Gujarati
Hindi
6.System of Particles and Rotational Motion
medium

A solid sphere of mass $1\ kg$ rolls on a table with linear speed $1\ m/s$. Its total kinetic energy is .......... $J$

A

$1$

B

$0.5$

C

$0.7$

D

$1.4$

Solution

$E=\frac{1}{2} M v^{2}+\frac{1}{2} I \omega^{2}$

$=\displaystyle \frac {1}{2}Mv^2$

$\displaystyle \frac {1}{2} \times \displaystyle \frac {2}{5}MR^2 \times \displaystyle \frac {V^2}{R^2}$

$=\displaystyle \frac {7}{10} Mv^2=0.7 \space J$ 

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.