JEE Advanced PYQ System of Particles and Rotational Motion – Miscellaneous Mixed Concepts Problems

Q 1 :

A football of radius $R$ is kept on a hole of radius $r (r < R)$ made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted making an angle $θ$ from the horizontal as shown in the figure below. The maximum value of $θ$ so that the football does not start rolling down the plank satisfies (figure is schematic and not drawn to scale) [2020]

$\sin θ = \frac{r}{R}$
$\tan θ = \frac{r}{R}$
$\sin θ = \frac{r}{2 R}$
$\cos θ = \frac{r}{2 R}$

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(1)

The maximum value of $q$ i.e., $q_{\max}$ , the football is about to roll, then $N_{2} = 0$ and all the forces $(m g and N_{1})$ must pass through contact point 'P'.

$∴$ $\cos (90 ° - θ_{\max}) = \frac{O Q}{O R} = \frac{r}{R} or, \sin θ_{\max} = \frac{r}{R}$

Q 2 :

A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed $ω$ , as shown in the figure. At time $t = 0$ , a small insect starts from O and moves with constant speed $v$ , with respect to the rod towards the other end. It reaches the end of the rod at $t = T$ and stops. The angular speed of the system remains $ω$ throughout. The magnitude of the torque $(| \vec{τ} |)$ about O, as a function of time is best represented by which plot? [2012]

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(2)

Angular momentum, $| L |$ or $L = I ω$ (about axis of rod)

Moment of inertia of the rod-insect system

$I = I_{rod} + m x^{2} = I_{rod} + m v^{2} t^{2}$

Here, $m$ = mass of insect

$∴$ $L = (I_{rod} + m v^{2} t^{2}) ω$

$Now | τ | = \frac{d L}{d t}$

$= (2 m v^{2} t ω) o r | τ | \propto t$

i.e., the graph is a straight line passing through origin.

After time $T, L$ = constant

$∴$ $| τ | or \frac{d L}{d t} = 0$

i.e., when the insect stops moving, L does not change and therefore $T$ becomes constant.

Q 3 :

A long horizontal rod has a bead which can slide along its length and initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular motion about $A$ with constant angular acceleration $α$ . If the coefficient of friction between the rod and the bead is $μ$ , and gravity is neglected, then the time after which the bead starts slipping is [2000]

$\sqrt{\frac{μ}{α}}$
$\frac{μ}{\sqrt{α}}$
$\frac{1}{\sqrt{μ α}}$
infinitesimal

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(1)

When we are giving an angular acceleration $(α)$ to the rod, the bead has instantaneous acceleration crinst $= L α .$ The bead has a tendency to move away from the centre. But due to the friction between the bead and the rod, this does not happen. If instantaneous angular velocity is $ω$ then

Here, necessary frictional force is provided by frictional force

$m L ω^{2} = μ (m a) \Rightarrow m L ω^{2} = μ m L α$

$\Rightarrow ω^{2} = μ α$

$Using ω = ω_{0} + α t,$

$\Rightarrow ω = α t$

$∴$ $α^{2} t^{2} = μ α \Rightarrow t = \sqrt{\frac{μ}{α}}$

Q 4 :

A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude $F = 0.5 N$ are applied simultaneously along the three sides of an equilateral triangle $X Y Z$ with its vertices on the perimeter of the disc (see figure).

One second after applying the forces, the angular speed of the disc in ${rad s}^{- 1}$ is [2014]

(2)

$3 [F \times r \times \frac{1}{2}] = I α$

$3 \times 0.5 \times 0.5 \times \frac{1}{2} = \frac{1}{2} \times 1.5 \times 0.5 \times 0.5 \times α$

$\Rightarrow α = 2 {rad s}^{- 1}$

$ω = ω_{0} + α t \Rightarrow ω = 0 + 2 \times 1 = 2 {rad s}^{- 1}$

Q 5 :

A pendulum consists of a bob of mass $m = 0.1 kg$ and a massless inextensible string of length $L = 1.0 m$ . It is suspended from a fixed point at height $H = 0.9 m$ above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse $P = 0.2 kg - m/s$ is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J kg - m^{2} / s$ . The kinetic energy of the pendulum just after the lift-off is $K$ Joules.

