JEE Advanced PYQ System of Particles and Rotational Motion – Angular Displacement, Velocity and Acceleration

Q 1 :

Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed $ω$ . The discs are in the same horizontal plane. At time $t = 0$ , the points $P$ and $Q$ are facing each other as shown in the figure. The relative speed between the two points $P$ and $Q$ is $v_{r}$ . In one time period (T) of rotation of the discs, $v_{r}$ as a function of time is best represented by [2012]

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(1)

At $t = 0, t = \frac{T}{2}$ and $t = T$ the relative velocity will be zero.

At $t = \frac{T}{4}$ and $t = \frac{3 T}{4}$ , the relative velocity will be maximum in magnitude.

Hence graph (1) correctly depicts $v_{r}$ versus $t$ graph.

Q 2 :

Consider a disc rotating in the horizontal plane with a constant angular speed $ω$ about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the $y - z$ plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1/8 rotation, (ii) their range is less than half the disc radius, and (iii) $ω$ remains constant throughout. Then [2012]

P lands in the shaded region and Q in the unshaded region.
P lands in the unshaded region and Q in the shaded region.
Both P and Q land in the unshaded region.
Both P and Q land in the shaded region.

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(3)

$At t = \frac{1}{8} \times \frac{2 π}{ω} = \frac{π}{4 ω}$

$x-coordinate of P = ω R (\frac{π}{4 ω}) = \frac{π R}{4} > R \cos 45 °$

$Therefore, both particles P and Q land in unshaded region.$

Q 3 :

A disc of mass M and radius R is free to rotate about its vertical axis as shown in the figure. A battery operated motor of negligible mass is fixed to this disc at a point on its circumference. Another disc of the same mass M and radius $\frac{R}{2}$ is fixed to the motor’s thin shaft. Initially, both the discs are at rest. The motor is switched on so that the smaller disc rotates at a uniform angular speed $ω$ . If the angular speed at which the large disc rotates is $\frac{ω}{n}$ , then the value of $n$ is ______. [2024]

(12)

Using conservation of angular momentum of the system (bigger + smaller disc) about the symmetric axis of large disc.

$\frac{1}{2} M {(\frac{R}{2})}^{2} ω - M (ω' R) . R - \frac{M R^{2}}{2} . ω' = 0$

$\Rightarrow \frac{ω}{8} = \frac{3 ω'}{2} \Rightarrow ω' = \frac{ω}{12}$

$Hence, the value of n = 12$

Q 4 :

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $ω$ is an example of non-inertial frame of reference. The relationship between the force ${\vec{F}}_{rot}$ experienced by a particle of mass $m$ moving on the rotating disc and the force ${\vec{F}}_{in}$ experienced by the particle in an inertial frame of reference is

${\vec{F}}_{rot} = {\vec{F}}_{in} + 2 m ({\vec{v}}_{rot} \times \vec{ω}) + m (\vec{ω} \times \vec{r}) \times \vec{ω} .$

where ${\vec{v}}_{rot}$ is the velocity of the particle in the rotating frame of reference and $\vec{r}$ is the position vector of the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed $ω$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis

perpendicular to the slot and the z-axis along the rotation axis $(\vec{ω} = ω \hat{k})$ . A small block of mass $m$ is gently placed in the slot at $\vec{r} = (\frac{R}{2}) \hat{i}$ at $t = 0$ and is constrained to move only along the slot.

Q. The distance $r$ of the block at time $t$ is [2016]

$\frac{R}{4} (e^{ω t} + e^{- ω t})$
$\frac{R}{2} \cos ω t$
$\frac{R}{4} (e^{2 ω t} + e^{- 2 ω t})$
$\frac{R}{2} \cos 2 ω t$

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(1)

Force on the block along slot $= m r ω^{2} = m a = m v \frac{d v}{d r}$

$∴$ $\int_{0}^{v} V d v = \int_{r / 2}^{r} ω^{2} r d r \Rightarrow v = ω \sqrt{r^{2} - \frac{R^{2}}{4}} = \frac{d r}{d t}$

$∴$ $\int_{R / 4}^{r} \frac{d r}{\sqrt{r^{2} - \frac{R^{2}}{4}}} = \int_{0}^{t} ω d t$

$On solving we get r + \sqrt{r^{2} - \frac{R^{2}}{4}} = \frac{R}{2} e^{ω t}$

$or r^{2} - \frac{R^{2}}{4} = \frac{R^{2}}{4} e^{2 ω t} + r^{2} - 2 r \cdot \frac{R}{2} e^{ω t}$

$∴$ $r = \frac{R}{4} (e^{ω t} + e^{- ω t})$

Q 5 :

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $ω$ is an example of non-inertial frame of reference. The relationship between the force ${\vec{F}}_{rot}$ experienced by a particle of mass $m$ moving on the rotating disc and the force ${\vec{F}}_{in}$ experienced by the particle in an inertial frame of reference is

${\vec{F}}_{rot} = {\vec{F}}_{in} + 2 m ({\vec{v}}_{rot} \times \vec{ω}) + m (\vec{ω} \times \vec{r}) \times \vec{ω} .$

where ${\vec{v}}_{rot}$ is the velocity of the particle in the rotating frame of reference and $\vec{r}$ is the position vector of the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed $ω$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis

perpendicular to the slot and the z-axis along the rotation axis $(\vec{ω} = ω \hat{k})$ . A small block of mass $m$ is gently placed in the slot at $\vec{r} = (\frac{R}{2}) \hat{i}$ at $t = 0$ and is constrained to move only along the slot.

Q. The net reaction of the disc on the block is [2016]

$\frac{1}{2} m ω^{2} R (e^{2 ω t} - e^{- 2 ω t}) \hat{j} + m g \hat{k}$
$\frac{1}{2} m ω^{2} R (e^{ω t} - e^{- ω t}) \hat{j} + m g \hat{k}$
$- m ω^{2} R \cos ω t \hat{j} - m g \hat{k}$
$m ω^{2} R \sin ω t \hat{j} - m g \hat{k}$

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(2)

${\vec{F}}_{rot} = {\vec{F}}_{in} + 2 m (V_{rot} \hat{i}) \times ω \hat{k} + m (ω \hat{k} \times r \hat{i}) \times ω \hat{k}$

$∴$ $m r ω^{2} \hat{i} = {\vec{F}}_{in} + 2 m V_{rot} ω (- \hat{j}) + m ω^{2} r \hat{i}$

${\vec{F}}_{in} = m r V_{rot} ω \hat{j} . . . (i)$

$But r = \frac{R}{4} (e^{ω t} + e^{- ω t})$

$∴$ $\frac{d r}{d t} = V_{r} = \frac{R}{4} (ω e^{ω t} - ω e^{- ω t}) . . . (i i)$

$From (i) and (ii)$

${\vec{F}}_{in} = 2 m \frac{R ω}{4} (e^{ω t} - e^{- ω t}) ω \hat{j}$

$∴$ ${\vec{F}}_{in} = \frac{m R ω^{2}}{2} (e^{ω t} - e^{- ω t}) \hat{j}$

$∴$ ${\vec{F}}_{reaction} = \frac{m R ω^{2}}{2} (e^{ω t} - e^{- ω t}) \hat{j} + m g \hat{k}$

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