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Angular momentum in case of rotation about a fixed axis - class-XI

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A particle of mass m travels with a speed v along positive direction of x-axis parallel to the line y=4. At t=0, the particle is at (0,4),. The angular momentum of the particle about the origin is 

  1. 0

  2. 4 mv directed along the positive z- axis

  3. 4 mv directed along the negative z- axis

  4. 4 mv directed along the positive y- axis


Correct Option: C
Explanation:
The correct option is C

We have,

$Mass=m,Speed=v$

A particle is moving along a straight line parallel to the y-axis.

So, position vector $(\hat r ) = 4 \hat j$ 

velocity of body $\hat V = v \hat i$

Since we know,

Angular momentum = $m(\hat r \times\hat V)$

$=m ( 4 \hat j \times v \hat i)$

$=-4mv\hat k$

Since,

$I \times j = k$ 
And, $j \times i = - k$

A spinning ice skater can increase his rate of rotation by bringing is arms and free leg closer to his body.
How does this procedure affect the skater's angular momentum and kinetic energy?

  1. Angular momentum remains the same while kinetic energy increases

  2. Angular momentum remains the same while kinetic energy decreases

  3. Both angular momentum and kinetic energy remain the same

  4. Angular momentum increases while kinetic energy remains the same


Correct Option: C

The angular momentum of a moving body remains constant if

  1. net external force is applied

  2. net pressure is applied

  3. net external torque is applied

  4. net external torque is not appled


Correct Option: D

Which of the following laws is not always true as per the present world?

  1. Law of conservation of Angular momentum

  2. Law of conservation of Charge 

  3. Law of conservation of linear momentum

  4. Law of conservation of Energy 


Correct Option: C

What should be the angular momentum of an electron in Bohr's hydrogen atom whose energy is -0.544 eV?

  1. $\large \frac{h}{\pi}$

  2. $\large \frac{3h}{2\pi}$

  3. $\large \frac{5h}{2\pi}$

  4. $\dfrac{2h}{2\pi}$


Correct Option: C

Angular momentum in $UCM$ is directed:

  1. Perpendicular to the plane of $UCM$

  2. Angular radius and away from centre

  3. Along radius and toward the centre

  4. Tangential to the $UCM$


Correct Option: C

In absence of external forces on a rigid system, which of the following quantities must remain constant?

  1. angular momentum

  2. linear momentum

  3. moment of inertia

  4. kinetic energy


Correct Option: A,B

A particle of mass 2kg located at the  position $\hat{i}+\hat{j}$     has a velocty$2(\hat{i}-\hat{j}+\hat{k})$      it s angular momentum about the x-axis in kg 

  1. Zero

  2. 8

  3. 12

  4. 4


Correct Option: B

Which parameters of all the particles of rotating fan are same :-

  1. Only angular $( \theta, \omega, \alpha)$

  2. Only linear (s,v,a)

  3. Both linear and angular

  4. None of them


Correct Option: A
Explanation:

In a rotational motion, only the angular parameters ($\theta, w,\alpha$) are same but not the linear parameters.

A man, sitting firmly over a rotating stool has his arms stretched. If he folds his arms, the work done by the man is :

  1. zero

  2. positive

  3. negative

  4. may be positive or negative


Correct Option: B
Explanation:

When the arms are stretched the rotational inertia is higher. When the man folds his arm he is basically decreasing his rotational inertia thereby gaining more angular velocity using the principle of conservation of angular momentum. This implies more angular distance is covered in the same time giving us positive work.

A 40 Kg mass hanging at the end of a rope of length L oscillates in a vertical plane with an angular amplitude $ \theta _o $. If the breaking strength of the rope is 80 Kg. Wt What is the maximum angular amplitude so that the mass can oscillates without the rope breaking ?

  1. $ 90^o $

  2. $ 60^o $

  3. $ 30^o $

  4. $ 45^o $


Correct Option: A

Assertion (A): Even though a planet revolve around the sun in an elliptical orbit, the angular momentum of planet is constant.
Reason (R) : Any force other than mutual gravitational force is absent between the planet and the sun.

  1. Both A and R are true and R is correct explanation of A

  2. Both A and R are true and R is not correct explanation of A

  3. A is true and R is false

  4. A is false and R is true


Correct Option: A
Explanation:

Angular momentum of planet revolving around the sun is constant because the line of force of the gravitational force passes through the  sun . Therefore torque about the sun is zero.
$\therefore$  Angular momentum will remain constant.  And there is no other force acting on the planet, which results into zero tangential force.