Q. The value of $J$ is ______ . [2021]

(0.18)

Q 6 :

A pendulum consists of a bob of mass $m = 0.1 kg$ and a massless inextensible string of length $L = 1.0 m$ . It is suspended from a fixed point at height $H = 0.9 m$ above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse $P = 0.2 kg - m/s$ is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J kg - m^{2} / s$ . The kinetic energy of the pendulum just after the lift-off is $K$ Joules.

Q. The value of $K$ is ______ . [2021]

(0.16)

Angular momentum $L = P \times r = P \times H$

$or, L = 0.2 \times 0.9 = 0.18 {kg m}^{2} / s$

$∴$ $J = 0.18$

There will be no velocity along the string just after the string becomes taut.

$∴$ $V_{⊥} = \frac{P \cos θ}{m} = \frac{0.2 \times 0.9}{1 \times 0.1} = 1.8 m/s$

$∴$ $Kinetic energy, K = \frac{1}{2} m V_{⊥}^{2} = \frac{1}{2} \times 0.1 \times {(1.8)}^{2} = 0.162 J$

Q 7 :

A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle $60 °$ with vertical? [2019]

[ $g$ is the acceleration due to gravity]

The angular speed of the rod will be $\sqrt{\frac{3 g}{2 L}}$
The radial acceleration of the rod's center of mass will be $\frac{3 g}{4}$
The normal reaction force from the floor on the rod will be $\frac{M g}{16}$
The angular acceleration of the rod will be $\frac{2 g}{L}$

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Select one or more options

(1, 2, 3)

The rod is released from rest so that it falls by rotating about its contact point with the floor without slipping.

$Gain in kinetic energy = loss in potential energy$

$\frac{1}{2} I ω^{2} = m g \frac{l}{2} (1 - \cos 60 °)$

$∴$ $\frac{m l^{2}}{3} ω^{2} = m g \frac{l}{2} \Rightarrow ω = \sqrt{\frac{3 g}{2 l}}$

$Now, τ = I α$

$∴$ $m g \times \frac{l}{2} \sin 60 ° = \frac{1}{3} m l^{2} α \Rightarrow α = \frac{3 \sqrt{3} g}{4 l}$

$Further, a_{t} = \frac{l}{2} α = \frac{3 \sqrt{3} g}{8}$

$Also a_{r} = ω^{2} \frac{l}{2} = \frac{3 g}{2 l} \times \frac{l}{2} = \frac{3 g}{4}$

For vertical motion of centre of mass

$m g - N = m (a_{r} \cos 60 ° + a_{t} \cos 30 °)$

$∴$ $m g - N = m [\frac{3 g}{4} \times \frac{1}{2} + \frac{3 \sqrt{3} g}{8} \times \frac{\sqrt{3}}{2}]$

$∴$ $N = \frac{M g}{16}$

Q 8 :

Two solid cylinders $P$ and $Q$ of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder $P$ has most of its mass concentrated near its surface, while $Q$ has most of its mass concentrated near the axis. Which statement(s) is(are) correct? [2012]

Both cylinders P and Q reach the ground at the same time.
Cylinder P has larger linear acceleration than cylinder Q.
Both cylinders reach the ground with same translational kinetic energy.
Cylinder Q reaches the ground with larger angular speed.

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(4)

As we know, acceleration of the center of mass of a cylinder rolling down an inclined plane

$a_{c} = \frac{g \sin θ}{1 + \frac{I}{M R^{2}}}$

In case of P the mass is concentrated away from the axis,

$So I_{P} > I_{Q}$

$∴$ $a_{P} < a_{Q} \Rightarrow v_{P} < v_{Q} \Rightarrow ω_{P} < ω_{Q}$

Q 9 :

A thin ring of mass 2 kg and radius 0.5 m is rolling without on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision: [2011]

the ring has pure rotation about its stationary CM.
the ring comes to a complete stop.
friction between the ring and the ground is to the left.
there is no friction between the ring and the ground.

