JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Rotational Mechanics

A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle

θ

with the horizontal. The acceleration of the will be

\frac{2}{b} g sin θ

where

b

is ____ (Round off to the Nearest Integer)
(

g =

acceleration due to gravity)

θ =

angle as shown in figure)

Answer: 3

Solution

For a body rolling down an incline without slipping, the acceleration of the centre is

a = \frac{g sin θ}{1 + \frac{I}{m a ^{2}}}

where

I

is the moment of inertia about the central axis. A solid disc has

I = \frac{1}{2} m a^{2}

, so

\frac{I}{m a ^{2}} = \frac{1}{2}

and

a = \frac{g sin θ}{1 + \frac{1}{2}} = \frac{2}{3} g sin θ

Comparing this with

\frac{2}{b} g sin θ

gives

b = 3

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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