Question

A bead of mass m = 0.2 kg lies at a distance r from the axis of rotation on a rigid wire as shown in figure. Coefficient of friction between wire and bead is $\mu$, and wire starts rotating from rest at an angular acceleration of $\alpha$ at t = 0. The whole arrangement is in gravity free region. r = 1m, $\mu =0.4$ & $\alpha =0.1r/s^2$

• A. Relative motion between bead and wire starts at t = 2 sec
• B. Power of friction force at t = 2 is 0.0016 watt
• C. Power of normal contact at t = 2 is 0.004 watt
• D. Power of net force on bead at t = 2 is 0.004 watt

Answer 1

Answer: (A), (C), (D)

The correct options are
A Relative motion between bead and wire starts at t = 2 sec
C Power of normal contact at t = 2 is 0.004 watt
D Power of net force on bead at t = 2 is 0.004 watt


$N=m{a}_t=mr\alpha$
$f_r=\mu N=\mu mr\alpha$

The bead starts sliding when friction ($f_r$) is less than centrifugal force.
$\Rightarrow f_r=m{\omega }^{2}r$
$\Rightarrow \mu mr\alpha =m{\omega }^{2}r$
$\mu \mathrm{m/}\mathrm{r/}\alpha =\mathrm{m/}{\omega }^2\mathrm{r/}$
${\omega }^2=\mu \alpha \text{and we know,}\omega =\alpha t$
${\alpha }^2{t}^2=\mu \alpha \Rightarrow t=\sqrt{\frac{\mu }{\alpha }}=\sqrt{\frac{0.4}{0.1}}=2sec$
$v=r\omega =r\alpha t\Rightarrow$ at t = 2, $v=1\times 0.1\times 2=0.2m/s$
$f_l=\mu N=\mu mr\alpha =0.4\times 0.2\times 1\times 0.1=0.008$
$N=mr\alpha =0.2\times 1\times 0.1=0.02$
Power of friction = $0$
Power of normal = $N\times v=0.004Watt$