arthham

JEE Advanced 2012 · System of Particles and Rotational Motion

The figure shows a system consisting of (i) a ring of outer radius 3R3R rolling clockwise without slipping on a horizontal surface with angular speed ω\omega and (ii) an inner disc of radius 2R2R rotating anti-clockwise with angular speed ω/2\omega/2. The ring and disc are separated by frictionless ball bearings. The point PP on the inner disc is at a distance RR from the origin, where OPOP makes an angle of 30∘30^\circ with the horizontal. Then with respect to the horizontal surface,

Answers: A. the point OO has linear velocity 3 Rωi^3\,R\omega\hat{i} · B. the point PP has linear velocity 114Rωi^+34Rωk^\frac{11}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}

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Written solution

Step 1: Find the velocity of the center of mass OO.

The ring of outer radius 3R3R rolls clockwise without slipping on the horizontal surface with angular speed ω\omega. The velocity of the center OO is directed horizontally to the right.

v⃗O=3Rω i^\vec{v}_O = 3R\omega\,\hat{i}

Step 2: Express the position vector of point PP relative to OO.

The point PP is at a distance RR from OO, and the line OPOP makes an angle of 30∘30^\circ with the horizontal.

OP→=Rcos⁡(30∘) i^+Rsin⁡(30∘) k^\overrightarrow{OP} = R\cos(30^\circ)\,\hat{i} + R\sin(30^\circ)\,\hat{k} OP→=R(32i^+12k^)\overrightarrow{OP} = R\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{k}\right)

Step 3: Express the angular velocity of the inner disc.

The inner disc rotates anti-clockwise with angular speed ω2\frac{\omega}{2}. In this coordinate system, the rotation vector points in the −j^-\hat{j} direction.

ω⃗d=−ω2 j^\vec{\omega}_d = -\frac{\omega}{2}\,\hat{j}

Step 4: Calculate the velocity of point PP relative to OO.

Use the relation for the velocity of a point on a rotating body.

v⃗P/O=ω⃗d×OP→\vec{v}_{P/O} = \vec{\omega}_d \times \overrightarrow{OP} v⃗P/O=(−ω2 j^)×R(32i^+12k^)\vec{v}_{P/O} = \left(-\frac{\omega}{2}\,\hat{j}\right) \times R\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{k}\right) v⃗P/O=−14Rω i^+34Rω k^\vec{v}_{P/O} = -\frac{1}{4}R\omega\,\hat{i} + \frac{\sqrt{3}}{4}R\omega\,\hat{k}

Step 5: Find the velocity of point PP with respect to the horizontal surface.

Combine the velocity of the center OO and the relative velocity of PP.

v⃗P=v⃗O+v⃗P/O\vec{v}_P = \vec{v}_O + \vec{v}_{P/O} v⃗P=3Rω i^+(−14Rω i^+34Rω k^)\vec{v}_P = 3R\omega\,\hat{i} + \left(-\frac{1}{4}R\omega\,\hat{i} + \frac{\sqrt{3}}{4}R\omega\,\hat{k}\right) v⃗P=114Rω i^+34Rω k^\vec{v}_P = \frac{11}{4}R\omega\,\hat{i} + \frac{\sqrt{3}}{4}R\omega\,\hat{k}

Answer: The velocity of the center OO is v⃗O=3Rω i^\vec{v}_O = 3R\omega\,\hat{i} and the velocity of point PP is v⃗P=114Rω i^+34Rω k^\vec{v}_P = \dfrac{11}{4}R\omega\,\hat{i} + \dfrac{\sqrt{3}}{4}R\omega\,\hat{k}.

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