A hollow cylinder and a solid cylinder start | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The time taken by a body to roll down an inclined plane is given by the formula $t = \sqrt{\frac{2L(1 + \frac{k^2}{R^2})}{g \sin 	heta}}$,where $

Vedclass · Accepted Answer

The hollow cylinder

Vedclass · Answer

(A) The time taken by a body to roll down an inclined plane is given by the formula $t = \sqrt{\frac{2L(1 + \frac{k^2}{R^2})}{g \sin 	heta}}$,where $k$ is the radius of gyration and $R$ is the radius of the object.For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $k^2 = \frac{1}{2}R^2$,which gives $\frac{k^2}{R^2} = 0.5$.For a hollow cylinder,the moment of inertia $I = MR^2$,so $k^2 = R^2$,which gives $\frac{k^2}{R^2} = 1$.Since the hollow cylinder has a larger value of $\frac{k^2}{R^2}$,it will have a larger value of $t$.Therefore,the hollow cylinder takes more time to reach the bottom.

$A$ hollow cylinder and a solid cylinder start rolling down an inclined plane. Which one will take more time to reach the bottom of the plane?

Similar Questions

How can a solid sphere and a hollow sphere of the same material and same size be distinguished without weighing them?

$A$ solid cylinder is placed on a rough surface of inclination $\theta$. The minimum value of the coefficient of static friction between the cylinder and the surface so that the cylinder rolls without slipping is:

$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls without slipping down an inclined plane of angle $\theta$. What is its acceleration?

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\,cm$. If the cylinder rolls without slipping,its speed upon reaching the bottom of the inclined plane is $...........\,ms^{-1}$. (Given $g = 10\,ms^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module