JEE Advanced PYQ System of Particles and Rotational Motion

Q 1 :

A bar of mass M = 1.00 kg and length L = 0.20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass m = 0.10 kg is moving on the same horizontal surface with $5.00 {m s}^{- 1}$ speed on a path perpendicular to the bar. It hits the bar at a distance L/2 from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity $ω$ . Which of the following statement is correct? [2023]

$ω = 6.98 {rad s}^{- 1}$ and $v = 4.30 {m s}^{- 1}$
$ω = 3.75 {rad s}^{- 1}$ and $v = 4.30 {m s}^{- 1}$
$ω = 3.75 {rad s}^{- 1}$ and $v = 10.0 {m s}^{- 1}$
$ω = 6.80 {rad s}^{- 1}$ and $v = 4.10 {m s}^{- 1}$

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(1)

About the hinge applying angular momentum conservation

$m v \frac{L}{2} + 0 = - m v \frac{L}{2} + \frac{M L^{2}}{3} ω . . . (i)$

After elastic collision,

$e = 1 = \frac{ω \frac{L}{2} + V}{u} \Rightarrow u = ω \frac{L}{2} + V . . . (i i)$

$Putting, m = 0.1 kg, M = 1 kg, L = 0.20 m and solving eq (i) & (ii) we get$

$ω \approx 6.98 {rad s}^{- 1} and v = 4.30 {m s}^{- 1}$

Q 2 :

A flat surface of a thin uniform disk A of radius R is glued to a horizontal table. Another thin uniform disk B of mass M and with the same radius R rolls without slipping on the circumference of A, as shown in the figure. A flat surface of B also lies on the plane of the table. The center of mass of B has fixed angular speed $ω$ about the vertical axis passing through the center of A. The angular momentum of B is $n M ω R^{2}$ with respect to the center of A. Which of the following is the value of $n$ ? [2022]

$2$
$5$
$\frac{7}{2}$
$\frac{9}{2}$

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(2)

$We have V = (2 R) ω$

$As B is rolling on circumference of A.$

$So, V = ω' R$

$\Rightarrow 2 R ω = ω' R$

$\Rightarrow ω' = 2 ω$

$So, \vec{L} w.r.t COM of A$

$= (I ω' + m (2 R) V) \hat{k}$

$= (\frac{1}{2} M R^{2} (2 ω) + m (2 R) (2 R ω)) \hat{k} = 5 M R^{2} ω \hat{k}$

$So, n = 5$

Q 3 :

A uniform wooden stick of mass 1.6 kg and length $l$ rests in an inclined manner on a smooth, vertical wall of height $h (< l)$ such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of $30 °$ with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio $h / l$ and the frictional force $f$ at the bottom of the stick are $(g = 10 {m s}^{- 2})$ . [2016]

$\frac{h}{l} = \frac{\sqrt{3}}{16}, f = \frac{16 \sqrt{3}}{3} N$
$\frac{h}{l} = \frac{3}{16}, f = \frac{16 \sqrt{3}}{3} N$
$\frac{h}{l} = \frac{3 \sqrt{3}}{16}, f = \frac{8 \sqrt{3}}{3} N$
$\frac{h}{l} = \frac{3 \sqrt{3}}{16}, f = \frac{16 \sqrt{3}}{3} N$

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(4)

$By vertical equilibrium,$

$N + N \sin 30 ° = 1.6 g$

$\Rightarrow N = \frac{3.2 g}{3} . . . (i)$

$By horizontal equilibrium,$

$f = N \cos 30 °$

$= \frac{\sqrt{3}}{2} N = \frac{16 \sqrt{3}}{3}$

$Torque about A,$

$1.6 g \times A B = N \times x$

$1.6 g \times \frac{ℓ}{2} \cos 60 ° = \frac{3.2 g}{3} \times x \Rightarrow \frac{3 ℓ}{8} = x . . . (i)$

$But \cos 30 ° = \frac{h}{x} \Rightarrow x = \frac{h}{\cos 30 °} . . . (i i)$

$From eq. (i) and (ii),$

$\frac{h}{\cos 30 °} = \frac{3 ℓ}{8} \Rightarrow \frac{h}{ℓ} = \frac{3 \sqrt{3}}{16}$

Q 4 :

A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed $ω$ . If the angular momentum of the system, calculated about O and P are denoted by ${\vec{L}}_{O}$ and ${\vec{L}}_{P}$ respectively, then [2012]

${\vec{L}}_{O}$ and ${\vec{L}}_{P}$ do not vary with time
${\vec{L}}_{O}$ varies with time while ${\vec{L}}_{P}$ remains constant
${\vec{L}}_{O}$ remains constant while ${\vec{L}}_{P}$ varies with time
${\vec{L}}_{O}$ and ${\vec{L}}_{P}$ both vary with time

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(3)

For all locations of $m$ , the angular momentum of the mass $m$ about O i.e., $L_{O}$ is $m r^{2} ω$ and is directed toward +z direction.

