Question

There are two blocks of masses $m_1$ and $m_2$ is placed on $m_2$ on a table which is rotating with an angular velocity $\omega$ about the vertical axis. The coefficients of friction between the blocks is ${\mu }_1$ and between $m_2$ and table is ${\mu }_2({\mu }_1<{\mu }_2)$. If the blocks are placed at distance R from the axis of rotation, for relative sliding between the surfaces in contact, find the:
a. Frictional force at the contacting surface
b. maximum angular speed $\omega$.

Answer 1

Free body diagram of upper block in shown in figure (1).

$N_1=m_{1}g$ [$\because$ vertical equilibrium]

Free body diagram of lower block in shown in figure (2).

For relative sliding, $F_{C1}=f_{12}=\mu N_1=\mu m_{1}g$

Also, bottom block must be at rest, i.e.,

$f_{21}=f_{12}+F_{C2}⟹{\mu }_2{N}_2={\mu }_1{N}_1+F_{C2}⟹{\mu }_2(m_1+m_2)=N_1{m}_{1}g+F_{C2}$

where, $F_{C1}=m_1{\omega }^2$ and $F_{C2}=m_2{\omega }^2$

For relative sliding,

$F_{C1}=\mu m_{1}g⟹m_1{\omega }^2=\mu m_{1}g⟹\omega =\sqrt{\frac{\mu g}{R}}$