Question

A hemispherical bowl of radius $R$ is rotating about its axis of symmetry which is kept vertical. A small ball kept in the bowl rotates with the bowl without slipping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the ball with the vertical is $\alpha$. Find the angular speed $\omega$ at which the bowl should rotate for this to happen.

• A. $\omega =\frac{g}{R\cos \alpha }$
• B. $\omega =\frac{g}{R\sin \alpha }$
• C. $\omega =\sqrt{\frac{g}{R\sin \alpha }}$
• D. $\omega =\sqrt{\frac{g}{R\cos \alpha }}$

Answer 1

The correct option is D $\omega =\sqrt{\frac{g}{R\cos \alpha }}$

We know for the maximum velocity, the block will try to move upwards as a result friction will be acting downwards.

we have given

$NCos\theta =FSin\theta +mg.$

$NCos\theta =\mu NSin\theta +mg.$

$N(Cos\theta -\mu Sin\theta )=mg.$

$\Rightarrow N=mg/(Cos\theta -\mu Sin\theta )$

Second equation,

$NSin\theta +\mu NCos\theta =m\omega ²rSin\theta$

$N(Sin\theta +\mu Cos\theta )=m\omega ²rSin\theta$

Substituting the above values of $N$ in this,

We will get,

$mg/(Cos\theta -\mu Sin\theta )\times (Sin\theta +\mu Cos\theta )=m\omega ²rSin\theta$

$\Rightarrow \omega ²=g(Sin\theta +\mu Cos\theta )/rSin\theta (Cos\theta -\mu Sin\theta )$

$\therefore \omega max.=\surd \left[g\right(Sin\theta +\mu Cos\theta \left)/rSin\theta \right(Cos\theta -\mu Sin\theta \left)\right]$

Now, Similarly, ωmin. can be find. For easy calculations, Substitute the Plus sign instead of minus sign and vice versa in max. so that we can get the min. angular velocity.

$\therefore \omega min.=\surd \left[g\right(Sin\theta -\mu Cos\theta \left)/rSin\theta \right(Cos\theta +\mu Sin\theta \left)\right]$

Hence, the option $D$ is the correct answer.