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) gsin θ1 - MR²/I (b) gsin θ1 + I/MR² (c) gsin θ1 + MR²/I (d) gsin θ1 - I/MR²

Answer: (b). Along incline: Mgsin θ - f = Ma; about centre: fR = Iα = I a/R. So f = Ia/R² and Mgsin θ = a(M + I/R²). a = (gsin θ)/(1 + I/MR²): answer (b).

A round uniform body of radius R, mass M and moment of inertia I… | JEE Main 2007 (AIEEE)

Q83JEE Main 2007 · AIEEEMedium

Rolling Motion

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

Show answer and solution

Answer: (b)

Along incline: $Mg\sin\theta - f = Ma$ ; about centre: $fR = I\alpha = I a/R$ .
So $f = \dfrac{Ia}{R^2}$ and $Mg\sin\theta = a\left(M + \dfrac{I}{R^2}\right)$ .
$a = \dfrac{g\sin\theta}{1 + I/MR^2}$ : answer (b).

Report a mistake

Rolling Motion question from JEE Main 2007 (AIEEE)

More questions on Rolling Motion