A) \[\frac{mg{{R}^{2}}}{l}\]
B) \[\frac{mg{{R}^{2}}}{l}\sin \,\left( \frac{l}{R} \right)\]
C) \[\frac{mg{{R}^{2}}}{l}\cos \,\left( \frac{l}{R} \right)\]
D) None of these
Correct Answer: B
Solution :
Here, \[\alpha =\frac{l}{R}\] Consider the element of the chain as shown in the figure below, its mass is, \[dm=\frac{m}{l}\times Rd\theta \]. Its PE w.r.t horizontal diagram as the reference position is. \[dU=dmg\times R\,\sin \,\theta \] \[\Rightarrow \] \[dU=\frac{m{{R}^{2}}g}{l}\,\sin \,\theta d\theta \] \[U=\,\int{dU}=\,\int_{\pi /2-\alpha }^{\pi /2}{\frac{m{{R}^{2}}g}{l}\sin \,\theta d\theta }\] \[=\frac{m{{R}^{2}}g}{l}\sin \,\alpha \] \[=\frac{m{{R}^{2}}g}{l}\,\sin \,\left( \frac{l}{R} \right)\]You need to login to perform this action.
You will be redirected in
3 sec