The law of conservation of angular momentum is obtained from Newton's II law in rotational motion when:

  1. External torque is maximum

  2. External torque is minimum

  3. External torque is zero

  4. External torque is constant


Correct Option: C
Explanation:

Newton's 2nd law says that External Force is directly proportional to the rate of change of momentum. Similarly, external torque is directly proportional to the rate of change of angular momentum. So if angular momentum is conserved i.e., the rate of change of angular momentum is zero then external torque is zero.

The energy of electron in an excited hydrogen atom is -3.4 eV. Its angular momentum according to Bohr's theory will be:

  1. $\cfrac{h}{\pi }$

  2. $\cfrac{h}{2\pi }$

  3. $\cfrac{3h}{2\pi }$

  4. $\cfrac{2h}{\pi }$


Correct Option: A
Explanation:

$E=\dfrac{-13.6}{n^{2}}eV$

$\dfrac{-13.6}{n^{2}}=\ 3.4 $
$\Rightarrow    n = 2$
So, its angular momentum $=\dfrac{nh}{2\pi }$ $=\dfrac{2h}{2\pi }$ $=\dfrac{h}{\pi }$

Under the action of a central force, there is a conservation of-

  1. Angular momentum only

  2. Mechanical energy only

  3. Angular momentum and mechanical energy

  4. Neither angular momentum nor mechanical energy


Correct Option: C
Explanation:

Under the action of a central force, there is a conservation ofAngular momentum and mechanical energy.

Four particles of masses in the ratio 1:2:3:4 are rotating in concentric circles of radii proportional to their masses with the same angular velocity. What is the angular momentum of the system

  1. $4mw^2r$

  2. $10mw^2r$

  3. $30mw^2r$

  4. $20mw^2r$


Correct Option: C
Explanation:

The angular momentum of a single particle is $mw^2r$. If there are 4 particles, we can write,
$L _{tot}=mw^2r+2mw^2(2r)+3mw^2(3r)+4mw^2(4r)=30mw^2r$

The correct option is (c)

Two identical particles of mass m move in a circle of radius r , 180 degrees out of phase at an angular speed  $\omega$ about the z-axis in a plane parallel to but a distance h above the x-y plane (Figure 19.14). Find the magnitude and the direction of the angular momentum L relative to the origin.

  1. $2mr^2w$

  2. $mr^2w$

  3. zero

  4. $mr^2w/2$


Correct Option: A

Two identical particles of mass m are separated by a distance of 1 metre each and are rotating with a constant speed of 5 m/s . What will be their angular momentum about an axis distant 0.8 m from the first particle on the line joining them

  1. 5m

  2. 3m

  3. 2m

  4. m


Correct Option: B

A particle is in uniform circular motion in a   horizontal plane. Its angular momentum is constant when the origin is taken at

  1. centre of the circle

  2. any point on the circumference of the circle.

  3. any point inside the circle.

  4. any point outside the circle.


Correct Option: A

A particle starts from the point 0.8 m and moves with a uniform velocity of 3 m/s. what is the angular momentum of the particle after 5 seconds about origin.mass of the particle is 1  kg.

  1. $12kgm^2$/s

  2. $24 kgm^2/s$

  3. $32 kgm^2/s$

  4. $44 kgm^2/s$


Correct Option: B
Explanation:

The correct option is B

Given,

$mass=1kg$

$t=5s$

$velocity=3m/s$

The angular momentum of the particle about a fixed position is given by:

$L = r \times mV$

So it is the product of momentum of a particle and its perpendicular distance from the fixed position.


Here particle is moving along the x-direction



So its momentum is along the x-direction

$P = mv = 1\times3 = 3 kgm/s$

Now, the perpendicular distance from origin to the momentum of particle is:

$r = 8 m$

Now, angular momentum is


$L = 8\times3 = 24 kg m^2/s$

So angular momentum is $ 24 kg m^2/s$

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t)=2t^{3}-6t^{2}$. The torque on the wheel becomes zero at :

  1. $t=1s$

  2. $t=0.5\ s$

  3. $t=0.25\ s$

  4. $t=2s$


Correct Option: A
Explanation:
Given, $\theta (t)=2t^3-6t^2$

We, know $T=\propto \dfrac{Id^2 \theta}{dt^2}=I\dfrac{d}{dt}(\dfrac{d \theta}{dt})=I\dfrac{d}{dt}(6t^2-12t)=I(12t-12)=0$