The angular momentum of mass $m$ about P, i.e., $L_{P}$ is $m v l$ and is directed for the given location of $m$ as shown in the figure.

For different location of $m$ , the direction of ${\vec{L}}_{P}$ remains changing.

Q 5 :

A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct? [2005]

$\vec{L}$ (angular momentum) is conserved about the centre
only direction of angular momentum $\vec{L}$ is conserved
It spirals towards the centre
its acceleration is towards the centre

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(2)

As particle is rotating in circular path and linear speed $V$ is decreasing, $L$ is not conserved in magnitude. Since the particle has two accelerations $a_{c}$ and $a_{t}$ therefore the net acceleration is not towards the centre.

The direction of $\vec{L}$ remains same even when the speed decreases.

Q 6 :

A block of mass $m$ is at rest under the action of force $F$ against a wall as shown in figure. Which of the following statement is incorrect? [2005]

$f = m g [f friction force]$
$F = N [N normal force]$
F will not produce torque
N will not produce torque

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(4)

The block is in equilibrium.

For translational equilibrium

$\sum F_{x} = 0 \Rightarrow F = N$

$\sum F_{y} = 0 \Rightarrow f = m g$

For rotational equilibrium

$\sum τ_{C} = 0$

Torque created by frictional force $(f)$ about $C = f \times a$ in clockwise direction.

There is another torque which should counter this torque.

The normal reaction N on the block will create a torque $N \times b$ in the anticlockwise direction.

Such that $f \times a = N \times b$

Hence N will produce torque.

Q 7 :

A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity $ω_{0}$ . A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is $K$ initially, its final kinetic energy will be [2004]

$2 K$
$\frac{K}{2}$
$K$
$\frac{K}{4}$

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(2)

${(K . E .)}_{rotational} = \frac{L^{2}}{2 I}$

$Here, L = constant$

$∴$ ${(K . E .)}_{_{rotational}} \times I = constant$

$Hence when I is doubled, {(K . E .)}_{rotational} becomes half.$

Q 8 :

A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remain conserved? [2003]

centre of the circle
on the circumference of the circle
inside the circle
outside the circle

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(1)

The net force acting on a particle undergoing uniform circular motion is centripetal force towards centre of the circle.

The torque due to this force about the centre is zero, therefore, angular momentum $\vec{L}$ about centre is conserved.

Q 9 :

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J = M V$ is imparted to the body at one of its ends, what would be its angular velocity? [2003]

$\frac{V}{L}$
$\frac{2 V}{L}$
$\frac{V}{3 L}$
$\frac{V}{4 L}$

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(1)

Applying change in angular momentum of the system = angular impulse given to the system about the centre of mass

${(Angular momentum)}_{f} - {(Angular momentum)}_{i}$

$= M v \times \frac{L}{2} = I_{C} ω . . . (i)$

Here $ω$ is the angular velocity of the rod.

Moment of Inertia of the system about its axis of rotation [centre of mass of the system]

$I_{C} = M {(\frac{L}{2})}^{2} + M {(\frac{L}{2})}^{2} = \frac{2 M L^{2}}{4} = \frac{M L^{2}}{2}$

$∴$ $\frac{M V L}{2} = (\frac{M L^{2}}{2}) ω \Rightarrow ω = \frac{V}{L}$

Q 10 :

A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity $ω_{0}$ . When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform $ω (t)$ will vary with time $t$ as [2002]

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(2)

There is no external torque, angular momentum remains conserved. As moment of inertia initially when the tortoise moves from A to B decreases and then increases when tortoise moves from B to C, so $ω$ will increase initially and then decreases. So the variation of $ω$ is nonlinear.

Q 11 :

An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down, one along AB and the other along AC as shown. Neglecting frictional effects, the quantities that are conserved as the beads slide down, are [2000]

angular velocity and total energy (kinetic and potential)
total angular momentum and total energy
angular velocity and moment of inertia about the axis of rotation
total angular momentum and moment of inertia about the axis of rotation

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(2)

Due to increase in perpendicular distance of the bead with the axis of rotation, $M . I .$ about the axis of rotation is not constant. So $I$ increases.

There is no external torque acting

$∴$ $τ_{ext} = \frac{d L}{d t} \Rightarrow L = constant \Rightarrow I ω = constant$

But, $I$ increases so $ω$ decreases.