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(3)

The angular impulse created by the frictional force between the ring and the ball tends to decrease the angular speed $ω$ of the ring about O.

After the collision $ω$ decreases but the ring remains rotating in the anticlockwise direction. Hence the friction between the ring and the ground at the point of contact is to the left.

Q 10 :

In the Column-I below, four different paths of a particle are given as functions of time. In these functions, $α$ and $β$ are positive constants of appropriate dimensions and $α \neq β$ . In each case, the force acting on the particle is either zero or conservative. In Column-II, five physical quantities of the particle are mentioned $\vec{p}$ is the linear momentum, $\vec{L}$ is the angular momentum about the origin, $K$ is the kinetic energy, $U$ is the potential energy and $E$ is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path. [2018]

Column-I Column-II

P. $\vec{r} (t) = α t \hat{i} + β t \hat{j}$ 1. $\vec{p}$

Q. $\vec{r} (t) = α \cos ω t \hat{i} + β \sin ω t \hat{j}$ 2. $\vec{L}$

R. $\vec{r} (t) = α (\cos ω t \hat{i} + \sin ω t \hat{j})$ 3. $K$

S. $\vec{r} (t) = α t \hat{i} + \frac{β}{2} t^{2} \hat{j}$ 4. $U$

5. $E$

P → 1, 2, 3, 4, 5; Q → 2, 5; R → 2, 3, 4, 5; S → 5
P → 1, 2, 3, 4, 5; Q → 3, 5; R → 2, 3, 4, 5; S → 2, 5
P → 2, 3, 4; Q → 5; R → 1, 2, 4; S → 2, 5
P → 1, 2, 3, 5; Q → 2, 5; R → 2, 3, 4, 5; S → 2, 5

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(1)

Q 11 :

Column-II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column-I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II. [2009]

Column-I Column-II

(A) The force exerted by X on Y has a magnitude (p)

Block Y of mass M left on a fixed inclined plane X, slides on it with a constant velocity.

(B) The gravitational potential energy of X is continuously increasing. (q)

Two ring magnets Y and Z, each of mass MMM, are kept in a frictionless vertical plastic stand so that they repel each other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity.

(C) Mechanical energy of the system X + Y is continuously decreasing. (r)

A pulley Y of mass m0m_0m0? is fixed to a table through a clamp X. A block of mass MMM hangs from a string that goes over the pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity.

(D) The torque of the weight of Y about point P is zero. (s)

A sphere Y of mass M is put in a non-viscous liquid X kept in a container at rest. The sphere is released and moves down in the liquid.

(t)

A sphere Y of mass M is falling with its terminal velocity in a viscous liquid X kept in a container.

$A \to (q, p); B \to (q, s, t); C \to (p, r, t); D \to (p, t)$
$A \to (p, t); B \to (q, s, t); C \to (p, r, t); D \to (q, p)$
$A \to (q, p); B \to (p, r, t); C \to (q, s, t); D \to (p, t)$
$A \to (p, r, t); B \to (q, p); C \to (q, s, t); D \to (p, t)$

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(2)

(p) As the velocity is constant

$∴$ $f = m g \sin θ$

$But f = μ N = μ m g \cos θ$

$∴$ $μ m g \cos θ = m g \sin θ \Rightarrow μ = \tan θ$

$The force by X on Y is the resultant of f and N .$

$\sqrt{f^{2} + N^{2}} = \sqrt{μ^{2} N^{2} + N^{2}} = \sqrt{μ^{2} + 1} N$

$= (\sqrt{\tan^{2} θ + 1}) m g \cos θ = \sec θ m g \cos θ = m g$

$= weight of Y .$

Again, due to the presence of frictional force between Y and X, the mechanical energy of the system (X + Y) decreases continuously as Y slides down.

(q) Lift moves up, X also moves up and therefore the gravitational energy of X is continuously increasing.

