JEE Main 2025 Apr 8 Shift 2, Physics Q24: Kinematics and Dynamics of Rotational Motion

JEE Main2025Apr 8, Shift 2Physics

A thin solid disk of $1$ kg is rotating along its diameter axis at the speed of $1800$ rpm. By applying an external torque of $25 π$ Nm for $40$ s, the speed increases to $2100$ rpm. The diameter of the disk is _____ m.

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Solution

Convert angular speeds: $ω_{i} = 1800 \cdot \frac{2 π}{60} = 60 π$ rad/s and $ω_{f} = 2100 \cdot \frac{2 π}{60} = 70 π$ rad/s.

Angular acceleration from kinematics: $α = \frac{ω _{f} - ω _{i}}{t} = \frac{10 π}{40} = \frac{π}{4}$ rad/s $^{2}$ .

Moment of inertia of a thin disk about a diameter: $I = \frac{m R ^{2}}{4}$ .

Newton's second law for rotation: $τ = I α$ , so $25 π = \frac{1 \cdot R ^{2}}{4} \cdot \frac{π}{4}$ .

Solving: $R^{2} = \frac{25 π \cdot 16}{π} = 400$ , so $R = 20$ m.

Diameter: $D = 2 R = 40$ m.

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JEE Main 2025 Apr 8 Shift 2, Physics Q24: Kinematics and Dynamics of Rotational Motion

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