Q 12 :

A particle of mass 1 kg is subjected to a force which depends on the position as $\vec{F} = - k (x \hat{i} + y \hat{j}) {kg ms}^{- 2}$ with $k = 1 {kg s}^{- 2}$ . At time $t = 0$ , the particle's position $\vec{r} = (\frac{1}{\sqrt{2}} \hat{i} + \sqrt{2} \hat{j}) m$ and its velocity $\vec{v} = (- \sqrt{2} \hat{i} + \sqrt{2} \hat{j} + \frac{2}{π} \hat{k}) {ms}^{- 1}$ . Let $v_{x}$ and $v_{y}$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z = 0.5 m$ , the value of $(x v_{y} - y v_{x})$ is ______ $m^{2} s^{- 1}$ . [2022]

(3)

Here, $\vec{F} = - k \vec{r}$ . So force passes through origin. So, $τ_{origin} = 0 \Rightarrow$ angular momentum about origin will be conserved

So, $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{\sqrt{2}} & \sqrt{2} & 0 \\ - \sqrt{2} & \sqrt{2} & \frac{2}{π} \end{matrix} |$ $=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & 0.5 \\ v_{x} & v_{y} & \frac{2}{π} \end{matrix} |$

$\Rightarrow \hat{k} [\frac{1}{\sqrt{2}} \times \sqrt{2} - (- \sqrt{2}) \times \sqrt{2}] = \hat{k} (x v_{y} - y v_{x})$

$\Rightarrow x v_{y} - y v_{x} = 3$

Q 13 :

A thin rod of mass $M$ and length $α$ is free to rotate in a horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass $M$ and of radius $a / 4$ is pivoted on this rod with its center at a distance $a / 4$ from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $Ω$ and the disc rotating about its vertical axis with angular velocity $4 Ω$ . The total angular momentum of the system about the point O is $(\frac{M a^{2} Ω}{48}) n .$ The value of $n$ is ____. [2021]

(49)

$Angular momentum of the system, L_{s} = L_{rod} + L_{disc}$

$= I_{rod} ω + I_{disc} ω + \vec{r} \times \vec{P}$

$= \frac{M a^{2}}{12} ω + \frac{M a^{2}}{32} ω + \frac{3 a}{16} \times \frac{3 a}{4} \times M \times ω$

$= \frac{M a^{2}}{12} \times 4 Ω + \frac{M a^{2}}{32} \times 4 Ω + \frac{3 a}{16} \times \frac{3 a}{4} \times M \times 4 Ω$

$∴$ $L_{s} = \frac{M a^{2}}{3} Ω + M {(\frac{3 a}{4})}^{2} Ω - \frac{M {(\frac{a}{4})}^{2} 4 Ω}{2}$

$or L_{s} = \frac{49}{48} M a^{2} Ω$ $∴$ $n = 49$

Q 14 :

A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of $9 {ms}^{- 1}$ with respect to the ground. The rotational speed of the platform in ${rad s}^{- 1}$ after the balls leave the platform is _______. [2014]

(4)

From conservation of angular momentum

$2 (m v r) = I ω$

$2 \times 0.05 \times 9 \times 0.25 = \frac{1}{2} \times 0.45 \times {(0.5)}^{2} \times ω$

$∴$ $ω = 4 {rad s}^{- 1}$

Q 15 :

A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of $10 {rad s}^{- 1}$ about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in ${rad s}^{- 1}$ ) of the system is [2013]

(8)

From conservation of angular momentum

$I_{1} ω_{1} = I_{2} ω_{2}$

$∴$ $ω_{2} = \frac{I_{1} ω_{1}}{I_{2}} = \frac{\frac{1}{2} M R^{2} \cdot ω_{1}}{[\frac{1}{2} M R^{2} + 2 (2 m r^{2})]}$

$Given: M = 50 kg, R = 0.4 m, ω_{1} = 10 rad/s,$ $m = 6.25 kg and r = 0.2 m$

$= \frac{\frac{1}{2} \times 50 \times {(0.4)}^{2} \times 10}{\frac{1}{2} \times 50 \times {(0.4)}^{2} + 2 [2 \times 6.25 \times {(0.2)}^{2}]}$

$= \frac{40}{4 + 1} = 8 rad/s$

Q 16 :

A binary star consists of two stars A (mass $2.2 M_{s}$ ) and B (mass $11 M_{s}$ ), where $M_{s}$ is the mass of the sun. They are separated by distance $d$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is [2010]

(6)

Let the center of mass of the binary star system be at the origin 'O'. Distance between $A O = x$ and $O B = (d - x)$

$0 = \frac{2.2 M_{s} (- x) + 11 M_{s} (d - x)}{2.2 M_{s} + 11 M_{s}}$

$\Rightarrow 0 = 2.2 M_{s} (- x) + 11 M_{s} (d - x) \Rightarrow x = \frac{5 d}{6}$

For a binary star system, angular speed $ω$ about the centre of mass 'O' is same for both the stars.