T of weight of Y about P as the perpendicular distance of the line of action of force from the point P is zero. Force exerted by X on Y $= M g + M g = 2 M g$

where $M g$ is wt. of Y and $M g$ is the force on Y due to Z.

(r)

In this case the force exerted by X on Y = force exerted by Y on X. The force on X due to Y is

$R = \sqrt{{(M g)}^{2} + [{(m_{0} + M) g]}^{2}} \neq M g$

The mechanical energy of the system (X + Y) is continuously decreasing as the system is coming down and its potential energy is decreasing, the kinetic energy remaining the same.

The torque of the weight of Y about $P \neq 0$

(s) Force on Y by X is = wt. of liquid displaced which cannot be equal to $M g$ as the density of Y > density of X (Y is sinking)

The gravitational potential energy of X increases continuously because as Y moves down, the centre of mass of X moves up.

(t) Sphere Y is moving with terminal velocity $V_{T}$

$∴$ Net force on Y is zero i.e. $M g = B + F_{v}$

The $B + F_{v}$ are exerted by X on Y.

The gravitational potential energy of X is continuously increasing because as Y moves down, the centre of mass of X moves up.

The mechanical energy of the system (X + Y) is continuously decreasing to overcome the viscous forces.

Q 12 :

One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $r$ . The finger rotates with an angular velocity $ω_{0}$ . The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is $μ$ and the acceleration due to gravity is $g$ . [2017]

Q. The total kinetic energy of the ring is

$M ω_{0}^{2} R^{2}$
$\frac{1}{2} M ω_{0}^{2} {(R - r)}^{2}$
$M ω_{0}^{2} {(R - r)}^{2}$
$\frac{3}{2} M ω_{0}^{2} {(R - r)}^{2}$

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(3)

Here $ω_{0} (R - r) = ω R$ $∴$ $ω = ω_{0} (\frac{R - r}{R})$

Total kinetic energy of the ring = (Kinetic rotational + kinetic energy translational)

${K . E}_{total} = \frac{1}{2} (2 M R^{2}) ω^{2} = M ω_{0}^{2} {(R - r)}^{2}$

Q 13 :

One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $r$ . The finger rotates with an angular velocity $ω_{0}$ . The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is $μ$ and the acceleration due to gravity is $g$ . [2017]

Q. The minimum value of $ω_{0}$ below which the ring will drop down is

$\sqrt{\frac{g}{μ (R - r)}}$
$\sqrt{\frac{2 g}{μ (R - r)}}$
$\sqrt{\frac{3 g}{2 μ (R - r)}}$
$\sqrt{\frac{g}{2 μ (R - r)}}$

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(1)

$μ M ω_{\min}^{2} (R - r) = M g$

$∴$ $ω_{\min} = \sqrt{\frac{g}{μ (R - r)}}$

Q 14 :

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.

These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed $ω$ , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed $ω$ in this case

Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to $x - z$ plane; Case (b) the disc with its face making an angle of $45 °$ with $x - y$ plane and its horizontal diameter parallel to $x$ -axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed $ω$ about the $z$ -axis. [2012]

Q. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?

It is vertical for both the cases (a) and (b).
It is vertical for case (a); and is at $45 °$ to the $x - z$ plane and lies in the plane of the disc for case (b).
It is horizontal for case (a); and is at $45 °$ to the $x - z$ plane and is normal to the plane of the disc for case (b).
It is vertical for case (a); and is $45 °$ to the $x - z$ plane and is normal to the plane of the disc for case (b).

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(1)

Axis of rotation is parallel to the z-axis. Hence for both the cases, instantaneous axis passing through is vertical.

Q 15 :

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.

These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed $ω$ , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the $z$ -axis and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed $ω$ in this case

Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to $x - z$ plane; Case (b) the disc with its face making an angle of $45 °$ with $x - y$ plane and its horizontal diameter parallel to $x$ -axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed $ω$ about the $z$ -axis. [2012]

Q. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?

