Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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) $\dfrac{{2g\sin \theta }}{3}$ (C) $\dfrac{{g\sin \theta }}{2}$ (D) zero

Last updated date: 07th Sep 2024
Total views: 77.4k
Views today: 1.77k
Verified
77.4k+ views
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.