$∴$ $\frac{L_{total}}{L_{B}} = \frac{2.2 M_{s} {(\frac{5 d}{6})}^{2} ω + 11 M_{s} {(\frac{d}{6})}^{2} ω}{11 M_{s} {(\frac{d}{6})}^{2} ω} = 6$

Q 17 :

Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the right one at 90.00 cm. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance $x_{R}$ from the center (50.00 cm) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.40 and 0.32, respectively, the value of $x_{R}$ (in cm) is ______ . [2020]

(25.6)

$Initially$

$N_{L} + N_{R} = M g, N_{L} = \frac{40}{90} M g = \frac{4 M g}{9}$

$Torque about centre (τ_{centre} = 0)$

$∴$ $N_{1} (50) = N_{2} (40), N_{R} = \frac{50}{90} M g = \frac{5 M g}{9}$

$5 N_{L} = 4 N_{R}$

$f_{1 k} = μ_{k} N_{L}, f_{1 L} = μ_{s} N_{L}$

$f_{1 k} = 0.32 N_{L}, f_{1 L} = 0.4 N_{L}$

$f_{2 k} = 0.32 N_{R}, f_{2 L} = 0.4 N_{R}$

If $X_{L}$ = distance of left finger from centre when right finger starts moving

${(τ_{n} = 0)}_{about centre} \Rightarrow N_{L} x_{L} = N_{R} (40)$

$f_{k 1} = f_{L 2} \Rightarrow 0.32 N_{L} = 0.40 N_{R}$

$4 N_{L} = 5 N_{R}$

$N_{L} x_{L} = \frac{4 N_{L}}{5} (40) \Rightarrow x_{L} = 32$

Now $x_{R}$ = distance when right finger stops and left finger starts moving

$N_{L} x_{L} = N_{R} x_{R} [Torque about centre, τ_{centre} = 0]$

$f_{L 1} = f_{k 2} \Rightarrow 0.4 N_{L} = 0.32 N_{R}$

$4 N_{L} = 4 N_{R}$

$\frac{4 N_{2}}{5} (32) = N_{R} x_{R}$

$∴$ $x_{R} = \frac{128}{5} = 25.6 cm$

Q 18 :

A particle of mass $M = 0.2 kg$ is initially at rest in the $x y$ -plane at a point $(x = - l, y = - h)$ , where $l = 10 m$ and $h = 1 m$ . The particle is accelerated at time $t = 0$ with a constant acceleration $a = 10 {m/s}^{2}$ along the positive $x$ -direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by $\vec{L}$ and $\vec{τ}$ , respectively. $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the positive $x$ , $y$ and $z$ -directions, respectively. If $\hat{k} = \hat{i} \times \hat{j}$ then which of the following statement(s) is(are) correct? [2021]

The particle arrives at the point $(x = l, y = - h)$ at time $t = 2 s$ .
$\vec{τ} = 2 \hat{k}$ when the particle passes through the point $(x = l, y = - h)$
$\hat{L} = 4 \hat{k}$ when the particle passes through the point $(x = l, y = - h)$
$\vec{τ} = \hat{k}$ when the particle passes through the point $(x = 0, y = - h)$

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Select one or more options

(1, 2, 3)

Particle is initially at rest in the $x y$ -plane at a point $x = - ℓ, y = - h$ where $ℓ = 10$ m and $h = 1 m$

$From s = u t + \frac{1}{2} a t^{2} \Rightarrow 20 = \frac{1}{2} \times 10 \times t^{2}$

$∴$ $t = 2 sec$

$Torque, \vec{τ} = \vec{r} \times \vec{F} and here {\vec{r}}_{B} = 10 \hat{i} - \hat{j}$

$\vec{F} = m \vec{a} = 0.2 \times 10 \hat{i} = 2 \hat{i}$ $∴$ $\vec{τ} = (10 \hat{i} - \hat{j}) \times (2 \hat{i}) = 2 \hat{k}$

$Angular momentum, \vec{L} = {\vec{r}}_{B} \times \vec{P} = {\vec{r}}_{B} \times m \vec{v}$

$From \vec{v} = \vec{u} + \vec{a} t = 10 \hat{i} \times 2 = 20 \hat{i}$

$\vec{L} = (0.2) [(10 \hat{i} - \hat{j}) \times 20 \hat{i}] = 4 \hat{k}$

$At point A (0, - 1)$

$\vec{τ} = {\vec{r}}_{A} \times \vec{F} = (- \hat{j}) \times 2 \hat{i} = 2 \hat{k} [∵ {\vec{r}}_{A} = - \hat{j}]$

Q 19 :

A rod of mass $m$ and length $L$ , pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $ω$ about the pivot.

