The rotational kinetic energy of a solid sphe | vedclass

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

Question

(A) The total energy at the top is potential energy,$PE = mgh$. At the bottom,this energy is converted into translational kinetic energy $(K_t)$ and r

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) The total energy at the top is potential energy,$PE = mgh$. At the bottom,this energy is converted into translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$. $mgh = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For a solid sphere,the moment of inertia is $I = \frac{2}{5}mR^2$ and $\omega = \frac{v}{R}$. Substituting these,$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$. Thus,$v^2 = \frac{10gh}{7}$. The rotational kinetic energy is $K_r = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2}) = \frac{1}{5}mv^2$. Substituting $v^2 = \frac{10gh}{7}$,we get $K_r = \frac{1}{5}m(\frac{10gh}{7}) = \frac{2}{7}mgh$. Given $m = 3 \; kg$,$g = 10 \; m/s^2$,and $h = 7 \; m$: $K_r = \frac{2}{7} 	imes 3 	imes 10 	imes 7 = 60 \; J$.

The rotational kinetic energy of a solid sphere of mass $3 \; kg$ and radius $0.2 \; m$ rolling down an inclined plane of height $7 \; m$ is (in $; J$)

Similar Questions

$A$ solid cylinder and a solid sphere,having same mass $M$ and radius $R$,roll down the same inclined plane from the top without slipping. They start from rest. The ratio of the velocity of the solid cylinder to that of the solid sphere,with which they reach the ground,will be:

What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination $\theta$?

$A$ solid sphere rolls on a surface with a translational velocity $v \ m/s$. It climbs up a curved surface without slipping. The minimum value of $v$ required to reach the height $h$ is:

$A$ uniform solid sphere of mass $m$ and radius $r$ rolls without slipping down an inclined plane,inclined at an angle $45^o$ to the horizontal. Find the minimum magnitude of the frictional coefficient required for rolling without slipping.

$A$ ball is rolling without slipping. The radius of gyration of the ball about an axis passing through its center of mass is $K$. If the radius of the ball is $R$,what fraction of the total energy is rotational kinetic energy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module