JEE Challenger
More from System of Particles and Rotational Motion
PhyMedAdv 2018 • Paper 1

Height of Inclined Plane from Rolling Ring and Disc Arrival Times

A uniform ring and a circular disc are released from rest simultaneously from the top of an inclined plane inclined at an angle of 60∘60^\circ to the horizontal. Both objects roll down without slipping along the line of greatest slope. If the difference in the time taken by them to reach the bottom is (2−3)/10 s(2 - \sqrt{3})/\sqrt{10}\text{ s}, find the vertical height of the incline in metres. (Take g=10 m s−2g = 10\text{ m s}^{-2}).

Official Numerical Answer0.75

Topics & Concepts

System of Particles and Rotational MotionRolling MotionMoment of Inertia

Step-by-Step Solution

To find the vertical height of the incline, we analyze the rolling motion of the ring and the disc.

For an object of mass mm, radius RR, and radius of gyration kk, rolling down an inclined plane of angle θ\theta without slipping, the linear acceleration along the plane is given by: a=gsin⁡θ1+k2R2a = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}

  1. For the uniform ring: The moment of inertia about its central axis is I=mR2I = mR^2, which gives k2R2=1\frac{k^2}{R^2} = 1. aring=gsin⁡60∘1+1=gsin⁡60∘2a_{\text{ring}} = \frac{g \sin 60^\circ}{1 + 1} = \frac{g \sin 60^\circ}{2}

  2. For the circular disc: The moment of inertia about its central axis is I=12mR2I = \frac{1}{2}mR^2, which gives k2R2=12\frac{k^2}{R^2} = \frac{1}{2}. adisc=gsin⁡60∘1+12=23gsin⁡60∘a_{\text{disc}} = \frac{g \sin 60^\circ}{1 + \frac{1}{2}} = \frac{2}{3} g \sin 60^\circ

Let hh be the vertical height of the inclined plane. The distance along the incline is: s=hsin⁡60∘s = \frac{h}{\sin 60^\circ}

Since both objects start from rest, the time taken to travel distance ss is: t=2sat = \sqrt{\frac{2s}{a}}

Substituting the accelerations:

  • For the ring: tring=2sgsin⁡60∘2=4sgsin⁡60∘=2sgsin⁡60∘t_{\text{ring}} = \sqrt{\frac{2s}{\frac{g \sin 60^\circ}{2}}} = \sqrt{\frac{4s}{g \sin 60^\circ}} = 2 \sqrt{\frac{s}{g \sin 60^\circ}}

  • For the disc: tdisc=2s23gsin⁡60∘=3sgsin⁡60∘=3sgsin⁡60∘t_{\text{disc}} = \sqrt{\frac{2s}{\frac{2}{3}g \sin 60^\circ}} = \sqrt{\frac{3s}{g \sin 60^\circ}} = \sqrt{3} \sqrt{\frac{s}{g \sin 60^\circ}}

The difference in time taken to reach the bottom is: Δt=tring−tdisc=(2−3)sgsin⁡60∘\Delta t = t_{\text{ring}} - t_{\text{disc}} = (2 - \sqrt{3}) \sqrt{\frac{s}{g \sin 60^\circ}}

Substituting s=hsin⁡60∘s = \frac{h}{\sin 60^\circ}: Δt=(2−3)hgsin⁡260∘=2−3sin⁡60∘hg\Delta t = (2 - \sqrt{3}) \sqrt{\frac{h}{g \sin^2 60^\circ}} = \frac{2 - \sqrt{3}}{\sin 60^\circ} \sqrt{\frac{h}{g}}

Since sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}: Δt=2(2−3)3hg\Delta t = \frac{2(2 - \sqrt{3})}{\sqrt{3}} \sqrt{\frac{h}{g}}

We are given that Δt=2−310 s\Delta t = \frac{2 - \sqrt{3}}{\sqrt{10}}\text{ s} and g=10 m s−2g = 10\text{ m s}^{-2}: 2(2−3)3h10=2−310\frac{2(2 - \sqrt{3})}{\sqrt{3}} \sqrt{\frac{h}{10}} = \frac{2 - \sqrt{3}}{\sqrt{10}}

Dividing both sides by 2−310\frac{2 - \sqrt{3}}{\sqrt{10}}: 23h=1\frac{2}{\sqrt{3}} \sqrt{h} = 1

h=32\sqrt{h} = \frac{\sqrt{3}}{2}

Squaring both sides: h=34=0.75 mh = \frac{3}{4} = 0.75\text{ m}

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