Question

Two blocks of mass $m_1=10kgand{m}_2=5kg$ 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 $m_{1}is0.5$ while there is no friction between $m_2$ and the table. The table is rotating with an angular velocity $10$ radian see about a vertical axis passing through its centre O. The masses are placed along a diameter of the table on either side of the centre O such that the mass m1 is at a distance of $0.124$m from O. The masses are observed to be at rest with rest with respect to an observer on the turn table.
Calculate the frictional force on $m_1$:

• A. $128$N
• B. $88$N
• C. $36$N
• D. $24$N

Answer 1

The correct option is C $36$N

frictional force $f$

$v=wr$

$v_2=10x_2$

$v_1=10x_1$

$\frac{m_2{v}^2}{x_2}=\frac{m_1{v}_1^2}{x_1}+f$

$m_2{w}^2{x}_2=m_1{w}^2{x}_1+f$

$5\times 100\times 0.176=10\times 100\times 0.124+f$

$f=88-124=-36N$

-ve sign indicates direction of friction force acts opposite.