Question

Consider a disc of radius $R$ rolling without sliding on a horizontal surface at a constant speed of $V$. Then,

• A. The speed of point A and B on the vertical diameter are $2V$ and $\frac{3V}{2}$ respectively.
• B. The speed of point A and B on the vertical diameter are $\frac{3V}{2}$ and $2V$ respectively.
• C. The time, after which for the first time speed of point A becomes equal to present speed (i.e., the speed at the instant shown in figure) of B is $\frac{R}{V}{\cos }^{-1}\left[\frac{4}{3}\right]$
• D. The time, after which for the first time speed of point A becomes equal to present speed (i.e., the speed at the instant shown in figure) of B is $\frac{2R}{v}{\cos }^{-1}\left[\frac{3}{4}\right]$

Answer 1

Answer: (A), (D)

The correct options are
A The speed of point A and B on the vertical diameter are $2V$ and $\frac{3V}{2}$ respectively.
D The time, after which for the first time speed of point A becomes equal to present speed (i.e., the speed at the instant shown in figure) of B is $\frac{2R}{v}{\cos }^{-1}\left[\frac{3}{4}\right]$

If disc is not sliding, then $V=\omega R$
$\therefore V_A=V+\omega R=2V\text{and}{V}_B=V+\frac{\omega R}{2}=\frac{3V}{2}$

Let speed of A become equal to $\frac{3V}{2}$ after the disc has rotated through an angle $\theta$ about its centre.
Then, $V_A=\sqrt{V^2+V^2+2V^2\cos \theta }\Rightarrow V_A=\frac{3V}{2}=V\sqrt{2(1+\cos \theta )}\Rightarrow \frac{3}{2}=2\cos \frac{\theta }{2}\Rightarrow \theta =2{\cos }^{-1}\left[\frac{3}{4}\right]\text{Time, t}=\frac{\theta }{\omega }=\frac{\theta R}{V}=\frac{2R}{V}{\cos }^{-1}\left[\frac{3}{4}\right]$