The maximum angular speed $ω_{M}$ is achieved for $x = x_{M}$ . Then [2020]

$ω = \frac{3 v x}{L^{2} + 3 x^{2}}$
$ω = \frac{12 v x}{L^{2} + 12 x^{2}}$
$x_{M} = \frac{L}{\sqrt{3}}$
$ω_{M} = \frac{v}{2 L} \sqrt{3}$

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Select one or more options

(1, 3, 4)

From angular momentum conservation about the pivoted point,

$m v x = (\frac{m L^{2}}{3} + m x^{2}) ω$

[As the combined system rotates with angular speed $ω$ about the pivot]

$∴$ $ω = \frac{m v x}{\frac{m L^{2}}{3} + m x^{2}} = \frac{3 v x}{L^{2} + 3 x^{2}}$

$∴$ $ω = \frac{3 v x}{L^{2} + 3 x^{2}}$

Hence option (1) is correct.

For maximum angular velocity, $\frac{d ω}{d x} = 0$

$\frac{d}{d x} (\frac{L^{2}}{x} + 3 x) = 0 \Rightarrow \frac{L^{2}}{x^{2}} + 3 = 0 \Rightarrow x = \frac{L}{\sqrt{3}}$

$∴$ $x_{m} = \frac{L}{\sqrt{3}}$

So option (3) is correct.

$ω_{m} = \frac{3 v x}{L^{2} + 3 x^{2}} = \frac{3 v \cdot \frac{L}{\sqrt{3}}}{L^{2} + 3 {(\frac{L}{\sqrt{3}})}^{2}} = \frac{\sqrt{3}}{2 L} V$

Hence option (4) is correct.

Q 20 :

Consider a body of mass 1.0 kg at rest at the origin at time $t = 0$ . A force $\vec{F} = (α t \hat{i} + β \hat{j})$ is applied on the body, where $α = 1.0 {Ns}^{- 1}$ and $β = 1.0 N$ . The torque acting on the body about the origin at time $t = 1.0 s$ is $\vec{τ}$ . Which of the following statements is (are) true? [2018]

$| \vec{τ} |$ $= \frac{1}{3} Nm$
The torque $\vec{τ}$ is in the direction of the unit vector $+ \hat{k}$
The velocity of the body at $t = 1 s$ is $\vec{v} = \frac{1}{2} (\hat{i} + 2 \hat{j}) {ms}^{- 1}$
The magnitude of displacement of the body at $t = 1 s$ is $\frac{1}{6} m$

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Select one or more options

(1, 3)

$Given \vec{F} = α t \hat{i} + β \hat{j} or \vec{F} = t \hat{i} + \hat{j}$ $(∵ α = 1 {Ns}^{- 1}, β = 1 N)$

$∴$ $\frac{m d \vec{v}}{d t} = t \hat{i} + \hat{j}$

$∴$ $d \vec{v} = t d t \hat{i} + d t \hat{j}$ $[∵ m = 1]$

$∴$ $\int_{0}^{v} d \vec{v} = \int_{0}^{t} t d t \hat{i} + \int_{0}^{t} d t \hat{j} \Rightarrow \vec{v} = \frac{t^{2}}{2} \hat{i} + t \hat{j}$

$At t = 1 s, \vec{v} = \frac{1}{2} \hat{i} + \hat{j} = \frac{1}{2} (\hat{i} + 2 \hat{j}) {ms}^{- 1}$

$Also, \vec{v} = \frac{d \vec{r}}{d t} = \frac{t^{2}}{2} \hat{i} + t \hat{j}$ $∴$ $d \vec{r} = \frac{t^{2}}{2} d t \hat{i} + t d t \hat{j}$

$or \int_{0}^{r} d \vec{r} = \int_{0}^{t} \frac{t^{2}}{2} d t \hat{i} + \int_{0}^{t} t d t \hat{j} \Rightarrow \vec{r} = \frac{t^{3}}{6} \hat{i} + \frac{t^{2}}{2} \hat{j}$

$At t = 1, \vec{r} = \frac{1}{6} \hat{i} + \frac{1}{2} \hat{j},$ $∴$ $| \vec{r} |$ $= \sqrt{\frac{1}{36} + \frac{1}{4}} = \sqrt{\frac{10}{36}}$

