A ball rolls without slipping. The radius of | vedclass

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

Question

(C) For a body rolling without slipping,the total kinetic energy $E_{	ext{total}}$ is the sum of translational kinetic energy $E_t$ and rotational ki

Vedclass · Accepted Answer

$\frac{K^2}{K^2 + R^2}$

Vedclass · Answer

(C) For a body rolling without slipping,the total kinetic energy $E_{	ext{total}}$ is the sum of translational kinetic energy $E_t$ and rotational kinetic energy $E_r$.$E_t = \frac{1}{2}mv^2$$E_r = \frac{1}{2}I\omega^2$Since $I = mK^2$ and $\omega = \frac{v}{R}$,we have $E_r = \frac{1}{2}(mK^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$.$E_{	ext{total}} = E_t + E_r = \frac{1}{2}mv^2 + \frac{1}{2}mv^2(\frac{K^2}{R^2}) = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$.The fraction of total energy associated with rotational kinetic energy is $\frac{E_r}{E_{	ext{total}}}$.$\frac{E_r}{E_{	ext{total}}} = \frac{\frac{1}{2}mv^2(\frac{K^2}{R^2})}{\frac{1}{2}mv^2(1 + \frac{K^2}{R^2})} = \frac{\frac{K^2}{R^2}}{\frac{R^2 + K^2}{R^2}} = \frac{K^2}{K^2 + R^2}$.

$A$ ball rolls 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 its total energy is associated with its rotational kinetic energy?

Similar Questions

Pseudocoelom develops from

$A$ particle falls from a height $h$ upon a fixed horizontal plane and rebounds. If $e$ is the coefficient of restitution,the total distance travelled before rebounding has stopped is

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $C$ is equal to (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module