Question

A block of mass $m$ is kept on a horizontal ruler. The friction coefficient between the ruler and the block in $\mu$. The ruler is fixed at one end and the block is at a distance $L$ from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. If the angular speed of the ruler is uniformly increased from zero at an angular acceleration $\alpha$, at what angular speed will the block slip?

• A. $\left[\right(\frac{\mu g}{L}{)}^2-{\alpha }^2{]}^{1/4}$
• B. $\left[\right(\frac{\mu g}{L}{)}^2+{\alpha }^2{]}^{1/4}$
• C. $\left[\right(\frac{\mu g}{L}{)}^2+{\alpha }^2]$
• D. None of these

Answer 1

The correct option is A $\left[\right(\frac{\mu g}{L}{)}^2-{\alpha }^2{]}^{1/4}$

The frictional force provides the required centripetal force here,

$f=mL{\omega }^2$ [until slipping]

where, $\omega$ is the angular velocity.

$\therefore \mu mg=mL{\omega }^2$

$\left(a\right)$ $⟹{\omega }_{max}=\sqrt{\frac{\mu g}{L}}$

$\left(b\right)$ ${\omega }_{max}-{\omega }_o=\alpha t$

$⟹{\omega }_{max}=\alpha t$

$⟹t=\frac{{\omega }_{max}}{\alpha }$

$⟹t=\frac{1}{\alpha }\sqrt{\frac{\mu g}{L}}$

The block will slip when the angular speed reaches $\sqrt{\frac{\mu g}{L}}$