Question

A solid sphere of mass 4 kg and radius 28 cm is on an inclined plane. If the acceleration of the sphere when it rolls down without sliding is \( 3.5 \, \text{m s}^{-2} \), then the acceleration of the sphere when it slides down without rolling is

• A. \( 2.5 \, \text{m s}^{-2} \)
• B. \( 3.5 \, \text{m s}^{-2} \)
• C. \( 1.7 \, \text{m s}^{-2} \)
• D. \( 4.9 \, \text{m s}^{-2} \)

Hint:

Acceleration of an object rolling down an inclined plane (angle \( \theta \)) without slipping: \( a_{roll} = \frac{g\sin\theta}{1 + k} \), where \( k = \frac{I}{MR^2} \). For a solid sphere, \( I = \frac{2}{5}MR^2 \), so \( k = 2/5 \). Acceleration of an object sliding down a smooth inclined plane (or when friction is insufficient for rolling and only sliding occurs due to \(Mg\sin\theta\)): \( a_{slide} = g\sin\theta \). The values of mass and radius are irrelevant if \(k\) is known or can be determined from the object's shape.

Answer 1

The Correct Option is D

Case 1: Rolling down without sliding.
Let the angle of inclination be \( \theta \).
The acceleration of a body rolling down an inclined plane without slipping is given by: \[ a_{roll} = \frac{g\sin\theta}{1 + \frac{I}{MR^2}} \] For a solid sphere, moment of inertia \( I = \frac{2}{5}MR^2 \).
So, \( \frac{I}{MR^2} = \frac{2}{5} \).
\[ a_{roll} = \frac{g\sin\theta}{1 + 2/5} = \frac{g\sin\theta}{7/5} = \frac{5}{7}g\sin\theta \] Given \( a_{roll} = 3.
5 \, \text{m s}^{-2} \).
\[ 3.
5 = \frac{5}{7}g\sin\theta \] We can find \( g\sin\theta \): \[ g\sin\theta = 3.
5 \times \frac{7}{5} = \frac{7}{2} \times \frac{7}{5} = \frac{49}{10} = 4.
9 \, \text{m s}^{-2} \] Case 2: Sliding down without rolling.
When the sphere slides down without rolling, friction is either absent or not sufficient to cause rolling (e.
g.
, a perfectly smooth plane, or kinetic friction if it's just sliding).
If it's "sliding without rolling", it typically means friction is negligible or we consider the case where only gravity component along the incline acts.
The force causing the sliding is the component of gravity along the incline, which is \( Mg\sin\theta \).
The acceleration when sliding down is \( a_{slide} = \frac{Mg\sin\theta}{M} = g\sin\theta \).
From Case 1, we found \( g\sin\theta = 4.
9 \, \text{m s}^{-2} \).
So, \( a_{slide} = 4.
9 \, \text{m s}^{-2} \).
The mass (4 kg) and radius (28 cm) are not directly needed to find \(g\sin\theta\) from \(a_{roll}\), and \(a_{slide}\) only depends on \(g\sin\theta\).
This matches option (4).