Q.43. A large drum having radius R is spinning around its axis with angular velocity , as shown in f

Q: What is the correct answer for JEE Main Physics Q43?

The correct answer is D. For the body of mass M to remain stuck to the inner wall of the drum, the static friction force must balance the gravitational force acting on the body.The forces acting on the body are:1. Gravitational force (weight) $mg$ acting downwards.2. Normal force N acting horizontally, towards the center of the drum. This normal force provides the centripetal force for circular motion.3. Static frictio

Q: Q.43. A large drum having radius R is spinning around its axis with angular velocity , as shown in figure. The minimum value of so that a body of mass M remains stuck to the inner wall of the drum,…

The correct answer is D. For the body of mass M to remain stuck to the inner wall of the drum, the static friction force must balance the gravitational force acting on the body.The forces acting on the body are:1. Gravitational force (weight) $mg$ acting downwards.2. Normal force N acting horizontally, towards the center of the drum. This normal force provides the centripetal force for circular motion.3. Static frictio

Q: What are the options for this JEE Main Physics question?

A. $$\sqrt{\frac{\mu R}{g}}$$ · B. $$\sqrt{\frac{2g}{\mu R}}$$ · C. $$\sqrt{\frac{g}{2\mu R}}$$ · D. $$\sqrt{\frac{g}{\mu R}}$$

Question

Q.43. A large drum having radius R is spinning around its axis with angular velocity , as shown in figure. The minimum value of  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 is , is : [Figure: Spinning drum]

FirstInTest · Accepted Answer

For the body of mass M to remain stuck to the inner wall of the drum, the static friction force must balance the gravitational force acting on the body.The forces acting on the body are:1.  Gravitational force (weight) $mg$ acting downwards.2.  Normal force N acting horizontally, towards the center of the drum. This normal force provides the centripetal force for circular motion.3.  Static friction force $f_s$ acting upwards, opposing the tendency to slide down.For the body to remain stuck, the centripetal force must be provided by the normal force: $N = M\omega^2 R$.The maximum static friction force is $f_{s,max} = \mu N$.For the body not to slide down, $f_s \ge mg$.So, $\mu N \ge mg$.Substitute $N = M\omega^2 R$: $\mu (M\omega^2 R) \ge mg$.$\mu M\omega^2 R \ge mg$.$\omega^2 \ge \frac{mg}{\mu MR}$.$\omega^2 \ge \frac{g}{\mu R}$.The minimum angular velocity is $\omega_{min} = \sqrt{\frac{g}{\mu R}}$.

Answer

$$\sqrt{\frac{\mu R}{g}}$$

Answer

$$\sqrt{\frac{2g}{\mu R}}$$

Answer

$$\sqrt{\frac{g}{2\mu R}}$$

Answer

$$\sqrt{\frac{g}{\mu R}}$$

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