A solid sphere having mass m and radius r rol | vedclass

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

Question

(B) For a solid sphere rolling without slipping,the moment of inertia is $I = \frac{2}{5}mr^2$ and the condition for rolling is $v = r\omega$,which im

Vedclass · Accepted Answer

$\frac{2}{7}$ rotational and $\frac{5}{7}$ translational

Vedclass · Answer

(B) For a solid sphere rolling without slipping,the moment of inertia is $I = \frac{2}{5}mr^2$ and the condition for rolling is $v = r\omega$,which implies $\omega = \frac{v}{r}$.The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} (\frac{2}{5}mr^2) (\frac{v}{r})^2 = \frac{1}{5}mv^2$.The translational kinetic energy is $K_{trans} = \frac{1}{2}mv^2$.The total kinetic energy is $K_{total} = K_{rot} + K_{trans} = \frac{1}{5}mv^2 + \frac{1}{2}mv^2 = \frac{7}{10}mv^2$.The fraction of rotational kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.The fraction of translational kinetic energy is $\frac{K_{trans}}{K_{total}} = \frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{1}{2} 	imes \frac{10}{7} = \frac{5}{7}$.Thus,the sphere has $\frac{2}{7}$ rotational and $\frac{5}{7}$ translational kinetic energy. Hence,option $(b)$ is correct.

$A$ solid sphere having mass $m$ and radius $r$ rolls down an inclined plane. Then its kinetic energy is

Similar Questions

$A$ spherical ball of mass $20 \ kg$ is stationary at the top of a hill of height $100 \ m$. It rolls down a surface to the ground,then climbs up another hill of height $30 \ m$ and finally rolls down to a horizontal base at a height of $20 \ m$ above the ground. The velocity attained by the ball is

An object slides down a smooth incline and reaches the bottom with velocity $v$. If the same mass is in the form of a ring and it rolls down an inclined plane of the same height and angle of inclination,then its velocity at the bottom of the inclined plane will be ............

$A$ disc of mass $3 \, kg$ rolls down an inclined plane of height $5 \, m$. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is .......... $J$. (Take $g = 10 \, m/s^2$)

$A$ solid homogeneous sphere rolls without slipping down an inclined plane of height $h$. The velocity of the sphere at the bottom is:

$A$ solid sphere rolls down from rest on an inclined plane. At the same time,a rectangular block slides down from rest on the same inclined plane. Which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module