Question

The rolling object rolls without slipping down an incline plane (angle of inclination $$\theta$$), then the minimum acceleration it can have is:
(A) $$g\sin \theta$$
(B) $${\displaystyle \frac{2g\sin \theta }{3}}$$
(C) $${\displaystyle \frac{g\sin \theta }{2}}$$
(D) zero

Answer 1

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: -
 $$a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{I}{m{R}^2}}}}$$
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.
 $$f\times R=I\times {\displaystyle \frac{a}{R}}$$ (using $$a=\alpha R$$ )
 $$f={\displaystyle \frac{Ia}{R^2}}$$ . . . (2)
Adding (1) and (2) we get: -
 $$mg\sin \theta =ma+{\displaystyle \frac{Ia}{R^2}}$$
 $$g\sin \theta =a+{\displaystyle \frac{Ia}{m{R}^2}}$$
 $$g\sin \theta =a(1+{\displaystyle \frac{I}{m{R}^2}})$$
 $$\Rightarrow a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{I}{m{R}^2}}}}$$
Now, in the above formula we see that acceleration will be minimum when the denominator = $$1+{\displaystyle \frac{I}{m{R}^2}}$$ 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: -
 $$a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{m{R}^2}{m{R}^2}}}}$$
 $$a={\displaystyle \frac{g\sin \theta }{1+1}}$$
 $$\Rightarrow a={\displaystyle \frac{g\sin \theta }{2}}$$

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.