Then, $12t-12=0\Rightarrow t=1s$

Let $A$ be the area swept by the line joining the earth and the sun during Feb 2012. The area swept by the same line during the first week of that month is

  1. $\dfrac{A}{4}$

  2. $\dfrac{7A}{29}$

  3. $A$

  4. $\dfrac{7A}{30}$


Correct Option: B

(i) Linear momentum is proportional to $1/n$
(ii) Radius is proportional to $n$
(iii) Kinetic energy is proportional to $1/n^{2}$
(iv) Angular momentum is proportional to $n$
Choose the correct option from the codes given below. 

  1. $(i),(iii),(iv)$ are correct

  2. $(i)$ is correct

  3. $(i),(ii)$ are correct

  4. $(iii)$ is correct


Correct Option: C

The position of a particle is given by $\overrightarrow { r } =(\hat{i} +2\hat{j} -\hat{k})$ and momentum $\overrightarrow { P } =(3\hat{i} +4\hat{j} -2\hat{k})$. The angular momentum is perpendicular to

  1. X-axis

  2. Y-axis

  3. Z-axis

  4. Line at equal angles to all the three axes


Correct Option: A
Explanation:

Given,

$\vec r=\hat i+2\hat j-\hat k$
$\vec P=3\hat i+4\hat j-2\hat k$
Angular momentum, $\vec L=\vec r \times \vec P$
$\vec L=(\hat i+2\hat j-\hat k)\times (3\hat i+4\hat j-2\hat k)$
$\vec L=0\hat i-\hat j-2\hat k$
Hence, The angular momentum is perpendicular to the X-axis.
The correct option is A.

A particle of mass $1\ kg$ is projected at an angle $ \theta $ with horizontal. Its co-ordinates at any instant are $(5m,5m)$ and itis having velocity components along $X- $axis and $Y-$axis as $8\ m/s$ and $4\ m/s$ respectively. Its angular momentum about the origin is 

  1. $- 20\ N-m$ $ \hat{k} $

  2. $+ 20\ N-m$ $ \hat{k} $

  3. $- 60\ N-m$ $ \hat{k} $

  4. $+ 60\ N-m$ $ \hat{k} $


Correct Option: B

A particle of mass m = 5 units is moving with a uniform speed v = $3\sqrt2$ units in the XY - plane along the line y = x + 4.The magnitude of the angular momentum about origin is  

  1. zero

  2. 60 unit

  3. 7.5 unit

  4. $40\sqrt2$ unit


Correct Option: B

The angular momentum of an electron in a hydrogen atom is proprotional to ( where r is redius of orbit) 

  1. $\frac{1}{\sqrt{r}}$

  2. $\frac{1}{r}$

  3. $r^{1/2}$

  4. $^r{2}$


Correct Option: C
Explanation:

Radius of the nth orbit rn n2/Z          n (r n)½

Angular momentum Ln = nh/(2π) (r n)½

$=r^{1/2}$

If a particle of mass m is moving with constant velocity V parallel to X-axis along $y=axis$. The angular momentum with respect to origin at any time 't' is

  1. $mV\hat{J}$

  2. $-mVa\hat{K}$

  3. $mV\hat{K}$

  4. $mVa\hat{J}$


Correct Option: A

The rotational kinetic energy of a hollow spherical shell 2.5 J. If its frequency of rotation is made 10 times, then new kinetic energy will be -

  1. $0.25 J$

  2. $2.5\times { 10 }^{ 2 }J$

  3. $2500 J$

  4. $2.5 J$


Correct Option: B
Explanation:
The rotational kinetic energy is given by
$K=\dfrac{1}{2}mr^2\omega ^2=2.5 J$. . . . .(1)
If frequency of rotation made 10 times , then the new rotational kinetic energy is
$K'=\dfrac{1}{2}mr^2 \times 10^2\omega ^2$
$K'=100K$ 
$K'=2.5\times 10^2J$
The correct option is B.

The angular momentum of an electron revolving in a circular orbit is J, What is its magnetic moment? 

