Laws of Motion

JEE Main level, single correct, medium

133 of 210

Question 133 of 210

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

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Answer: (D) gμR\sqrt{\frac{g}{\mu R}}

Consider the mass MM in the frame of the rotating drum. The forces acting on the mass are:
1. Weight
MgMg acting vertically downwards.
2. Normal reaction
NN from the wall acting towards the center, which provides the centripetal force: N=Mω2RN = M\omega^2 R.
3. Frictional force
ff acting vertically upwards to balance the weight.
For the mass to remain stuck to the wall, the frictional force must balance the weight:
f=Mgf = Mg.
The maximum available static friction is
fmax=μN=μMω2Rf_{max} = \mu N = \mu M\omega^2 R.
To prevent slipping, we must have
fmax≥Mgf_{max} \ge Mg.
μMω2R≥Mg\mu M\omega^2 R \ge Mg
ω2≥gμR\omega^2 \ge \frac{g}{\mu R}
ω≥gμR\omega \ge \sqrt{\frac{g}{\mu R}}
The minimum value of
ω\omega is gμR\sqrt{\frac{g}{\mu R}}.

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