JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

A large drum having radius R is spinning around its axis with angular velocity $ω$ , as shown in figure. The minimum value of $ω$ 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 $μ$ , is :

A
$\frac{μg}{R}$
B
$\frac{2 g}{μ R}$
C
$\frac{g}{2 μ R}$
D
$\frac{g}{μ R}$

Solution

The wall pushes inward (normal force N) and supplies the centripetal acceleration: $N = M ω^{2} R$ .

Friction acts vertically upward and supports the block's weight. At the threshold (minimum $ω$ ), friction is at its limit: $μ N = M g$ .

Combine: $μ M ω^{2} R = M g \Rightarrow ω = \frac{g}{μ R}$ .

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