A round uniform body of radius R , mass M and moment of inertia I , rolls down (without slipping) an inclined plane making an angle θ with the horizon - Physics MCQ AIEEE 2007 2026 | ExamTest.live | ExamTest.live

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A round uniform body of radius R, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is:

A $\frac{g s i n θ}{1 + I / M R ^{2}}$
B $\frac{g s i n θ}{1 + M R ^{2} / I}$
C $\frac{g s i n θ}{1 - I / M R ^{2}}$
D $\frac{g s i n θ}{1 - M R ^{2} / I}$

Solution & Step-by-step Explanation

For a body rolling without slipping down an incline:Linear motion: M g sin θ - f = M a Rotational motion: τ = f R = I α No slipping condition: a = R α ⟹ α = a / R Substituting f from (2) into (1): M g sin θ - \frac{I α}{R} = M a M g sin θ - \frac{I a}{R ^{2}} = M a M g sin θ = a (M + \frac{I}{R ^{2}}) a = \frac{M g sin θ}{M + I / R ^{2}} = \frac{g sin θ}{1 + I / M R ^{2}}

Practice this question

Try it yourself before checking the explanation above.

A round uniform body of radius R, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is:

A

\frac{g s i n θ}{1 + I / M R ^{2}}

B

\frac{g s i n θ}{1 + M R ^{2} / I}

C

\frac{g s i n θ}{1 - I / M R ^{2}}

D

\frac{g s i n θ}{1 - M R ^{2} / I}

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Rotational Mechanics Rolling Motion

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