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**Hint:**We will first apply Newton’s laws of motion to calculate acceleration on an inclined plane for pure rolling. We can also directly use the formula for acceleration: -

\[a = \dfrac{{g\sin \theta }}{{1 + \dfrac{I}{{m{R^2}}}}}\]

Where,

a = acceleration of the rolling body on the incline. (pure rolling)

\[\theta \] = inclination of the plane from horizontal.

g = acceleration due to gravity.

I = moment of inertia of the body about its centre of mass.

R = radius of the rolling body.

After calculating this acceleration, we will check for its minimum possible value.

Complete step by step solution

Figure 1/Pure Rolling on an inclined Plane

We will apply Newton’s Law of motion along the incline: -

\[\sum F = m\overrightarrow a \]

Where,

\[\sum F = \] Net Force acting on the body.

M = mass of the body.

a = acceleration of the body in the direction of force.

\[mg\sin \theta - f = ma\] . . . (1)

We will apply newton’s Law for rotational mechanics: -

\[\sum \tau = I\alpha \]

Where,

\[\sum \tau \] = Net Torque acting on the body.

I = Inertia of the body about centre of mass.

\[\alpha \] = Angular Acceleration of the body.

\[f \times R = I \times \dfrac{a}{R}\] (using \[a = \alpha R\] )

\[f = \dfrac{{Ia}}{{{R^2}}}\] . . . (2)

Adding (1) and (2) we get: -

\[mg\sin \theta = ma + \dfrac{{Ia}}{{{R^2}}}\]

\[g\sin \theta = a + \dfrac{{Ia}}{{m{R^2}}}\]

\[g\sin \theta = a(1 + \dfrac{I}{{m{R^2}}})\]

\[ \Rightarrow a = \dfrac{{g\sin \theta }}{{1 + \dfrac{I}{{m{R^2}}}}}\]

Now, in the above formula we see that acceleration will be minimum when the denominator = \[1 + \dfrac{I}{{m{R^2}}}\] will be maximum. The denominator will be maximum for the maximum value of I.

I have its maximum possible value for a ring which is ‘mR$^2$’. Thus, we take \[I = m{R^2}\] .

Now, acceleration becomes: -

\[a = \dfrac{{g\sin \theta }}{{1 + \dfrac{{m{R^2}}}{{m{R^2}}}}}\]

\[a = \dfrac{{g\sin \theta }}{{1 + 1}}\]

\[ \Rightarrow a = \dfrac{{g\sin \theta }}{2}\]

**Hence, option(c) is correct.**

**Note**

(A) In the above question, it is stated that the object rolls without slipping. This is the case for pure rolling. Hence, frictional force will be self-adjusting and not equal to maximum value of friction, that is, \[{\mu _s}N\] . Thus, do not keep f = \[{\mu _s}N\] in the above question to calculate acceleration using eq-(1).

(B) However, this is exactly what is done to calculate acceleration when the rolling motion is not pure. In that case, acceleration is given by the following equation: -

\[a = g\sin \theta - {\mu _k}\cos \theta \] (for impure rolling motion)

(C) Also, remember that the relation: - “ \[a = \alpha R\] ” is only valid in case of pure rolling motion and that is why we were able to use this relation in the above question.

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