Question

A small particle of mass $0.36\text{g}$ rests on horizontal turntable at a distance $25\text{cm}$ from the axis of spindle. The turntable is accelarated at a rate of $\alpha =\frac{1}{3}{\text{rad/s}}^2$. The frictional force that the table exerts on the particle $2\text{s}$ after the startup is

• A. $40\mu N$
• B. $30\mu N$
• C. $50\mu N$
• D. $60\mu N$

Answer 1

The correct option is C $50\mu N$

Given:
Mass $m=0.36\text{g}\text{or}0.00036\text{kg}$

Radius $r=25\text{cm}\text{or}0.25\text{m}$

Angular accelaration $\alpha =\frac{1}{3}{\text{rad/s}}^2$

Initial velocity = $0$

Angular speed after $2s$ , $\omega ={\omega }_0+2t$
$\Rightarrow \omega =0+\frac{1}{3}\times 2=\frac{2}{3}\text{rad/s}$

Tangential accelaration $a_t=r\alpha$

$a_t=0.25\times \frac{1}{3}=\frac{25}{300}{\text{m/s}}^2$

Total accelaration , $a=\sqrt{a_c^2+a_t^2}=\sqrt{(r{\omega }^2{)}^2+(a_t{)}^2}$

=$\sqrt{{[\left(\frac{25}{100}\right)\times {\left(\frac{2}{3}\right)}^2]}^2+{\left(\frac{25}{300}\right)}^2}$

=$\sqrt{\frac{1}{81}+\frac{1}{144}}$

$=0.1389{\text{m/s}}^2$

Frictional force provides required centipetal and tangential
accelarations.

$\therefore f_s=ma=0.00036\times 0.1389=50\times {10}^{-6}\text{N}$

$=50\mu \text{N}$

Hence, option (c) is the correct answer.

Why this question: Concept: Only frictional force is acting on the particle along the surface of the table, so it is responsible for both circular motion as well tangential acceleration.