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QJEEAdv12Ph1270: JEE Advanced 2012 Physics - Class 11 JEEAdv12Ph1270 Solution: The figure shows a system consisting of (i) a ring of outer…

The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω\omega and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed ω2{\omega \over 2} . The ring and disc are separated by frictionless ball bearings. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30∘30^\circ with the horizontal. Then with respect to the horizontal surface,
IIT-JEE 2012 Paper 2 Offline Physics - Rotational Motion Question 54 English

Multiple Choice Options

  • A the point O has linear velocity 3Rωi^3R\omega \widehat i
  • B the point P has linear velocity 114Rωi^+34Rωk^{{11} \over 4}R\omega \widehat i + {{\sqrt 3 } \over 4}R\omega \widehat k
  • C the point P has linear velocity 134Rωi^−34Rωk^{{13} \over 4}R\omega \widehat i - {{\sqrt 3 } \over 4}R\omega \widehat k
  • D the point P has linear velocity (3−34)Rωi^+14Rωk^\left( {3 - {{\sqrt 3 } \over 4}} \right)R\omega \widehat i + {1 \over 4}R\omega \widehat k

Step-by-Step Verified Solution

The outer ring rolls without slipping, meaning the point on the ring in contact with the ground (denoted as point C) is stationary, i.e., v⃗C=0→\vec{v}_{\mathrm{C}} = \overrightarrow{0} . Therefore, the velocity of point O is :

v⃗O=v⃗C+ω⃗o×r⃗CO=0→+ωȷ^×3Rk^=3Rωı^\vec{v}_{\mathrm{O}} = \vec{v}_{\mathrm{C}} + \vec{\omega}_{\mathrm{o}} \times \vec{r}_{\mathrm{CO}} = \overrightarrow{0} + \omega \hat{\jmath} \times 3R \hat{k} = 3R\omega \hat{\imath}

Point P is located on the inner disc, which has an angular velocity ω⃗i=−ω/2ȷ^\vec{\omega}_{\mathrm{i}} = -\omega / 2 \hat{\jmath} . The position vector from O to P is given by :

r⃗OP=Rcos⁡30∘ı^+Rsin⁡30∘k^=3R2ı^+R2k^\vec{r}_{\mathrm{OP}} = R \cos 30^{\circ} \hat{\imath} + R \sin 30^{\circ} \hat{k} = \frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k}

Thus, the velocity of point P is calculated as follows :

v⃗Pamp;=v⃗O+ω⃗i×r⃗OPamp;=3Rωı^+(−ω2ȷ^)×(3R2ı^+R2k^)amp;=11ωR4ı^+3ωR4k^.\begin{aligned} \vec{v}_{\mathrm{P}} & = \vec{v}_{\mathrm{O}} + \vec{\omega}_{\mathrm{i}} \times \vec{r}_{\mathrm{OP}} \\ & = 3R\omega \hat{\imath} + \left(-\frac{\omega}{2} \hat{\jmath}\right) \times \left(\frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k}\right) \\ & = \frac{11\omega R}{4} \hat{\imath} + \frac{\sqrt{3} \omega R}{4} \hat{k} . \end{aligned}

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