Question

Two blocks of mass $m_1=10kg$ and $m_2=5kg$ connected to each other by a massless inextensible string of length $0.3m$ are placed along a diameter of a 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 $10rad/sec$ about a vertical axis passing through its centre. The masses are placed along the diameter of the table on either side of the centre $O$ such that $m_1$ is at a distance of $0.124m$ from $O$. The masses are observed to be at rest with respect to an observer on the turn table.
(i) Calculate the frictional force on $m_1$
(ii) What should be the minimum angular speed of the turn table so that the masses will slip from this position.

• A. 36, 11.7
• B. 45, 13
• C. 55, 16
• D. 78, 19

Answer 1

The correct option is A 36, 11.7

Given mass $m_{1}and{m}_2$ 10 kg and 5 kg respectively length of the string l=0.3 , the coefficient of friction ${\mu }_1$ = 0.5 and angular velocity $\omega$= 10 rad/sec.

(1) Frictional force on $m_1$=Centrifugal force on $m_1$

- centrifugal force on $m_2$

Centrifugal force on $m_1$

$F_1=m_1{\omega }^2$

centrifugal force on $m_2$

$F_1=m_2{\omega }^2$

Thus frictional force on $m_1$

$F_f=F_2-F_1={\omega }^2\left\{m_2\right(L-l)-m_{1}l\}={10}^2\left\{5\right(0.3-0.124)-10\times 0.124\}=36N$

(2) For the mass to slip from the position it must satisfy the equation

${\text{F}}_f^k={\text{F}}_f{m}_{1}g{\mu }_1={\omega }^2\left\{m_2\right(L-l)-m_{1}l\}\omega =\sqrt{\frac{m_{1}g{\mu }_1}{\left\{m_2\right(L-l)-m_{1}l\}}}=\sqrt{\frac{10\times 10\times 0.5}{\left\{5\right(0.3-0.124)-10(0.124)}}=11.78rad/sec$