JEE Challenger
More from System of Particles and Rotational Motion
PhyMedAdv 2012 • Paper 2

Relative Speed Between Opposite Points on Two Rotating Discs

Two identical circular discs of radius RR rotate about their central axes in opposite directions with identical constant angular speed ω\omega. Both discs lie in the same horizontal plane. At initial time t=0t = 0, points PP and QQ on the respective circumferences directly face each other as depicted in the figure. Let vrv_r be the magnitude of the relative velocity between points PP and QQ. Which of the following graphs best represents vrv_r as a function of time over one full period of rotation TT?

Question Diagram 1

Options

A
Option A
Correct
B
Option B
C
Option C
D
Option D

Topics & Concepts

System of Particles and Rotational MotionRelative MotionKinematics

Step-by-Step Solution

To determine the correct graph representing the magnitude of the relative velocity vr(t)v_r(t) between points PP and QQ as a function of time tt, we can set up a Cartesian coordinate system with the horizontal axis aligned along the line connecting the centers of the two discs, and the vertical axis perpendicular to it.

Let the center of the left disc be C1C_1 and the center of the right disc be C2C_2.

1. Kinematics of Points PP and QQ

At t=0t = 0:

  • Point PP lies on the rightmost edge of the left disc. Its initial angular position relative to C1C_1 is θ1(0)=0\theta_1(0) = 0.
  • Point QQ lies on the leftmost edge of the right disc. Its initial angular position relative to C2C_2 is θ2(0)=π\theta_2(0) = \pi.

The two discs rotate in opposite directions with constant angular speed ω\omega. Without loss of generality, assume the left disc rotates counter-clockwise and the right disc rotates clockwise.

  • For point PP on the left disc (counter-clockwise rotation): r⃗P(t)=C1+Rcos⁡(ωt)i^+Rsin⁡(ωt)j^\vec{r}_P(t) = C_1 + R \cos(\omega t)\hat{i} + R \sin(\omega t)\hat{j} Differentiating with respect to time tt, the velocity vector of point PP is: v⃗P(t)=−Rωsin⁡(ωt)i^+Rωcos⁡(ωt)j^\vec{v}_P(t) = -R\omega \sin(\omega t)\hat{i} + R\omega \cos(\omega t)\hat{j}

  • For point QQ on the right disc (clockwise rotation): r⃗Q(t)=C2+Rcos⁡(π−ωt)i^+Rsin⁡(π−ωt)j^=C2−Rcos⁡(ωt)i^+Rsin⁡(ωt)j^\vec{r}_Q(t) = C_2 + R \cos(\pi - \omega t)\hat{i} + R \sin(\pi - \omega t)\hat{j} = C_2 - R \cos(\omega t)\hat{i} + R \sin(\omega t)\hat{j} Differentiating with respect to time tt, the velocity vector of point QQ is: v⃗Q(t)=Rωsin⁡(ωt)i^+Rωcos⁡(ωt)j^\vec{v}_Q(t) = R\omega \sin(\omega t)\hat{i} + R\omega \cos(\omega t)\hat{j}

2. Relative Velocity and Relative Speed

layout The relative velocity vector v⃗r(t)\vec{v}_r(t) between points PP and QQ is: v⃗r(t)=v⃗P(t)−v⃗Q(t)=(−Rωsin⁡(ωt)−Rωsin⁡(ωt))i^+(Rωcos⁡(ωt)−Rωcos⁡(ωt))j^\vec{v}_r(t) = \vec{v}_P(t) - \vec{v}_Q(t) = \left( -R\omega \sin(\omega t) - R\omega \sin(\omega t) \right)\hat{i} + \left( R\omega \cos(\omega t) - R\omega \cos(\omega t) \right)\hat{j}

v⃗r(t)=−2Rωsin⁡(ωt)i^\vec{v}_r(t) = -2R\omega \sin(\omega t)\hat{i}

The magnitude of the relative velocity (relative speed) vr(t)v_r(t) is given by: vr(t)=∣v⃗r(t)∣=2Rω∣sin⁡(ωt)∣v_r(t) = |\vec{v}_r(t)| = 2R\omega |\sin(\omega t)|

3. Behavior Over One Full Period TT

Since the time period of rotation is T=2πωT = \frac{2\pi}{\omega}, we evaluate vr(t)v_r(t) at key intervals:

  • At t=0t = 0: vr(0)=2Rω∣sin⁡(0)∣=0v_r(0) = 2R\omega |\sin(0)| = 0
  • At t=T4=π2ωt = \frac{T}{4} = \frac{\pi}{2\omega}: vr(T4)=2Rω∣sin⁡(π2)∣=2Rωv_r\left(\frac{T}{4}\right) = 2R\omega \left|\sin\left(\frac{\pi}{2}\right)\right| = 2R\omega
  • At t=T2=πωt = \frac{T}{2} = \frac{\pi}{\omega}: vr(T2)=2Rω∣sin⁡(π)∣=0v_r\left(\frac{T}{2}\right) = 2R\omega |\sin(\pi)| = 0
  • At t=3T4=3π2ωt = \frac{3T}{4} = \frac{3\pi}{2\omega}: vr(3T4)=2Rω∣sin⁡(3π2)∣=2Rωv_r\left(\frac{3T}{4}\right) = 2R\omega \left|\sin\left(\frac{3\pi}{2}\right)\right| = 2R\omega
  • At t=T=2πωt = T = \frac{2\pi}{\omega}: vr(T)=2Rω∣sin⁡(2π)∣=0v_r(T) = 2R\omega |\sin(2\pi)| = 0

Thus, vr(t)v_r(t) consists of two positive semi-sinusoidal lobes (humps) over one full period TT, starting and ending at zero. This profile is correctly represented in graph (A).

Correct Option: A

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