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 \omega and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed \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^\circ with the horizontal. Then with respect to the horizontal surface,

the point O has linear velocity 3R\omega \widehat i, the point P has linear velocity 13 \over 4R\omega \widehat i - \sqrt 3 \over 4R\omega \widehat k

the point O has linear velocity 3R\omega \widehat i

the point P has linear velocity 11 \over 4R\omega \widehat i + \sqrt 3 \over 4R\omega \widehat k

the point P has linear velocity 13 \over 4R\omega \widehat i - \sqrt 3 \over 4R\omega \widehat k

the point P has linear velocity ( 3 - \sqrt 3 \over 4 )R\omega \widehat i + 1 \over 4R\omega \widehat k

Rotational Motion PYQ — JEE Advanced 2012 · Shift 2 · Q48 with Solution

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Correct answer: A, C

Velocity of the center $O$ of the rolling ring

The outer ring has radius $3R$ and rolls clockwise without slipping with angular speed $\omega$ .

For pure rolling on a horizontal surface, $v_O = (3R)\omega$ and since clockwise rolling on the ground means motion toward the right, $\vec v_O = 3R\omega\,\hat i$

So Option A is correct.

Velocity of point $P$ on the inner disc

The inner disc is mounted on frictionless bearings, so its center is also at $O$ . Hence the absolute velocity of any point $P$ on the disc is $\vec v_P = \vec v_O + \vec v_{P/O}$ where $\vec v_{P/O}$ is due to the disc’s own rotation.

The inner disc has radius $2R$ and rotates anticlockwise with angular speed $\omega_d = \frac{\omega}{2}$

Point $P$ is at distance $R$ from $O$ , with $OP$ making angle $30^\circ$ with the horizontal. Thus

= \frac{\sqrt3 R}{2}\,\hat i + \frac{R}{2}\,\hat k$$ Take anticlockwise angular velocity along $+\hat j$ (out of the plane convention with motion in the $ik$-plane): $$\vec \omega_d = \frac{\omega}{2}\,\hat j$$ Then $$\vec v_{P/O} = \vec \omega_d \times \vec r_{P/O}$$ Using $$\hat j \times \hat i = -\hat k, \qquad \hat j \times \hat k = \hat i$$ we get $$\vec v_{P/O} = \frac{\omega}{2}\hat j \times \left(\frac{\sqrt3 R}{2}\hat i + \frac{R}{2}\hat k\right)$$ $$= \frac{\omega}{2}\left(\frac{\sqrt3 R}{2}(\hat j\times \hat i) + \frac{R}{2}(\hat j\times \hat k)\right)$$ $$= \frac{\omega}{2}\left(\frac{\sqrt3 R}{2}(-\hat k) + \frac{R}{2}\hat i\right)$$ $$= \frac{R\omega}{4}\hat i - \frac{\sqrt3 R\omega}{4}\hat k$$ Now add translational velocity of the center: $$\vec v_P = 3R\omega\hat i + \left(\frac{R\omega}{4}\hat i - \frac{\sqrt3 R\omega}{4}\hat k\right)$$ $$\vec v_P = \frac{13}{4}R\omega\hat i - \frac{\sqrt3}{4}R\omega\hat k$$ So **Option C is correct**. --- 3. **Check all options** - **A:** $\vec v_O = 3R\omega\hat i$ ✅ Correct - **B:** $\dfrac{11}{4}R\omega\hat i + \dfrac{\sqrt3}{4}R\omega\hat k$ ❌ Incorrect - **C:** $\dfrac{13}{4}R\omega\hat i - \dfrac{\sqrt3}{4}R\omega\hat k$ ✅ Correct - **D:** $\left(3-\dfrac{\sqrt3}{4}\right)R\omega\hat i + \dfrac14 R\omega\hat k$ ❌ Incorrect --- 4. **Final derived answer** The correct options are: $$\boxed{A,\ C}$$ --- 5. **Comparison with stored answer** Stored correct answer: **A, B** My derived answer: **A, C** These do **not** match. The likely issue in the stored answer is the rotational velocity of point $P$ on the inner disc: for anticlockwise rotation, the velocity at a point in the first quadrant must have a positive $\hat i$ component and negative $\hat k$ component, which matches option C, not B.

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View written solutionFree

Correct answer: A, C

Velocity of the center $O$ of the rolling ring

The outer ring has radius $3R$ and rolls clockwise without slipping with angular speed $\omega$ .

For pure rolling on a horizontal surface, $v_O = (3R)\omega$ and since clockwise rolling on the ground means motion toward the right, $\vec v_O = 3R\omega\,\hat i$

So Option A is correct.

Velocity of point $P$ on the inner disc

The inner disc has radius $2R$ and rotates anticlockwise with angular speed $\omega_d = \frac{\omega}{2}$

Point $P$ is at distance $R$ from $O$ , with $OP$ making angle $30^\circ$ with the horizontal. Thus

= \frac{\sqrt3 R}{2}\,\hat i + \frac{R}{2}\,\hat k$$ Take anticlockwise angular velocity along $+\hat j$ (out of the plane convention with motion in the $ik$-plane): $$\vec \omega_d = \frac{\omega}{2}\,\hat j$$ Then $$\vec v_{P/O} = \vec \omega_d \times \vec r_{P/O}$$ Using $$\hat j \times \hat i = -\hat k, \qquad \hat j \times \hat k = \hat i$$ we get $$\vec v_{P/O} = \frac{\omega}{2}\hat j \times \left(\frac{\sqrt3 R}{2}\hat i + \frac{R}{2}\hat k\right)$$ $$= \frac{\omega}{2}\left(\frac{\sqrt3 R}{2}(\hat j\times \hat i) + \frac{R}{2}(\hat j\times \hat k)\right)$$ $$= \frac{\omega}{2}\left(\frac{\sqrt3 R}{2}(-\hat k) + \frac{R}{2}\hat i\right)$$ $$= \frac{R\omega}{4}\hat i - \frac{\sqrt3 R\omega}{4}\hat k$$ Now add translational velocity of the center: $$\vec v_P = 3R\omega\hat i + \left(\frac{R\omega}{4}\hat i - \frac{\sqrt3 R\omega}{4}\hat k\right)$$ $$\vec v_P = \frac{13}{4}R\omega\hat i - \frac{\sqrt3}{4}R\omega\hat k$$ So **Option C is correct**. --- 3. **Check all options** - **A:** $\vec v_O = 3R\omega\hat i$ ✅ Correct - **B:** $\dfrac{11}{4}R\omega\hat i + \dfrac{\sqrt3}{4}R\omega\hat k$ ❌ Incorrect - **C:** $\dfrac{13}{4}R\omega\hat i - \dfrac{\sqrt3}{4}R\omega\hat k$ ✅ Correct - **D:** $\left(3-\dfrac{\sqrt3}{4}\right)R\omega\hat i + \dfrac14 R\omega\hat k$ ❌ Incorrect --- 4. **Final derived answer** The correct options are: $$\boxed{A,\ C}$$ --- 5. **Comparison with stored answer** Stored correct answer: **A, B** My derived answer: **A, C** These do **not** match. The likely issue in the stored answer is the rotational velocity of point $P$ on the inner disc: for anticlockwise rotation, the velocity at a point in the first quadrant must have a positive $\hat i$ component and negative $\hat k$ component, which matches option C, not B.

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