IIT-JEE 2012 · Paper 2 · Physics Q8

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PhysicsSingle correct+3 / -1Question 8 of 60

Two identical discs of same radius $R$ are rotating about their axes in opposite directions with the same constant angular speed $\omega$ . The discs are in the same horizontal plane. At time $t=0$ , the points $P$ and $Q$ are facing each other as shown in the figure. The relative speed between the two points $P$ and $Q$ is $v_r$ . In one time period $(T)$ of rotation of the discs, $v_r$ as a function of time is best represented by

Two equal discs of radius R side by side in a horizontal plane, rotating in opposite senses with angular speed ω; point P on the right rim of the left disc faces point Q on the left rim of the right disc.

v_r against t over one period T: two full arches, zero at t = 0, T/2 and T.

v_r against t: starts at a non-zero value, with dips to zero at T/4 and 3T/4 (one full arch and two half arches).

v_r against t: four arches in one period, zero at t = 0, T/4, T/2, 3T/4, T.

v_r against t: four arches offset so that v_r is non-zero at t = 0.

Answer and worked solution

Official answer: A

Velocities. Take P on the left disc (counter-clockwise) and Q on the right disc (clockwise, the mirror image). At angle $\omega t$ from their starting position,
$\vec v_P=R\omega(-\sin\omega t,\cos\omega t),\qquad\vec v_Q=R\omega(\sin\omega t,\cos\omega t).$

Relative speed.
$v_r=|\vec v_P-\vec v_Q|=2R\omega|\sin\omega t|.$
It is zero at $t=0$ , $T/2$ and $T$ , with two equal arches in one period.

(A)

Topics:Relative velocity Uniform circular motion

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IIT-JEE 2012 · Paper 2, Physics question 8