A large drum having radius $R$ is spinning around its axis with angular velocity $\omega$, as shown in figure. The minimum value of $\omega$ so that a body of mass $M$ remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass $M$ as $\mu$, is :

Question

A curve is given between the potential energy $U$ of a particle and its position on the $x$-axis as shown. Given: \[ \tan\theta_1 = 1,\qquad \tan\theta_2 = 3,\qquad \tan\theta_3 = -\frac{1}{2} \] If $F_{AB}$ is the force acting on the particle during motion from $A$ to $B$, similarly $F_{BC}$, $F_{CD}$ and $F_{DE}$ are the forces during $B$ to $C$, $C$ to $D$ and $D$ to $E$ respectively, arrange the magnitudes of these forces in decreasing order.

Answer 1

To find the minimum angular velocity \omega required for the body of mass M to remain stuck to the inner wall of the drum, we need to analyze the forces acting on the body.

The forces at play are:

Centripetal Force: Provided by the normal force, N, acting radially inwards due to the circular motion of mass M.
Frictional Force: Acts vertically upwards to balance the gravitational force.
Gravitational Force: Acts vertically downwards, equal to Mg.

For the body to remain stuck, the static frictional force must be equal to the gravitational force, i.e., f = Mg, where f \leq \mu N.

The normal force N is the centripetal force required to keep the mass rotating:

N = M \omega^2 R

Thus, the frictional force can be written as:

\mu N = \mu M \omega^2 R

Setting the frictional force equal to the gravitational force for minimum condition:

\mu M \omega^2 R = Mg

Solving for \omega gives:

\omega^2 = \frac{g}{\mu R}

\omega = \sqrt{\frac{g}{\mu R}}

Therefore, the minimum value of \omega is \sqrt{\frac{g}{\mu R}}, which corresponds to option B.

Answer 2