A uniform solid sphere of radius R and radius | vedclass

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

Question

(C) In pure rolling motion,the total kinetic energy is the sum of translational and rotational kinetic energy.
Translational $KE = \frac{1}{2}mv^2$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In pure rolling motion,the total kinetic energy is the sum of translational and rotational kinetic energy.
Translational $KE = \frac{1}{2}mv^2$.
Rotational $KE = \frac{1}{2}I\omega^2 = \frac{1}{2}(mK^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$.
Total $KE = \frac{1}{2}mv^2 + \frac{1}{2}mv^2(\frac{K^2}{R^2}) = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2}) = \frac{1}{2}mv^2(\frac{R^2+K^2}{R^2})$.
The fraction of total energy associated with rotation is $\frac{Rotational \ KE}{Total \ KE} = \frac{\frac{1}{2}mv^2(\frac{K^2}{R^2})}{\frac{1}{2}mv^2(\frac{R^2+K^2}{R^2})} = \frac{K^2}{K^2+R^2}$.

$A$ uniform solid sphere of radius $R$ and radius of gyration $K$ about an axis passing through the centre of mass,is rolling without slipping. Then the fraction of total energy associated with its rotation will be

Similar Questions

$A$ sphere of mass $0.5 \ kg$ and diameter $1 \ m$ is rolling without slipping with a constant velocity of $5 \ m/s$. The ratio of its rotational kinetic energy to its total kinetic energy is:

When a rolling body enters onto a smooth horizontal surface,it will ............

$A$ uniform solid sphere of mass $m$ and radius $R$ is being pulled on a horizontal surface with a force $F$ parallel to the surface,applied at its topmost point. If the acceleration of the center of mass of the sphere is $a$ and it is rolling without slipping on the surface,the value of $F$ is:

$A$ sphere of mass $50 \, g$ and diameter $20 \, cm$ is rolling without slipping with a velocity of $5 \, cm/s$. Its total kinetic energy is:

$A$ uniform sphere of mass $500\,g$ rolls without slipping on a plane surface such that its centre moves at a speed of $0.02\,m/s$. The total kinetic energy of the rolling sphere would be (in $J$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module