  1. $\frac{mJ}{2e}$

  2. $\frac{eJ}{2m}$

  3. $\frac{2m}{eJ}$

  4. $\frac{emJ}{2}$


Correct Option: B
Explanation:

According to Bohr modal of hydrogen like atoms, negatively charged electrons revolves around the positively charged nucleus. This uniform circular motion of electrons is equivalent to a current loop which possesses a magnetic dipole moment =IA

I:- current in the loop

A:-area

Consider an electron revolving anticlockwise around a nucleus in an orbit of radius r with speed v and time period T.

Equivalent current,

I= charge/time

=e/t = e/2πr/v

=ev/2πr

Area of current loop ,A=πr2

So , orbital magnetic moment of electron is=IA

=(ev/2πr). πr2

=evr/2

As J is angular momentum. J=mvr

m- mass of electron

So orbital magnetic moment = eJ/2m

$E _n$ and $J _n$ denote the total energy magnitude and the  angular momentum of an electron in the nth allowed orbit of the Both atom .Then:

  1. $E _n \pi J _n$

  2. $E _n \pi \frac {1} {J}$

  3. $E _n \pi { J } _{ n }^{ 2 }$

  4. $E _n \pi \frac {1} {J^2}$


Correct Option: C

A wooden block of mass $2\ m$ is hung with the help of a light string of length $l$ in the vertical plane.
A bullet of mass $\dfrac{m}{4}$ moving horizontally with velocity $v _{0}\left(v _{0}=\sqrt{5gl}\right)$ penetrates the block and comes out with velocity $\dfrac{v _{0}}{2}$. The maximum height reached by the block is (Assume string remains vertical till bulled passes through the block)

  1. $\dfrac{{v} _{0}^{2}}{256\ g}$

  2. $\dfrac{{v} _{0}^{2}}{128\ g}$

  3. $\dfrac{{v} _{0}^{2}}{512\ g}$

  4. $\dfrac{{v} _{0}^{2}}{32\ g}$


Correct Option: B

A person standing on a rotating platform has his hands lowered. He suddenly outstretch his arms. The angular momentum.

  1. Becomes Zero

  2. Increases

  3. Decreases

  4. Remains the same


Correct Option: B

If a person sitting on a rotating stool with his hands outstretched, suddenly lowers his hands, then his :

  1. Kinetic energy will decrease.

  2. Moment of inertia will decrease.

  3. Angular momentum will increase.

  4. Angular velocity will remain constant.


Correct Option: B
Explanation:

According to conservation of Angular Momentum,
$I \omega$ = constant
Hence, when a person sitting on a rotating stool suddenly  lowers his hands, then his angular velocity will increase and moment of inertia will decrease. 

A man spinning in free space changes the shape of his body, eg. by spreading his arms or curling up. By doing this, he can change his :

  1. moment of inertia

  2. angular momentum

  3. angular velocity

  4. rotational kinetic energy


Correct Option: A,C,D
Explanation:

The moment of inertia can be increased or angular velocity can be decreased by stretching hands outside. Also, smaller the moment of inertia means less resistance to rotation (some rapid corrections have to made to achieve rotational equilibrium), hence change in moment of inertia leads the change in rotational kinetic energy. While angular momentum remains conserved.

Two particles are initially moving with angular momentum $\vec{L _{1}}$ and $\vec{L _{2}}$ in a region of space with no external torque. A constant external torque $\vec{\tau}$ then acts on one particle, but not on the other particle, for a time interval $\Delta{t}$. What is the change in the total angular momentum of the two particles?

  1. $\vec\Delta L=\vec {L _{1}}-\vec{L _{2}}$

  2. $\Delta L=\dfrac{1}{2}(\vec {L _{1}}-\vec{L _{2}})$

  3. $\vec\Delta L=\tau \Delta t$

  4. $\vec\Delta L$ is not applicable for this system.


Correct Option: C

uniform disc of mass M and radius R is rotating about its centre of mass (the centre of mass is at rest )with an angular speed $\omega $.the angular momentum of disc about a point A (as shown)will be.