$\vec{τ} = \vec{r} \times \vec{F} = (\frac{1}{6} \hat{i} + \frac{1}{2} \hat{j}) \times (\hat{i} + \hat{j})$ $(at t = 1 s)$

$or \vec{τ} = - \frac{1}{3} \hat{k}$ $∴$ $| \vec{τ} |$ $= \frac{1}{3} Nm$

Q 21 :

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point O is given by $V (r) = \frac{k r^{2}}{2}$ , where $k$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $R$ about the point O. If $v$ is the speed of the particle and $L$ is the magnitude of its angular momentum about O, which of the following statements is (are) true? [2018]

$v = \sqrt{\frac{k}{2 m}} R$
$v = \sqrt{\frac{k}{m}} R$
$L = \sqrt{m k} R^{2}$
$L = \sqrt{\frac{m k}{2}} R^{2}$

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Select one or more options

(2, 3)

Given : potential energy, $V (r) = \frac{K r^{2}}{2}$

$Applying,$ $| F |$ $= \frac{d V}{d r} = \frac{d}{d r} [\frac{k r^{2}}{2}] = k r$

$For r = R, F = k R$

$Also F = \frac{m v^{2}}{R}$ $(∵ particle is moving in circular orbit)$

$or, \frac{m v^{2}}{R} = k R \Rightarrow v = \sqrt{\frac{k}{m}} R$

$And angular momentum, L = m v R = m (\sqrt{\frac{k}{m}} R) R = \sqrt{k m} R^{2}$

Q 22 :

A wheel of radius $R$ and mass $M$ is placed at the bottom of a fixed step of height $R$ as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque $τ$ about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct? [2017]

If the force is applied at point P tangentially then decreases continuously as the wheel climbs
If the force is applied normal to the circumference at point X then $τ$ is constant
If the force is applied normal to the circumference at point P then $τ$ is zero
If the force is applied tangentially at point S then $τ \neq 0$ but the wheel never climbs the step

1

2

3

4

(3)

If the force $(F)$ is applied at $P$ tangentially then the $τ$ remains constant and $τ = F \times 2 R .$

If force is applied normal to $X$ , then as the wheel climbs, the perpendicular distance of force from Q will go on changing. Initially the perpendicular is QM, later it becomes QM'.

If the force $(F)$ is applied normal to the circumference at point P then $τ = 0$ .

If the force $(F)$ is applied tangentially at point S then $τ = F \times R$ and the wheel climbs.

Q 23 :

A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure).

At some instant of time, the angle made by the bar with the vertical is $θ$ . Which of the following statements about its motion is/are correct? [2017]

The midpoint of the bar will fall vertically downward
The trajectory of the point A is a parabola
Instantaneous torque about the point in contact with the floor is proportional to $\sin θ$
When the bar makes an angle $θ$ with the vertical, the displacement of its midpoint from the initial position is proportional to $(1 - \cos θ)$

1

2

3

4

Select one or more options

(1, 3, 4)

Force acting on COM $F_{x} = 0,$ $∴$ $a_{x} = 0 .$ Therefore the force acting in vertical direction will move the mid point or COM of the bar fall vertically downwards.

When the bar makes an angle $θ$ the height of its COM $= \frac{L}{2} \cos θ$

$∴$ Displacement of its mid point from the initial position $= \frac{L}{2} - \frac{L}{2} \cos θ = \frac{L}{2} (1 - \cos θ)$

Instantaneous torque about the point of contact P

$τ = m g \times \frac{L}{2} \sin θ$

$Now x = \frac{L}{2} \sin θ, y = L \sin (90 ° - θ) = L \cos θ$

$∴$ ${(\frac{2 x}{L})}^{2} + {(\frac{y}{L})}^{2} = 1 or \frac{4 x^{2}}{L^{2}} + \frac{y^{2}}{L^{2}} = 1$

Thus path of A is an ellipse not parabola.

Q 24 :

The position vector $\vec{r}$ of a particle of mass $m$ is given by the following equation $\vec{r} (t) = α t^{3} \hat{i} + β t^{2} \hat{j},$

where $α = \frac{10}{3} {ms}^{- 3}$ , $β = 5 {ms}^{- 2}$ and $m = 0.1 kg$ . At $t = 1 s$ , which of the following statement(s) is(are) true about the particle? [2016]

The velocity $\vec{v}$ is given by $\vec{v} = (10 \hat{i} + 10 \hat{j}) {ms}^{- 1}$
The angular momentum $\vec{L}$ with respect to the origin is given by $\vec{L} = - \frac{5}{3} \hat{k} N m s$
The force $\vec{F}$ is given by $\vec{F} = (\hat{i} + 2 \hat{j}) N$
The torque $\vec{τ}$ with respect to the origin is given by $\vec{τ} = - \frac{20}{3} \hat{k} N m$

