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JEE Advanced 2022 · Paper 1, Physics question 6

A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination θ=30∘\theta = 30^\circ from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance r=0.5r = 0.5 m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is _____ m s−2\mathrm{m\,s^{-2}}. (Take g=10 m s−2g = 10\ \mathrm{m\,s^{-2}}.)

A sphere on an incline of angle θ (incline descending to the right). Two forces of 1 N, both parallel to the incline, act at points a perpendicular distance r = 0.5 m from the centre on opposite sides of it: one on the upper-left side pointing down the incline, the other on the lower-right side pointing up the incline. They form a couple.
Answer and worked solution

Official answer: 2.8

The couple. The two 1 N forces are equal and opposite, so there is no net force from them. Their lines of action are 2r=12r=1 m apart, giving a couple τ=1\tau=1 N m.

From the figure, the up-slope force acts on the side away from the incline and the down-slope force on the side nearer it. So the couple opposes the rolling rotation.

Equations (friction ff up the slope, I=25mR2I=\frac25mR^2):
ma=mgsin⁡θ−f,IaR=fR−τ.ma=mg\sin\theta-f,\qquad I\frac aR=fR-\tau.

Solve. Eliminating ff:
a(m+IR2)=mgsin⁡θ−τR  ⟹  a=5−11.4=207≈2.86 m s−2.a\left(m+\frac{I}{R^2}\right)=mg\sin\theta-\frac\tau R\implies a=\frac{5-1}{1.4}=\frac{20}7\approx2.86\ \text{m s}^{-2}.

Topics:Rolling motionTorque and rotational dynamics
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