The figure shows a system consisting of (i) a ring of outer radius $3 R$ rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and (ii) an inner disc of radius $2 R$ rotating anti-clockwise with angular speed $\omega / 2$. The ring and disc are separated by frictionless ball bearings. The system is in the $x-z$ plane. The point $P$ on the inner disc is at a distance $R$ from the origin, where $O P$ makes an angle of $30^{\circ}$ with the horizontal. Then with respect to the horizontal surface,

(A) the point $O$ has a linear velocity $3 R \omega \hat{i}$.; (B) the point $P$ has a linear velocity $\frac{11}{4} R \omega \hat{i}+\frac{\sqrt{3}}{4} R \omega \hat{k}$.

(C) the point $P$ has a linear velocity $\frac{13}{4} R \omega \hat{i}-\frac{\sqrt{3}}{4} R \omega \hat{k}$.

(D) the point $P$ has a linear velocity $\left(3-\frac{\sqrt{3}}{4}\right) R \omega \hat{i}+\frac{1}{4} R \omega \hat{k}$.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Rotational Mechanics

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The figure shows a system consisting of (i) a ring of outer radius

3 R

rolling clockwise without slipping on a horizontal surface with angular speed

ω

and (ii) an inner disc of radius

2 R

rotating anti-clockwise with angular speed

ω /2

. The ring and disc are separated by frictionless ball bearings. The system is in the

x - z

plane. The point

P

on the inner disc is at a distance

R

from the origin, where

O P

makes an angle of

3 0^{\circ}

with the horizontal. Then with respect to the horizontal surface,

(A)the point $O$ has a linear velocity $3 R ω \hat{i}$ .✓ Correct
(B)the point $P$ has a linear velocity $\frac{11}{4} R ω \hat{i} + \frac{3}{4} R ω \hat{k}$ .✓ Correct
(C)the point $P$ has a linear velocity $\frac{13}{4} R ω \hat{i} - \frac{3}{4} R ω \hat{k}$ .
(D)the point $P$ has a linear velocity $(3 - \frac{3}{4}) R ω \hat{i} + \frac{1}{4} R ω \hat{k}$ .

Answer: (A), (B)

Solution

The ring rolls without slipping, so the velocity of its center

O

v_{O} = 3 R ω \hat{i}

. The disc rotates anti-clockwise with angular speed

ω /2

about

O

. Point

P

is at distance

R

from

O

30°

to horizontal. The velocity of

P

relative to

O

v_{P, O} = \frac{ω}{2} \times R \times (perpendicular direction) = \frac{R ω}{2} (- sin 30° \hat{i} + cos 30° \hat{k}) = - \frac{R ω}{4} \hat{i} + \frac{R ω 3}{4} \hat{k}

. Therefore,

v_{P} = v_{O} + v_{P, O} = 3 R ω \hat{i} - \frac{R ω}{4} \hat{i} + \frac{R ω 3}{4} \hat{k} = \frac{11 R ω}{4} \hat{i} + \frac{R ω 3}{4} \hat{k}

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JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Rotational Mechanics

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