
BETA
