Rotational Motion Question 221
Question: The spherical body and block is m. Moment of inertia of the spherical body about centre of mass is $2mR^{2}$ . The spherical body rolls on the horizontal surface. There is no slipping at any surfaces in contact. The ratio of kinetic energy of the spherical body to that of block is
Options:
A) 3/4
B) 1/3
C) 2/3
D) 1/2
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let v be the linear velocity of centre of mass of the spherical body and w its angular velocity about centre of mass.
Then $\omega =\frac{v}{2R}$
KE of spherical body
$K _{1}=\frac{1}{2}mv^{2}+\frac{1}{2}I{{\omega }^{2}}$
$K _{1}=\frac{1}{2}mv^{2}+\frac{1}{2}{{(2mR)}^{2}}\left( \frac{v^{2}}{4R^{2}} \right)=\frac{3}{4}mv^{2}$?.
(i) Speed of the block will be
$v’=(\omega )(3R)=3R\omega =(3R)\left( \frac{v}{2R} \right)=\frac{3}{2}v$
$\therefore $KE of block $K _{2}=\frac{1}{2}mv^{2}$
$=\frac{1}{2}m{{\left( \frac{3}{2}v \right)}^{2}}=\frac{9}{8}mv^{2}$
From equations (i) and (ii), $
$\frac{K _{1}}{K _{2}}=\frac{2}{3}$
