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← IIT-JEE 2012 · Paper 2
PhysicsMultiple correct+4Question 20 of 60

IIT-JEE 2012 · Paper 2, Physics question 20

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 system is in the xx-zz plane. The point PP on the inner disc is at a distance RR from the origin, where OPOP makes an angle of 30° with the horizontal. Then with respect to the horizontal surface,

A ring of outer radius 3R rolling clockwise (ω) on a horizontal surface, with an inner disc of radius 2R on ball bearings rotating anti-clockwise at ω/2; point P on the inner disc is at distance R from the centre O, with OP at 30° above the +x axis.
Choose one or more options
Answer and worked solution

Official answer: A, B

Centre. The ring (radius 3R3R) rolls clockwise without slipping, so v⃗O=3Rω i^\vec v_O=3R\omega\,\hat i. (A) true.

Point P. P belongs to the inner disc, which turns anticlockwise at ω/2\omega/2 about O (the bearings let it rotate independently). With OP→=R(cos⁡30°,sin⁡30°)\overrightarrow{OP}=R(\cos30°,\sin30°) in the xx-zz plane, anticlockwise rotation gives
v⃗P/O=ω2R(−sin⁡30°,cos⁡30°)=Rω(−14,34).\vec v_{P/O}=\frac\omega2R(-\sin30°,\cos30°)=R\omega\left(-\frac14,\frac{\sqrt3}4\right).

Total.
v⃗P=(3−14)Rω i^+34Rω k^=114Rω i^+34Rω k^.\vec v_P=\left(3-\frac14\right)R\omega\,\hat i+\frac{\sqrt3}4R\omega\,\hat k=\frac{11}4R\omega\,\hat i+\frac{\sqrt3}4R\omega\,\hat k.
(B) true; (C) and (D) false.

(A), (B)

Topics:Rolling motionRelative velocity
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