Question

A disc with linear velocity $V$ and angular velocity $\omega$ is placed on the ground.
Suppose $a$ and $\alpha$ be the magnitudes of linear and angular acceleration due to friction. Then match the statements in Column $I$ with statements in column $II$


$\begin{array}{cc}ColumnI& ColumnII\\ \left(A\right)\text{When}V=R\omega & \left(p\right)a=R\alpha (a\ne 0)\\ \text{(B)When}V=\frac{R\omega }{2}& \left(q\right)a>R\alpha \\ \left(C\right)\text{When}V=2R\omega & \left(r\right)a<R\alpha \\ & \left(s\right)\text{None of these}\end{array}$

• A. $A\to \left(q\right),B\to \left(t\right),C\to \left(p\right)$
• B. $A\to \left(p\right),B\to \left(r\right),C\to \left(r\right)$
• C. $A\to \left(s\right),B\to \left(r\right),C\to \left(r\right)$
• D. $A\to \left(s\right),B\to \left(r\right),C\to \left(s\right)$

Answer 1

The correct option is C $A\to \left(s\right),B\to \left(r\right),C\to \left(r\right)$

Speed of point of contact can be considered as $V_P=V-\omega R$

If $V=R\omega ,V_P=0$, speed of point of contact is zero (No slipping). So, force of friction will be zero. Therefore, $a$ and $\alpha$ both are zero.

When $V=\frac{R\omega }{2},V_P=-\frac{R\omega }{2}$
$\Rightarrow$ Disc is slipping backward . So, force of friction$\left(f\right)$ will be maximum and in forward direction.
$a=\frac{f}{m}=\frac{\mu mg}{m}=\mu g\alpha =\frac{fR}{I}=\frac{\mu mgR}{\frac{1}{2}m{R}^2}=\frac{2mg}{R}$

$\Rightarrow R\alpha =2\mu g$
$\therefore a<R\alpha$

When $V=2R\omega ,V_P=R\omega$

$\Rightarrow$ Disc is slipping forward . So, force of friction$\left(f\right)$ will be maximum and in backward direction.
$a=\frac{f}{m}=\frac{\mu mg}{m}=\mu g\alpha =\frac{fR}{I}=\frac{\mu mgR}{\frac{1}{2}m{R}^2}=\frac{2mg}{R}$

$\Rightarrow R\alpha =2\mu g$
$\therefore a<R\alpha$