JEE Main 2025 Apr 3 Shift 1, Physics Q10: Rolling Motion

JEE Main2025Apr 3, Shift 1Physics

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

A
$3.5 m/s^{2}$
B
$0.35 m/s^{2}$
C
$2.5 m/s^{2}$
D
$0.25 m/s^{2}$

Solution

Take torque about the point of contact (instantaneous axis of rotation). The applied force at the top has moment arm $2 r$ . The moment of inertia of a solid sphere about a tangent at the contact point (parallel-axis) is $I_{contact} = \frac{2}{5} m r^{2} + m r^{2} = \frac{7}{5} m r^{2} .$

$F \cdot 2 r = I_{contact} α \Rightarrow 49 \cdot 2 r = \frac{7}{5} (20) r^{2} α \Rightarrow α = \frac{49 \cdot 2}{\frac{7}{5} \cdot 20 r} = \frac{14}{4 r} rad/s^{2} .$

Rolling without slipping: $a = r α = \frac{14}{4} = 3.5 m/s^{2} .$

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JEE Main 2025 Apr 3 Shift 1, Physics Q10: Rolling Motion

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