question_answer

Two particles each of mass m are attached to ends of a light string of length a and placed on a horizontal turntable at distances a and \[2a\] from centre of rotation Let turntable rotates with a maximum angular speed \[\omega ,\] such that particle remains over it without any slip. Necessary friction coefficient must be

A) \[\frac{a{{\omega }^{2}}}{2g}\]

B) \[\frac{2a{{\omega }^{2}}}{g}\]

C) \[\frac{3a{{\omega }^{2}}}{2g}\]

D) \[\frac{a{{\omega }^{2}}}{3g}\]

Correct Answer: C

Solution :

Let O is centre of rotation, then

In above figure,

T =tension in string when masses about to slip.

Accelerations of A and B are

\[{{a}_{A}}=a{{\omega }^{2}}\]

\[{{a}_{B}}=2a{{\omega }^{2}}\]

Frictions of A and B are

\[{{f}_{A}}=\mu mg\]

\[{{f}_{B}}=\mu mg\]

Net force on A is

\[{{F}_{A}}=\mu mg-T=ma{{\omega }^{2}}\]

Net force on B is

\[{{F}_{B}}=\mu mg+T=m(2a){{\omega }^{2}}\]

For no slip; \[{{F}_{A}}+{{F}_{B}}\]

\[{{f}_{A}}+{{f}_{B}}=3am{{\omega }^{2}}\]\[\Rightarrow \]\[\mu =\frac{3a{{\omega }^{2}}}{2g}\]