1

2

3

4

Select one or more options

(1, 2, 4)

$Given : \vec{r} = α t^{3} \hat{i} + β t^{2} \hat{j}$

$\vec{r} = \frac{10}{3} t^{3} \hat{i} + 5 t^{2} \hat{j} m$

$∴$ $\vec{v} = \frac{d \vec{r}}{d t} = 10 t^{2} \hat{i} + 10 t \hat{j} {ms}^{- 1}$

$and \vec{a} = \frac{d \vec{v}}{d t} = 20 t \hat{i} + 10 \hat{j} {ms}^{- 2}$

$At t = 1 s, {\vec{r}}_{t = 1} = \frac{10}{3} \hat{i} + 5 \hat{j} m$

${\vec{v}}_{t = 1} = 10 \hat{i} + 10 \hat{j} {ms}^{- 1}, {\vec{a}}_{t = 1} = 20 \hat{i} + 10 \hat{j} {ms}^{- 2}$

${\vec{p}}_{t = 1} = \hat{i} + \hat{j} {kg ms}^{- 1}$

$\vec{L} = \vec{r} \times \vec{p} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{10}{3} & 5 & 0 \\ 1 & 1 & 0 \end{matrix} |$ $= \hat{k} [\frac{10}{3} - 5] = - \frac{5}{3} \hat{k} {kg ms}^{- 1}$

$\vec{F} = m \vec{a} = (2 \hat{i} + \hat{j}) N$

$\vec{τ} = \vec{r} \times \vec{F} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{10}{3} & 5 & 0 \\ 2 & 1 & 0 \end{matrix} |$ $= \hat{k} [\frac{10}{3} - 10] = - \frac{20}{3} \hat{k} Nm$

Q 25 :

Two thin circular discs of mass m and 4m, having radii $a$ and $2 a$ , respectively, are rigidly fixed by a massless, rigid rod of length $l = \sqrt{24} a$ through their centres. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $ω$ . The angular momentum of the entire assembly about the point 'O' is $\vec{L}$ (see the figure). Which of the following statement(s) is(are) true? [2016]

The centre of mass of the assembly rotates about the z-axis with an angular speed of $ω / 5$
The magnitude of angular momentum of centre of mass of the assembly about the point O is $81 m a^{2} ω$
The magnitude of angular momentum of the assembly about its centre of mass is $\frac{17}{2} m a^{2} ω$
The magnitude of the z-component of $\vec{L}$ is $55 m a^{2} ω$

1

2

3

4

Select one or more options

(1, 3)

Position of CM on the axis of rod.

$x_{C M} = \frac{m (0) + 4 m (l)}{m + 4 m} = \frac{4 l}{5}$

$\cos θ = \frac{1}{\sqrt{l^{2} + a^{2}}} = \frac{\sqrt{24 a}}{\sqrt{a^{2} + 24 a^{2}}} = \frac{\sqrt{24}}{5}$ $[∵ l = \sqrt{24 a} given]$

$O A = \sqrt{{(2 l)}^{2} + {(2 a)}^{2}} = \sqrt{96 a^{2} + 4 a^{2}} = 10 a$

Let complete system rotates about z-axis with a constant angular velocity $ω'$

$∴$ $\frac{ω'}{ω} = \frac{2 π (2 a)}{2 π (10 a)} \Rightarrow ω' = \frac{ω}{5}$

Magnitude of angular momentum of the system about its center of mass

$L_{C M} = I_{C M} ω = [\frac{m a^{2}}{2} + \frac{4 m {(2 a)}^{2}}{2}] ω = \frac{17}{2} m a^{2} ω$

Magnitude of angular momentum of CM of system about point O.

$L' = 5 m \times \frac{9 l ω}{5} \times \frac{9 a}{5} = \frac{81 l m ω a}{5} = \frac{81 \sqrt{24} m ω a^{2}}{5}$

Magnitude of z-component of angular momentum of system about point O

$L_{z} = L' \cos θ - L_{C M} \sin θ$

$= \frac{81 \sqrt{24} m ω a^{2}}{5} \times \frac{\sqrt{24}}{5} - \frac{17}{5} m a^{2} ω \times \frac{1}{5}$

$= m a^{2} ω (\frac{1944}{25} - \frac{17}{10})$

Q 26 :

A ring of mass M and radius R is rotating with angular speed $ω$ about a fixed vertical axis passing through its centre O with two point masses each of mass $\frac{M}{8}$ at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is $\frac{8}{9} ω$ and one of the masses is at a distance of $\frac{3}{5} R$ from O. At this instant the distance of the other mass from O is [2015]