  1. $M{R^2}\omega + MhR\omega $

  2. $\frac{1}{2}M{R^2}\omega $

  3. $\frac{1}{2}M{R^2}\omega + MhR\omega $

  4. $\frac{1}{2}M{R^2}\omega + 1/2MhR\omega $


Correct Option: A

when a mass is rotating in a plane about a fixed point, its angular momentum is directed along

  1. radius

  2. the tangent to the orbit

  3. a line perpendicular to the plane of rotation

  4. none of above


Correct Option: C

The angular momentum of a system of particles is conserved 

  1. When no external force acts upon the system

  2. When no external torque acts upon the system

  3. When no external impulse acts upon the system

  4. none of these 


Correct Option: B

A ballet dancer spins about a vertical axis at $120 rpm$ with arms stretched with her arms folded the moment of inertia about the axis of rotation decreases by $40\%$ calculate new rate of rotation

  1. $100 rpm$

  2. $150 rpm$

  3. $200 rpm$

  4. $250 rpm$


Correct Option: C

A particle of mass $300 g$ is moving with a speed of $20 ms-1$ along the straight line $y= x-4\sqrt { 2 }$. The angular momentum of the particle about the origin is (where y & x are in metres)

  1. $ 24 kg m^2s^{-1}$

  2. $24\sqrt{ 2} kg m^2s^{-1}$

  3. $12 kg m^2s^{-1}$

  4. $6\sqrt{ 2} kg m^2s^{-1}$


Correct Option: C

 A rod of mass M and length L is placed on a smooth horizontal table and is hit by a ball moving horizontally and perpendicular to length of rod and sticks to it.Then conservation of angular momentum can be applied 

  1. About any point on the rod

  2. About a point at the centre of the rod

  3. About end point of the rod

  4. None


Correct Option: C

Two spherical bodies of equal mass (M) revolve about their centre of mass. The distance between the centre of the two masses is r. The angular momentum of each about their centre of mass is

  1. $2 \sqrt {GM^3 r}$

  2. $\frac{1}{2} \sqrt {GM^3 r}$

  3. $\frac{1}{2} \sqrt {2GM^3 r}$

  4. $\frac{1}{2} \sqrt {\frac{GM^3 r}{2}}$


Correct Option: D

Choose the INCORRECT statements

  1. If Linear momentum of system is conserved then angular momentum of system must be also conserved.

  2. If angular momentum is conserved then angular velocity must be also conserved about the same axis.

  3. If not torque about any axis is zero then force must also be zero.

  4. If a rigid body is in translational equilibrium then its linear velocity is constant but angular velocity may be varying


Correct Option: D

A satellite with a mass of $M$ moves in a circular orbit of radius $R$ at a constant speed of $v$. Which of the following must be true?
(I) The net force on the satellite is equal to MR and is directed toward the centre of the orbit,
(II) The net work done on the satellite by gravity in one revolution is zero.
(III) The angular momentum of the satellite is constant.

  1. I only

  2. III only

  3. I and II only

  4. II and III only

  5. I, II and III


Correct Option: C

A circular ring of mass $1$ Kg and radius $0.2$ m executes $10$ revolutions per sec. Its angular momentum would be -$( kg-m^2/sec)$

  1. $0.025$

  2. $0.25$

  3. $2.5$

  4. $25$


Correct Option: B

The total angular momentum of a body is equal to the angular momentum  of its center of mass if the body has:

  1. only rotational motion

  2. only translational motion

  3. both rotational and translational motion

  4. no motion at all


Correct Option: C

 Two bodies of different masses have same K.E. The one having more momentum is

  1. Heavier body

  2. lighter body

  3. both none

  4. both


Correct Option: B
Explanation:

$K.E.$ for a given momentum is inversely proportional to the mass$.$

So$,$ the lighter mass has greater kinetic energy$.$ For two bodies having same kinetic energy$,$ the heavier one has greater momentum$.$
Hence,
option $(B)$ is correct answer.  

A force $\vec F = \alpha \hat i + 3\hat j + 6\hat k$ is acting at a point $\vec r = 2\hat i - 6\hat j - 12\hat k$. The value of $\alpha$ for which angular momentum about the origin is conserved is:-

  1. $1$

  2. $-1$

  3. $2$

  4. zero


Correct Option: B
Explanation:

Given,
$\vec r=2\hat i-6\hat j-12\hat k$
$\vec F=\alpha\hat i +3\hat j+6\hat k$
For the conservation of angular momentum about origin, the torgue acting on the particle will be zero.
$\tau=\vec r\times \vec F$
$\vec r\times \vec F=0$
$(\alpha \hat i+3\hat j+6\hat k)\times (2\hat i-6\hat j-12\hat k)=0\hat i+0\hat j+0\hat k$
$\hat i(-36+36)-\hat j (12+12\alpha)+\hat k(6+6\hat k)=0\hat i+0\hat j+\hat k$
By equating it's coefficient, we get
$12+12\alpha =0$,   $6+6\alpha =0$
$12=-12\alpha$
$\alpha =-1$
The correct option is B.

