Question

A solid sphere and a thin uniform circular disc of same radius are rolling down an inclined plane without slipping. If the acceleration of the sphere is 3 ms\(^{-2}\), then the acceleration of the disc is

• A. 4 ms\(^{-2}\)
• B. 2.8 ms\(^{-2}\)
• C. 3 ms\(^{-2}\)
• D. 3.2 ms\(^{-2}\)

Hint:

The acceleration of a body rolling down an incline depends on its moment of inertia factor \( \beta = I/(mR^2) \) via the formula \( a = \frac{g\sin\theta}{1+\beta} \). Smaller \( \beta \) means larger acceleration. A sphere (\(\beta=2/5\)) will always roll faster than a disc (\(\beta=1/2\)), which will roll faster than a ring (\(\beta=1\)).

Answer 1

The Correct Option is B

The formula for the acceleration of an object rolling down an inclined plane without slipping is:
\( a = \frac{g \sin\theta}{1 + \frac{I}{mR^2}} \), where I is the moment of inertia about the center of mass.
The term \( k^2 = I/m \) is the square of the radius of gyration, so the formula can be written as \( a = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}} \).
For a solid sphere, the moment of inertia is \( I_{sphere} = \frac{2}{5}mR^2 \).
The factor \( \frac{I}{mR^2} \) for the sphere is \( \frac{2}{5} \).
So, the acceleration of the sphere is \( a_{sphere} = \frac{g \sin\theta}{1 + 2/5} = \frac{g \sin\theta}{7/5} = \frac{5}{7}g \sin\theta \).
For a thin uniform circular disc, the moment of inertia is \( I_{disc} = \frac{1}{2}mR^2 \).
The factor \( \frac{I}{mR^2} \) for the disc is \( \frac{1}{2} \).
So, the acceleration of the disc is \( a_{disc} = \frac{g \sin\theta}{1 + 1/2} = \frac{g \sin\theta}{3/2} = \frac{2}{3}g \sin\theta \).
We are given that \( a_{sphere} = 3 \) ms\(^{-2}\).
From the sphere's acceleration, we can find the value of the term \( g \sin\theta \).
\( 3 = \frac{5}{7}g \sin\theta \implies g \sin\theta = \frac{3 \times 7}{5} = \frac{21}{5} = 4.2 \).
Now, use this value to find the acceleration of the disc.
\( a_{disc} = \frac{2}{3} (g \sin\theta) = \frac{2}{3} \times 4.2 = 2 \times 1.4 = 2.8 \) ms\(^{-2}\).