$\frac{2}{3} R$
$\frac{1}{3} R$
$\frac{3}{5} R$
$\frac{4}{5} R$

1

2

3

4

(4)

Applying conservation of angular momentum about the axis

$M R^{2} ω = M R^{2} \times \frac{8 ω}{9} + \frac{M}{8} \times \frac{9 R^{2}}{25} \times \frac{8 ω}{9} + \frac{M}{8} r^{2} \times \frac{8 ω}{9}$

Solving we get

$\Rightarrow r = \frac{4 R}{5}$

Q 27 :

In the figure, a ladder of mass $m$ is shown leaning against a wall. It is in static equilibrium making an angle $θ$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $μ_{1}$ and that between the floor and the ladder is $μ_{2}$ . The normal reaction of the wall on the ladder is $N_{1}$ and that of the floor is $N_{2}$ . If the ladder is about to slip, then [2014]

$μ_{1} = 0, μ_{2} \neq 0 and N_{2} \tan θ = \frac{m g}{2}$
$μ_{1} \neq 0, μ_{2} = 0 and N_{1} \tan θ = \frac{m g}{2}$
$μ_{1} \neq 0, μ_{2} \neq 0 and N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}$
$μ_{1} = 0, μ_{2} \neq 0 and N_{1} \tan θ = \frac{m g}{2}$

1

2

3

4

Select one or more options

(3, 4)

$When μ_{1} \neq 0 and μ_{2} \neq 0$

Horizontal equilibrium, $N_{1} = μ_{2} N_{2}$

Vertical equilibrium, $m g = N_{2} + μ_{1} N_{1}$

Solving the above equations we get

$N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}$

$When μ_{1} = 0; Torque about P$

$m g \times \frac{l}{2} \cos θ = N_{1} \times l \sin θ$

$\Rightarrow N_{1} \tan θ = \frac{m g}{2}$

Q 28 :

If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that [2009]

linear momentum of the system does not change in time
kinetic energy of the system does not change in time
angular momentum of the system does not change in time
potential energy of the system does not change in time

1

2

3

4

(1)

$Given \sum {\vec{F}}_{ext} = 0$

$\sum {\vec{F}}_{ext} = \frac{d {\vec{p}}_{system}}{d t} = 0 \Rightarrow {\vec{p}}_{system} = constant$

i.e., Linear momentum of the system does not change in time.

Due to internal forces acting in the system, the kinetic and potential energy may change with time.

Also zero external force may create a torque. Thus the torque will change the angular momentum of the system in time.

Topic Question Set

Q 5 :

A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct? [2005]

Q 6 :

A block of mass $m$ is at rest under the action of force $F$ against a wall as shown in figure. Which of the following statement is incorrect? [2005]

Q 8 :

A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remain conserved? [2003]

Q 9 :

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J = M V$ is imparted to the body at one of its ends, what would be its angular velocity? [2003]

Q 23 :

A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure).

At some instant of time, the angle made by the bar with the vertical is $θ$ . Which of the following statements about its motion is/are correct? [2017]

Q 28 :

If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that [2009]

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Topic Question Set

Q 5 : A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct? [2005]

Q 6 : A block of mass m is at rest under the action of force F against a wall as shown in figure. Which of the following statement is incorrect? [2005]

Q 8 : A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remain conserved? [2003]

Q 9 : Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J=MV is imparted to the body at one of its ends, what would be its angular velocity? [2003]

Q 20 : Consider a body of mass 1.0 kg at rest at the origin at time t=0. A force F→=(αti^+βj^) is applied on the body, where α=1.0 Ns-1 and β=1.0 N. The torque acting on the body about the origin at time t=1.0 s is τ→. Which of the following statements is (are) true? [2018]

Q 23 : A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is θ. Which of the following statements about its motion is/are correct? [2017]

Q 24 : The position vector r→ of a particle of mass m is given by the following equation r→(t)=αt3i^+βt2j^, where α=103ms-3, β=5 ms-2 and m=0.1 kg. At t=1 s, which of the following statement(s) is(are) true about the particle? [2016]

Q 28 : If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that [2009]

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Q 5 :

A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct? [2005]

Q 6 :

A block of mass $m$ is at rest under the action of force $F$ against a wall as shown in figure. Which of the following statement is incorrect? [2005]

Q 8 :

A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remain conserved? [2003]

Q 9 :

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J = M V$ is imparted to the body at one of its ends, what would be its angular velocity? [2003]

Q 23 :

A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure).

At some instant of time, the angle made by the bar with the vertical is $θ$ . Which of the following statements about its motion is/are correct? [2017]

Q 28 :

If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that [2009]