Question

A disc is given an initial angular velocity ${\omega }_0$ and placed on a rough horizontal surface as shown in figure. The quantities which will not depend on the coefficient of friction is/are

• A. the time until rolling begins
• B. the displacement of the disc until rolling begins
• C. the velocity when rolling begins
• D. the work done by the force of friction

Answer 1

Answer: (C), (D)

The correct options are
C the velocity when rolling begins
D the work done by the force of friction

As initially the point A tends to slide backwards, thus the kinetic frictional force will act in forward direction. When pure rolling starts, disc will move with linear velocity $v$ and rotate with angular velocity $w$.

$f=\mu mg$

$\tau =I\alpha$

$fR=\frac{1}{2}M{R}^2\alpha ⟹\alpha =\frac{2f}{MR}$

$w=w_o-\alpha t$

$\frac{wR}{2}=\frac{w_{o}R}{2}-\frac{f}{M}t$ ..........(1)

Also, $a=\frac{f}{M}$ i.e $a$ depends on $\mu$

$v=at=\frac{f}{M}t$ .........(2)

Adding (1) and (2), $v+\frac{wR}{2}=\frac{w_{o}R}{2}$

As $v=Rw$ (pure rolling)

$⟹v=\frac{w_{o}R}{3}$

Hence $v$ is independent of $\mu .$

From (2) $v=\mu gt⟹t$ depends on $\mu$

Work-energy theorem, $W_f=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}^2$

Now $v$ and $w$ do not depend on $\mu$ and hence $W_f$ is independent of coefficient of friction

$s=\frac{1}{2}a{t}^2=\frac{1}{2}\mu g\times \frac{v^2}{(\mu g{)}^2}$

$⟹s=\frac{2v^2}{\mu g}$

Thus $s$ also depends on $\mu$