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QJEEM24Pheb6a: JEE Main 2024 Physics - Class 11 JEEM24Pheb6a Solution: A circular disc reaches from top to bottom of an inclined plane …

A circular disc reaches from top to bottom of an inclined plane of length ll . When it slips down the plane, if takes t st \mathrm{~s} . When it rolls down the plane then it takes (α2)1/2t s\left(\frac{\alpha}{2}\right)^{1 / 2} t \mathrm{~s} , where α\alpha is _________.

Step-by-Step Verified Solution

To find the value of α\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.

  1. Equation for Time in Slipping Mode:

The acceleration aa of the disc while slipping is given by:

a=gsin⁡θa = g \sin \theta

where gg is the acceleration due to gravity and θ\theta is the angle of the inclined plane.

The time tt to travel down the incline of length ll with this acceleration from rest is:

l=12at2⇒t=2la=2lgsin⁡θl = \frac{1}{2} a t^2 \Rightarrow t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{2l}{g \sin \theta}}

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.

  1. Equation for Time in Rolling Mode:

For a solid disc, the moment of inertia II is 12MR2\frac{1}{2} MR^2 , where MM is the mass and RR is the radius of the disc. The acceleration aa when rolling down without slipping is reduced due to the rotational inertia:

a=gsin⁡θ1+IMR2=gsin⁡θ1+12=2gsin⁡θ3a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2g \sin \theta}{3}

The time to travel the same distance ll is:

troll=2laroll=2l2gsin⁡θ3=3lgsin⁡θt_{\text{roll}} = \sqrt{\frac{2l}{a_{\text{roll}}}} = \sqrt{\frac{2l}{\frac{2g \sin \theta}{3}}} = \sqrt{\frac{3l}{g \sin \theta}}

Compare Times:

Given in the problem is the relation:

troll=(α2)1/2tt_{\text{roll}} = \left(\frac{\alpha}{2}\right)^{1/2} t

From the derived formulas:

3lgsin⁡θ=(α2)1/22lgsin⁡θ\sqrt{\frac{3l}{g \sin \theta}} = \left(\frac{\alpha}{2}\right)^{1/2} \sqrt{\frac{2l}{g \sin \theta}}

Solving for α\alpha :

3=(α2)1/22\sqrt{3} = \left(\frac{\alpha}{2}\right)^{1/2} \sqrt{2}

3=α2×23 = \frac{\alpha}{2} \times 2

3=α3 = \alpha

Conclusion:

Thus, α\alpha is 3.

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