JEE Advanced 2022 — Paper 1 — PHY-I solution

JEE Advanced2022Paper 1PHY-I

A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination $θ = 3 0^{\circ}$ from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance $r = 0.5$ m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ______ ms $^{- 2}$ . (Take $g = 10 ms^{- 2}$ .)

Your answer

Solution

Taking torque about the instantaneous contact point (no slipping). Moment of inertia of solid sphere about contact point: $I_{co n t a c t} = \frac{2}{5} m R^{2} + m R^{2} = \frac{7}{5} m R^{2}$ .

Net torque about contact point:

Gravity component along incline: $m g sin 3 0^{\circ} \cdot R = 1 \cdot 10 \cdot 0.5 \cdot 1 = 5$ N·m (driving rotation down-the-incline).

The two applied 1 N forces act at $r = 0.5$ m from centre. With one force directed up the incline and the other down the incline (on opposite sides of the centre), their torques about the contact point partially cancel. Net opposing torque: $1 \cdot 1 = 1$ N·m.

Net torque: $τ = 5 - 1 = 4$ N·m.

$α = \frac{τ}{I _{co n t a c t}} = \frac{4}{( 7/5 ) ( 1 ) ( 1 ) ^{2}} = \frac{20}{7}$ rad/s $^{2}$ .

$a_{c m} = α R = \frac{20}{7} \approx 2.86$ m/s $^{2}$ .

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