Question

A disc of radius R rolls on a horizontal surface with linear velocity $\overrightarrow{v}=v\hat{i}$ and angular velocity $\overrightarrow{\omega }=-\omega \hat{k}$ there is a particle P on the circumference of the disc which has velocity in vertical direction. the height of that particle from the ground will be

• A. $R+\frac{v}{\omega }$
• B. $R-\frac{v}{\omega }$
• C. $R+\frac{R}{2}$
• D. $R-\frac{R}{2}$

Answer 1

Answer: (B)

$R-\frac{v}{\omega }$

$\begin{array}{c}\text{Let there be a particle at point}p\text{with angular}\\ \text{separation as shown in the figure-}\\ \text{sinu it is given that only vertical velocity of the}\\ \text{partide is there-}\end{array}$

$\begin{array}{cc}\Rightarrow \text{Horizontal velocity}& =0\\ \Rightarrow v=Rw\cos \theta & \\ \Rightarrow \cos \theta =\frac{v}{R\omega }& \end{array}$

$\begin{array}{c}\text{From the Figure we get -}\\ \begin{array}{c}h=R(1-\cos \theta )=R(1-\frac{v}{Rw})\\ h=R-\frac{v}{\omega }\end{array}\\ \text{Hence, option (B) is correct.}\end{array}$