JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Rotational Mechanics

A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of

1 ms^{- 1}

goes up an inclined plane which makes an angle of

3 0^{0}

with the horizontal plane, without slipping. How long will the sphere take to return to the starting point

A

(A)0.60 s
(B)0.57 s✓ Correct
(C)0.52 s
(D)0.80 s

Answer: (B)

Solution

For a solid sphere rolling without slipping,

I = \frac{2}{5} m R^{2}

, and the retardation along the incline is

a = \frac{g sin θ}{1 + \frac{I}{m R ^{2}}} = \frac{g sin 3 0 ^{\circ}}{1 + \frac{2}{5}} = \frac{5 g sin 3 0 ^{\circ}}{7} = \frac{5 ( 9.8 ) ( 0.5 )}{7} = 3.5 m s^{- 2}

Rolling is maintained on the way down as well, so the acceleration has the same magnitude and the descent takes as long as the climb. With an initial speed of

1 m s^{- 1}

t = \frac{2 v}{a} = \frac{2 ( 1 )}{3.5} = 0.57 s

The sphere is back at

A

after

0.57

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Rotational Mechanics

Solution

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