Question 401: Centre of Mass and Rotational Motion

Question

A circular disc reaches from top to bottom of an inclined plane of length $$l$$. When it slips down the plane, if takes $$t \mathrm{~s}$$. When it rolls down the plane then it takes $$\left(\frac{\alpha}{2}\right)^{1 / 2} t \mathrm{~s}$$, where $$\alpha$$ is _________.

Accepted Answer

To find the value of $ \alpha $ from the given problem, we need to analyze the motion of a circular disc moving down an inclined plane in two different modes: slipping and rolling.
Slipping:
When the disc slips without rolling, it is primarily subjected to kinetic friction and gravity, without any rolling friction or torque affecting rotational motion. The motion can be considered as purely translational.
Equation for Time in Slipping Mode
:
The acceleration $ a $ of the disc while slipping is given by:
$a = g \sin 	heta$
where $ g $ is the acceleration due to gravity and $ 	heta $ is the angle of the inclined plane.
The time $ t $ to travel down the incline of length $ l $ with this acceleration from rest is:
$l = \frac{1}{2} a t^2 \Rightarrow t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{2l}{g \sin 	heta}}$
Rolling:
When the disc rolls, both translational and rotational motions are involved, and the rolling motion means that there is a rotational inertia factor that affects the acceleration.
Equation for Time in Rolling Mode
:
For a solid disc, the moment of inertia $ I $ is $ \frac{1}{2} MR^2 $, where $ M $ is the mass and $ R $ is the radius of the disc. The acceleration $ a $ when rolling down without slipping is reduced due to the rotational inertia:
$a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}} = \frac{g \sin 	heta}{1 + \frac{1}{2}} = \frac{2g \sin 	heta}{3}$
The time to travel the same distance $ l $ is:
$t_{	ext{roll}} = \sqrt{\frac{2l}{a_{	ext{roll}}}} = \sqrt{\frac{2l}{\frac{2g \sin 	heta}{3}}} = \sqrt{\frac{3l}{g \sin 	heta}}$
Compare Times:
Given in the problem is the relation:
$t_{	ext{roll}} = \left(\frac{\alpha}{2}ight)^{1/2} t$
From the derived formulas:
$\sqrt{\frac{3l}{g \sin 	heta}} = \left(\frac{\alpha}{2}ight)^{1/2} \sqrt{\frac{2l}{g \sin 	heta}}$
Solving for $ \alpha $:
$\sqrt{3} = \left(\frac{\alpha}{2}ight)^{1/2} \sqrt{2}$
$3 = \frac{\alpha}{2} 	imes 2$
$3 = \alpha$
Conclusion:
Thus, $ \alpha $ is
3
.

Centre of Mass and Rotational Motion

A circular disc reaches from top to bottom of an inclined plane of length $l. When it slips down the plane, if takes t \mathrm{~s}. When it rolls down the plane then it takes \left(\frac{\alpha}{2}\right)^{1 / 2} t \mathrm{~s}, where \alpha$ is _________.

Choose the correct answer: