A cylinder of mass M and radius R rolls on an | vedclass

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

Question

(C) For a cylinder rolling without slipping,the total kinetic energy $(K)$ is the sum of translational kinetic energy and rotational kinetic energy.$K

Vedclass · Accepted Answer

$\frac{3}{4}Mv^2$

Vedclass · Answer

(C) For a cylinder rolling without slipping,the total kinetic energy $(K)$ is the sum of translational kinetic energy and rotational kinetic energy.$K = K_{	ext{trans}} + K_{	ext{rot}}$$K = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$ and the rolling condition is $v = R\omega$ (or $\omega = v/R$).Substituting these into the equation:$K = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$$K = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2$$K = \frac{3}{4}Mv^2$Thus,the gain in kinetic energy is $\frac{3}{4}Mv^2$.

$A$ cylinder of mass $M$ and radius $R$ rolls on an inclined plane. The gain in kinetic energy is

Similar Questions

$A$ loop rolls down on an inclined plane. The fraction of its total kinetic energy that is associated with the rotational motion is

$A$ solid sphere is rolling down an inclined plane. What is the ratio of its translational kinetic energy to its rotational kinetic energy?

The ratio of the accelerations for a solid sphere (mass $m$ and radius $R$) rolling down an incline of angle $\theta$ without slipping and slipping down the incline without rolling is

Write the condition for rolling without slipping from an inclined plane.

An inclined plane makes an angle $30^{\circ}$ with the horizontal. $A$ solid sphere rolls down from the top of the inclined plane from rest without slipping. Its linear acceleration along the plane is equal to (where $g$ is acceleration due to gravity and $\sin 30^{\circ} = 0.5$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module