Solve the following :

Solution:

(a)

For block not to slip

ff $\geq \mathrm{mL} \omega^{2}$

$\mu \mathrm{N} \geq \mathrm{mL} \omega^{2}$

$\mu \mathrm{mg} \geq \mathrm{mL} \omega^{2}$

$\omega^{2} \leq \frac{\mu g}{L}$

$\omega_{\max }=\sqrt{\frac{\mu g}{L}}$

(b)

Now, ruler makes uniformly accelerated circular motion at angular acceleration of $\alpha, \mathbf{s o}$ tangential

acceleration

$\mathbf{A}_{\mathrm{t}}=\mathrm{L} \alpha$

$F_{t}=m a_{t}$

$F_{t}=m L a$

$F_{t}=m L \alpha$

Radial force $F_{c}=m L \omega^{\prime 2}$

Where $\omega^{\prime}$ is maximum angular speed at which it slips 

$\mu \mathrm{N}=\sqrt{\left(F_{c}^{2}+F_{t}^{2}\right)}$

$\mu \mathrm{mg}=\sqrt{\left.\left\{m L w^{\prime}\right)^{2}+\left\{m L \alpha^{\prime}\right)^{2}\right\}}$

squaring and solving,

$\omega=\left[\left(\frac{u g}{L}\right)^{2} \cdot \alpha^{2}\right]^{1 / 4}$