Question

Two blocks of mass $m_1=10\,kg$ and $m_2=5\,kg$ connected to each other by a massless inextensible string of length $0.3\, m$ are placed along a diameter of turn table. The coefficient of friction between the table and $m_1$ is $0.5$ while there is no friction between $m_2$ and the table. The table is rotating with an angular velocity of $10 \,rad/s$ about a vertical axis passing through its centre $O$. The masses are placed along the diameter of table on either side of the centre $O$ such that the mass $m_1$ is at a distance of $0.124 \,m$ from $O$. The masses are observed to be at rest with respect to an observer on the turn table.
(a) Calculate the frictional force on $m_1$
(b) What should be the minimum angular speed of the turn table, so that the masses will slip from this position?
(c) How should the masses be placed with the string remaining taut so that there is no frictional force acting on the mass $m_1$?

Answer 1

Given, $m_1 = 10 \,kg,$
$m_2 = 5\,kg, \omega = 10\, rad/s $
$r=0.3\,m, r_1=0.124\,m$
$\therefore r_=r-r_1=0.124\, m$
(a) Masses $m_1$ and $m_2$ are at rest with respect to rotating table.
Let $f$ be the friction between mass $m_1$ and table.

Free body diagram of $m_1 $ and $m_2$ respect to ground

$T = m_2r_2\omega^2 .....$ (i)
Since, $m_2 r_2 \omega^2 < m_1 r_1 \omega^2$
$Therefore, m_1 r_1 \omega^2 > T$
and friction on $m_1$ will be inward (toward centre)
$f+T=m_1 r_1\omega^2 ...(iii)$
From Eqs. (i) and (ii), we get
$ f=m_1 r_1\omega^2-m_2 r_2\omega^2 ...(iii)$
$ =(m_1 r_1-m_2 r_2)\omega^2$
$= (10 \times 0.124 - 5 \times 0.176) (10)^2 N$
$=36 \,N$
Therefore, frictional force on $m_1$ is $36\, N$ (inwards)
(b) From Eq. (iii)
$ f=(m_1 r_1-m_2 r_2)\omega^2$
Masses will start slipping when this force is greater than $f_{max} or$
$f=(m_1 r_1-m_2 r_2)\omega^2 > f_{max} > \mu m_1 g$
$\therefore $ Minimum values of $\omega$ is
$\omega_{min}=\sqrt{\frac{\mu m_1g}{m_1 r_1-m_2 r_2}}=\sqrt{\frac{0.5\times10\times9.8}{10\times0.124-5\times0.176}}$
(c) From Eq. (iii), frictional force $f = 0$
where, $ m_1 r_1 = m_2 r_2$
or $ \frac{r_1}{r_2}=\frac{m_2}{m_1}=\frac{5}{10}=\frac{1}{2}$
and $ r=r_1+r_=2=0.3\,m$
$\therefore r_1=0.1\,m$ and $r_2=0.2\,m$
i.e. mass $m_2$ should placed at $0.2 \,m$ and $m_1$ at $0. 1\,m$ from the centre $O$.