It is $\sqrt{2} ω$ for both the cases
It is $ω$ for case (a); and $ω / \sqrt{2}$ for case (b)
It is $ω$ for case (a); and $\sqrt{2 ω}$ for case (b)
It is $ω$ for both the cases

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(4)

For a rigid body $ω$ is same for any point of the body.

Q 16 :

A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in horizontal plane.

The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity ${\vec{V}}_{0} = V_{0} \hat{i}$ . The coefficient of friction is $μ$ . [2008]

Q. The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is

$- k x$
$- 2 k x$
$- \frac{2 k x}{3}$
$- \frac{4 k x}{3}$

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(4)

$2 k x - f = M a and a = R α$

$∴$ $\frac{2 k x - f}{M} = R [\frac{f R}{\frac{1}{2} M R^{2}}]$

$= R [\frac{f R}{\frac{1}{2} M R^{2}}]$

Solving this equation, we get

$∴$ $| F_{net} |$ $= 2 k x - f = 2 k x - \frac{2 k x}{3} = \frac{4 k x}{3}$

This is opposite to displacement

$f = \frac{2 k x}{3}$

$∴$ $F_{net} = - \frac{4 k x}{3} directed towards the equilibrium$

Q 17 :

A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in horizontal plane.

The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity ${\vec{V}}_{0} = V_{0} \hat{i}$ . The coefficient of friction is $μ$ . [2008]

Q. The centre of mass of the disk undergoes simple harmonic motion with angular frequency $ω$ equal to –

$\sqrt{\frac{k}{M}}$
$\sqrt{\frac{2 k}{M}}$
$\sqrt{\frac{2 k}{3 M}}$
$\sqrt{\frac{4 k}{3 M}}$

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(4)

$F_{net} = 1 - (\frac{4 k x}{3}) x$

$∴$ $a = \frac{F_{net}}{M} = - (\frac{4 k}{3 M}) x = - ω^{2} x \Rightarrow ω = \sqrt{\frac{4 k}{3 M}}$

Q 18 :

A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in horizontal plane.

The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity ${\vec{V}}_{0} = V_{0} \hat{i}$ . The coefficient of friction is $μ$ . [2008]

Q. The maximum value of $V_{0}$ for which the disk will roll without slipping is –

$μ g \sqrt{\frac{M}{k}}$
$μ g \sqrt{\frac{M}{2 k}}$
$μ g \sqrt{\frac{3 M}{k}}$
$μ g \sqrt{\frac{5 M}{2 k}}$

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(3)

Mechanical energy is conserved in case of pure rolling motion

$∴$ $\frac{1}{2} M v_{0}^{2} + \frac{1}{2} (\frac{1}{2} M R^{2}) {(\frac{v_{0}}{R})}^{2} = 2 [\frac{1}{2} k x_{\max}^{2}]$

$∴$ $x_{\max} = \sqrt{\frac{3 M}{4 k}} v_{0}$

$F_{\max} = μ M g = \frac{2 k x_{\max}}{3} = \frac{2 k}{3} \sqrt{\frac{3 M}{4 k} v_{0}}$

$∴$ $v_{0} = μ g \sqrt{\frac{3 M}{k}}$

Q 19 :

Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2 I$ respectively about the common axis. Disc A is imparted an initial angular velocity $2 ω$ using the entire potential energy of a spring compressed by a distance $x_{1}$ . Disc B is imparted an angular velocity $ω$ by a spring having the same spring constant and compressed by a distance $x_{2}$ . Both the discs rotate in the clockwise direction. [2007]

Q. The loss of kinetic energy in the above process is

$\frac{I ω^{2}}{2}$
$\frac{I ω^{2}}{3}$
$\frac{I ω^{2}}{4}$
$\frac{I ω^{2}}{6}$

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(2)

$Loss in kinetic energy = {(K . E .)}_{initial} - {(K . E .)}_{final}$

$= [\frac{1}{2} I {(2 ω)}^{2} + \frac{1}{2} (2 I) ω^{2}] - [\frac{1}{2} (I + 2 I) {(\frac{4}{3} ω)}^{2}]$

