Question

A small particle of mass $0.36\text{g}$ rests on a horizontal turntable at a distance $25\text{cm}$ from the axis of the spindle. The turntable is accelerated 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:
(Here, $\mu$ in the options represents micro)

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

Answer 1

The correct option is C $50\mu \text{N}$

Mass of particle, $m=0.36\text{gm}=36\times {10}^{-5}\text{kg}$


$\alpha =\frac{1}{3}{\text{rad/s}}^2,r=25\text{cm}=\frac{1}{4}\text{m}$
So tangential acceleration of particle:
$a_t=r\alpha =\frac{1}{4}\times \frac{1}{3}=\frac{1}{12}{\text{m/s}}^2$
Centripetal acceleration of particle:
$a_c={\omega }^{2}r$ ... (i)

Now, at $t=2\text{sec}$, finding the $\omega$:
$\alpha =\frac{d\omega }{dt}\Rightarrow d\omega =\alpha dt$
$\underset{0}{\overset{\omega }{\int }}d\omega =\alpha \underset{t=0}{\overset{t=2}{\int }}dt\Rightarrow \omega =\frac{1}{3}{\left[t\right]}_0^2$
$\therefore \omega =\frac{2}{3}\text{rad/s}$

From Eq. (i):
$a_c={\omega }^{2}r=\frac{4}{9}\times \frac{1}{4}=\frac{1}{9}{\text{m/s}}^2$

For the particle, net force will be acting in the direction of net acceleration $\left(a_{net}\right)$ and frictional force will provide net force.


$\Rightarrow a_{net}=\sqrt{a_c^2+a_t^2}=\sqrt{\frac{1}{81}+\frac{1}{144}}$
$\therefore a_{net}=\sqrt{\frac{144+81}{({12}^2\times 9^2)}}=\frac{5}{36}{\text{m/s}}^2$
Hence
$f=F_{\text{net}}$
$\Rightarrow f=m{a}_{net}$
$=(36\times {10}^{-5})\times \frac{5}{36}$
or $f=50\times {10}^{-6}\text{N}=50\mu \text{N}$