Question

If a disc slides from top to bottom of an inclined plane, it takes time $t_1$. If it rolls, it takes time $t_2$. Now $\frac{t_2^2}{t_1^2}$ is:

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $\frac{2}{5}$

Answer 1

Answer: (B)

$\frac{2}{3}$

For rolling the acceleration of body along inclined plane in downward direction is

$a_r=\frac{gsin\theta }{1+\left(\frac{I}{m{R}^2}\right)}$

Here,

I=Moment of inertia of body about centre of mass

R=Radius of the body

m=mass of the body

$\theta =$ angle of incloination of plane

So,the time taken to reach the bottom of plane is

$t_2=\sqrt{\frac{2l}{a_r}}$

$⟹t_2^2=\frac{2l}{sin\theta }(1+\frac{I}{m{R}^2})$

For sliding motion along inclined plane the acceleration of centre of mass in downward direction is $gsin\theta$.

So, the time taken by the body to reach the bottom is

$t_1=\sqrt{\frac{2l}{gsin\theta }}$

$⟹t_1^2=\frac{2l}{gsin\theta }$

So, $\frac{t_1^2}{t_2^2}=\frac{m{R}^2}{m{R}^2+I}$

For disk $I=\frac{1}{2}m{R}^2$

hence, $\frac{t_1^2}{t_2^2}=\frac{m{R}^2}{\frac{3}{2}m{R}^2}=\frac{2}{3}$