$= 3 I ω^{2} - \frac{8}{3} I ω^{2} = \frac{I ω^{2}}{3}$

Q 20 :

Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2 I$ respectively about the common axis. Disc A is imparted an initial angular velocity $2 ω$ using the entire potential energy of a spring compressed by a distance $x_{1}$ . Disc B is imparted an angular velocity $ω$ by a spring having the same spring constant and compressed by a distance $x_{2}$ . Both the discs rotate in the clockwise direction. [2007]

Q. When disc B is brought in contact with disc A, they acquire a common angular velocity in time $t$ . The average frictional torque on one disc by the other during this period is

$\frac{2 I ω}{3 t}$
$\frac{9 I ω}{2 t}$
$\frac{9 I ω}{4 t}$
$\frac{3 I ω}{2 t}$

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(1)

When disc B is brought in contact with disc A

Let $ω$ be the common velocity. From conservation of angular momentum for the two disc system

$I (2 ω) + 2 I (ω) = (I + 2 I) ω' \Rightarrow ω' = \frac{4}{3} ω$

Torque on disc A

$τ_{A} = \frac{Δ L_{A}}{t} = \frac{L_{f} - L_{i}}{t} = \frac{I \times \frac{4}{3} ω - I \times 2 ω}{t} = \frac{- 2 I ω}{3 t}$

Here negative sign indicates that the torque creates angular retardation.

Q 21 :

Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2 I$ respectively about the common axis. Disc A is imparted an initial angular velocity $2 ω$ using the entire potential energy of a spring compressed by a distance $x_{1}$ . Disc B is imparted an angular velocity $ω$ by a spring having the same spring constant and compressed by a distance $x_{2}$ . Both the discs rotate in the clockwise direction. [2007]

Q. The ratio $\frac{x_{1}}{x_{2}}$ is

$2$
$\frac{1}{2}$
$\sqrt{2}$
$\frac{1}{\sqrt{2}}$

1

2

3

4

(3)

$For disc A$

$\frac{1}{2} k x_{1}^{2} = \frac{1}{2} I {(2 ω)}^{2} \Rightarrow k x_{1}^{2} = 2 I ω^{2}$

$For disc B$

$\frac{1}{2} k x_{2}^{2} = \frac{1}{2} \times 2 I ω^{2} \Rightarrow k x_{2}^{2} = I ω^{2}$

$∴$ $\frac{k x_{1}^{2}}{k x_{2}^{2}} = \frac{2 I ω^{2}}{I ω^{2}} \Rightarrow \frac{x_{1}}{x_{2}} = \sqrt{2}$

Topic Question Set

(1)

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	Column-I		Column-II
P.	$\vec{r} (t) = α t \hat{i} + β t \hat{j}$	1.	$\vec{p}$
Q.	$\vec{r} (t) = α \cos ω t \hat{i} + β \sin ω t \hat{j}$	2.	$\vec{L}$
R.	$\vec{r} (t) = α (\cos ω t \hat{i} + \sin ω t \hat{j})$	3.	$K$
S.	$\vec{r} (t) = α t \hat{i} + \frac{β}{2} t^{2} \hat{j}$	4.	$U$
		5.	$E$

	Column-I		Column-II
(A)	The force exerted by X on Y has a magnitude	(p)	Block Y of mass M left on a fixed inclined plane X, slides on it with a constant velocity.
(B)	The gravitational potential energy of X is continuously increasing.	(q)	Two ring magnets Y and Z, each of mass MMM, are kept in a frictionless vertical plastic stand so that they repel each other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity.
(C)	Mechanical energy of the system X + Y is continuously decreasing.	(r)	A pulley Y of mass m0m_0m0? is fixed to a table through a clamp X. A block of mass MMM hangs from a string that goes over the pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity.
(D)	The torque of the weight of Y about point P is zero.	(s)	A sphere Y of mass M is put in a non-viscous liquid X kept in a container at rest. The sphere is released and moves down in the liquid.
		(t)	A sphere Y of mass M is falling with its terminal velocity in a viscous liquid X kept in a container.