Question

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. $\frac{1}{\alpha }{[\left(\frac{\mu g}{L}\right)-{\alpha }^2]}^{1/4}$
• B. $\frac{1}{\alpha }{[\left(\frac{\mu g}{L}\right)-{\alpha }^2]}^{1/3}$
• C. $\frac{1}{\alpha }{[\left(\frac{\mu g}{2L}\right)-{\alpha }^2]}^{1/4}$
• D. $\frac{1}{\alpha }{[\left(\frac{\mu g}{2L}\right)-{\alpha }^2]}^{1/3}$

Answer 1

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

Let the time at which block slips be $t$.

Tangential acceleration is, $a_t=\alpha L$

Radial acceleration is, $a_r=\frac{v^2}{L}=\frac{{\omega }^2{L}^2}{L}={\alpha }^2{t}^{2}L$

Net acceleration of block is, $a=\sqrt{a_t^2+a_r^2}=\alpha L\sqrt{1+{\alpha }^2{t}^4}$

The limiting frictional force on block is, $f=\mu mg$

Block starts to slip when $f=ma⟹\alpha L\sqrt{1+{\alpha }^2{t}^4}=\mu g$

$⟹\frac{1}{\alpha }[(\frac{\mu g}{L}{)}^2-{\alpha }^2{]}^{1/4}$

Corresponding Angular speed is,

$\omega =\alpha t=[(\frac{\mu g}{L}{)}^2-{\alpha }^2{]}^{1/4}$