Question

A symmetrical body of mass $M$, radius $R$ and radius of gyration $k$ is rolling on a horizontal surface without slipping. If linear velocity of centre of mass is $v_c$ and angular velocity is $\omega$; then:

• A. the total KE of body is $\frac{1}{2}m{v}_c^2(1+\frac{k^2}{R^2})$
• B. the rotational KE is $\frac{1}{2}M{R}^2{\omega }^2$
• C. the translational KE is $\frac{1}{2}M{v}_c^2$
• D. the moment of inertia of body is $M{k}^2$

Answer 1

Answer: (A), (C), (D)

The correct options are
A the total KE of body is $\frac{1}{2}m{v}_c^2(1+\frac{k^2}{R^2})$
C the translational KE is $\frac{1}{2}M{v}_c^2$
D the moment of inertia of body is $M{k}^2$

$Z=m{k}^2$
$KE=\frac{1}{2}m{V}_c^2+\frac{1}{2}I{w}^2$ ( total energy)
$V_c=wR$
$K{P}_2=\frac{1}{2}m{V}_c^2+\frac{1}{2}\frac{m{K}^2{V}_c^2}{k^2}$
$=\frac{1}{2}m{V}_c^2(1+\frac{k^2}{k^2})$
Rotational KE=$\frac{1}{2}I{w}^2$
$=\frac{1}{2}m{k}^2{w}^2$
translation K E =$\frac{1}{2}M{V}_c^2$