question_answer

A round uniform body of radius R, mass M and moment of inertia\[x=2\text{ }cm\]rolls down (without slipping) an inclined plane making an angle \[\theta \] with the horizontal. Then, its acceleration is       AIEEE  Solved  Paper-2007

A)  \[\frac{g\sin \theta }{1+l/M{{R}^{2}}}\]

B)         \[\frac{g\sin \theta }{1+M{{R}^{2}}/l}\]

C)         \[\frac{g\sin \theta }{1-l/M{{R}^{2}}}\]

D)         \[\frac{g\sin \theta }{1-M{{R}^{2}}/l}\]

Correct Answer: A

Solution :

While rolling a body on a incline, has three types of energy. Kinetic energy, rotational kinetic energy and potential energy when there is no external force acting on the system and all the external forces are conservative, the total mechanical energy of the system will remain conserved. Assuming that no energy is used up against friction, the loss in potential energy is equal to the total gain in the kinetic energy. Thus, from work-energy theorem, \[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{OB}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{2}\] or    \[\frac{1}{2}{{v}^{2}}\,(M+l/{{R}^{2}})=Mgh\] or    \[{{v}^{2}}=\frac{2Mgh}{M+l/{{R}^{2}}}\,=\frac{2gh}{1+l/M{{R}^{2}}}\] If s is the distance covered along the plane, then \[I={{I}_{0}}(1-{{e}^{-t/\tau }})\] \[{{v}^{2}}=\frac{2gs\,\sin \theta }{1+l/M{{R}^{2}}}\] Now,     \[{{v}^{2}}=2\,as\] \[2\,as\,=\frac{2\,gs\,\sin \theta }{1+l/M{{R}^{2}}}\,\,or\,\,a=\frac{g\sin \theta }{1+l/M{{R}^{2}}}\]