Question

What is the ratio of rolling kinetic energy and rotational kinetic energy in the motion of the disc

• A. 1:1
• B. 2:7
• C. 2:1
• D. 3:1

Answer 1

The correct option is C 2:1

The rolling kinetic energy is given by

$K_t=\frac{1}{2}m{v}^2$

Rolling starts when, $v=\omega r$

$K_t=\frac{1}{2}m{r}^2{\omega }^2$. . . . . . .(1)

The rotational kinetic energy is given by,

$K_r=\frac{1}{2}I{\omega }^2$

where, $I=$ moment of inertia about the axis passing through the center of the disc.

$I=\frac{1}{2}m{r}^2$

$K_r=\frac{1}{2}.\frac{1}{2}m{r}^2{\omega }^2=\frac{1}{4}m{r}^2{\omega }^2$. . . . .(2)

Ratio, $\frac{K_t}{K_r}=\frac{m{r}^2{\omega }^2/2}{m{r}^2{\omega }^2/4}=\frac{2}{1}$

$K_t:K_r=2:1$