Rolling Motion

Question

A solid sphere is released from rest and rolls without slipping down a rough incline that makes an angle of 30 degrees with the horizontal. Taking g as 10 m/s^2, what is the linear acceleration of the sphere's centre of mass?

RankGuru · Accepted Answer

For a body rolling without slipping down an incline, applying Newton's second law along the slope together with the torque equation about the centre yields $a = \frac{g \sin	heta}{1 + \frac{I}{MR^2}}$. The static friction needed to prevent slipping reduces the acceleration below the frictionless value. A solid sphere has $I = \frac{2}{5}MR^2$, so the factor $\frac{I}{MR^2} = \frac{2}{5} = 0.4$. Substituting, $a = \frac{10 	imes \sin 30^\circ}{1 + 0.4} = \frac{10 	imes 0.5}{1.4} = \frac{5}{1.4} \approx 3.57	ext{ m/s}^2$. The value 5 m/s$^2$ ignores rotational inertia and is just $g\sin	heta$, valid only for frictionless sliding. The value 2.5 m/s$^2$ halves the correct result without basis. The value 7.14 m/s$^2$ wrongly multiplies rather than divides by the inertia factor. This follows the NCERT analysis of rolling bodies on an incline. As a plausibility check, a rolling sphere must accelerate more slowly than a frictionless block ($3.57 < 5$), since part of the gravitational drive goes into rotation, and the answer respects this bound. It is also notable that this acceleration is independent of the sphere's mass and radius, depending only on the incline angle and the dimensionless shape factor, so a tiny marble and a large solid ball would slide-free roll down the same slope with identical acceleration.

Rolling Motion

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