A unit mass at position vector $\vec { r } = ( 3 \hat { i } + 4 \hat { j } )$ is moving with a velocity $\vec { v } = ( 5 \hat { i } - 6 \hat { j } )$ What is the angular momentum of the body about the origin

  1. 2 units along $z$ -axis

  2. 38 units along $x$ -axis

  3. 38 units along $y$ -axis

  4. 38 units along $Z$ -axis


Correct Option: C

A man standing on a platform holds weights in his outstretched arms. The system rotates freely about a central vertical axis. If he now draws the weights inwards close to his body, 

  1. the angular velocity of the system will increase.

  2. the angular momentum of the system will decrease.

  3. the kinetic energy of the system will increase.

  4. he will have to expend some energy to draw the weights in.


Correct Option: A,C,D
Explanation:

When the person contracts his hands he is decreasing his moment of inertia thereby increasing his angular velocity using the principle of conservation of angular momentum $I _1\omega _1=I _2\omega _2$
As the angular velocity increases, its kinetic energy also increases. Also energy would be required to draw the weights inwards from its position.

A particle performs uniform circular motion with an angular momentum L. If the frequency of particle's motion is doubled and its kinetic energy halved, the angular momentum becomes:

  1. $2L$

  2. $4L$

  3. $ \mathrm{L} / 2 $

  4. $ \mathrm{L} / 4 $


Correct Option: C

What is the magnitude of vertical force required to produced a moment of 20 Nm at point A (1 m, 1 m) if the force is acting at point B (2m,2m)

  1. 40 N

  2. 30 N

  3. 20 N

  4. 10 N


Correct Option: C

A stone tied to one end of the string is revolved round a rod in such a way that the string winds over the rod and get shortened. In this process which of the following quantities remain constant?

  1. Moment of inertia

  2. Momentum

  3. Angular momentum

  4. Kinetic energy


Correct Option: C
Explanation:

As the distance between the stone and the rotational axis is changing the moment of inertia thus changes. Momentum keeps on changing around the circular path. As the moment of inertia  changes we see that the kinetic energy changes too. The angular momentum is the only conserved quantity since there is no external torque acting on the stone.

A swimmer while jumping into water from a height easily forms a loop in air, if

  1. He pulls his arms and leg in

  2. He spreads legs and his arms

  3. He keeps himself straight

  4. His body is so formed


Correct Option: A
Explanation:

by pulling his arms and legs in he decreases his moment of inertia, thereby increasing his angular velocity using the principle of conservation of angular momentum.

If no external force acts on a system

  1. Velocity of centre of mass remains constant

  2. Total kinetic energy of the system remains constant

  3. Angular momentum of system may change

  4. Total mechanical energy of the system remains constant


Correct Option: A
Explanation:

No force means no acceleration implying constant velocity. Also as velocity is constant kinetic energy remains constant. The angular momentum may change due to the change in moment of inertia. Also the total mechanical energy may change due the change in potential energy.

Two loops P and Q are made from a uniform wire. The radii of P and Q are $r _1$ and $r _2$ respectively, and their moments of inertia about their own axises are $I _1$ and $I _2$ respectively. If $I _2 = 4I _1$, the $r _1/r _2$ equals:

  1. $4^{2/3}$

  2. $4^{1/3}$

  3. $4^{7/3}$

  4. $4^{-1/3}$


Correct Option: C

A particle moves on a circular path with decreasing speed. Choose the correct statement.

  1. Angular momentum remains constant.

  2. Acceleration $\vec{a}$ is towards the centre.

  3. Particle moves in a spiral path with decreasing radius.

  4. The direction of angular momentum remains constant.


Correct Option: C

A particle of mass 2 kg located at the position $ \left( \hat { i } +\hat { j }  \right) m $ has a velocity $ 2\left( \hat { i } -\hat { j } +\hat{ k } \right) ms^{-1} $ . Its angular momentum along z-axis in $ kgm^2 s^{-1}  $ is

  1. $-8$

  2. $+8$

  3. $-4$

  4. zero


Correct Option: A

Angular momentum of a system a particles changes, when

  1. Force acts on a body

  2. Torque acts on a body

  3. Direction of velocity changes

  4. None of these


Correct Option: B
Explanation:

If we apply a torque on a body, then angular momentum of the body changes according to the relation
$\bar{\tau} = \displaystyle\frac{d\bar{L}}{dt} \Rightarrow if \space \bar{\tau} = 0$ then, $\overrightarrow{L} = constant$

A gymnast takes turns with her arms & legs stretched. When she pulls her arms & legs in

  1. The angular velocity decreases

  2. The moment of inertia decreases

  3. The angular velocity stays constant

  4. The angular momentum increases


Correct Option: B
Explanation:

Since no external torque act on gymnast, so angular momentum $(L =

I\omega)$ is conserved. After pulling her arms & legs, the angular velocity increases but moment of inertia of gymnast, decreases in, such a way that angular momentum remains constant.

If a running boy jumps on a rotating table, which of the following is conserved?

  1. Linear momentum

  2. K.E

  3. Angular momentum

  4. None of these


Correct Option: C
Explanation:

The boy does not exert a torque to rotating table by jumping, so angular momentum is conserved i.e., 
$\displaystyle\frac{d\bar{L}}{dt} = 0 \Rightarrow \bar{L} = constant$

In which of the following case(s), the angular momentum is conserved ?

  1. The planet Neptune moves in elliptical orbit around the sun with sun at one focus

  2. A solid sphere rolling on an inclined plane

  3. An electron revolving around the nucleus in an elliptical orbit

  4. An $\alpha $ particle approaching a heavy nucleus from sufficient distance.


Correct Option: A,C,D
Explanation:

Angular momentum is conserved if net external torque is zero
(A) torque about sun is zero , $\because$ line of focus passer though it and no external force is present.
(B) Torque of friction is present 
(C) Force on the electron is internal , $\therefore$ torque is zero 
(D) No external force is present which may produce changes in angular momentum.

When tall buildings are constructed on earth, the duration of day night 

  1. slightly increases

  2. slightly decreases

  3. has no change

  4. none of these


Correct Option: A
Explanation:

Moment of inertia $I$ inerease. Therefore, $\omega $ decreases (as $I \omega = $ constant). Hence time period increases
$\displaystyle \left( as: T=\frac{2 \pi}{\omega} \right) $

If the ice at the poles melts and flows towards the equator, how will it affect the duration of day-night?

  1. Increase

  2. Decrease

  3. Remain same

  4. none of these


Correct Option: A
Explanation:

 Mass will move towards equator (away from axis) So, I will increase. Therefore $\omega$ will decrease
(as $I \omega =$ constant) Hence, time period will increase
$\displaystyle \left( as T=\frac{2 \pi}{\omega} \right) $
Duration of day-night increase

A cylinder is rolling down a rough inclined plane. Its angular momentum about the point of contact remains constant. Is this statement true or false?

  1. True

  2. False


Correct Option: B
Explanation:

$acceleration\quad of\quad the\quad cylinder\quad is\quad given\quad by\ mgsin\theta -f=ma....(1)\ fr=I\alpha ....(2)\ and\quad a=\alpha r...(3)\ acceleration\quad is\quad not\quad always\quad zero\quad so\quad the\quad velocity\quad of\quad the\quad \ cyclinder\quad is\quad aloways\quad changing\quad and\quad the\quad angular\quad momentum\quad \ about\quad the\quad point\quad of\quad contact\quad is\quad mvr,\quad which\quad is\quad not\quad always\ constant.\ $

A uniform thin circular ring of mass 'M' and radius 'R' is rotating about its fixed axis, passing through its centre and perpendicular to its plane of rotation, with a constant angular velocity $\omega$. Two objects each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity of :

  1. $\displaystyle \frac{\omega\, M}{(M\, +\, m)}$

  2. $\displaystyle \frac{\omega M}{(M\, +\, 2m)}$

  3. $\displaystyle \frac{\omega M}{M\, -\, 2m}$

  4. $\displaystyle \frac{\omega(M\, +\, 3m)}{M}$


Correct Option: B
Explanation:

Moments of inertia (MOI) of the ring before attaching the masses $I$= $ MR^2$,

MOI of the ring after attaching the masses $I^{'}= (M+2m) R^2$
Let angular momentum after the attaching the masses $\omega ^{'}$
Since there is no external torque,  so we use conservation of angular momentum.
$I \times \omega= I ^ {'} \times \omega ^ {'} $
$\Rightarrow MR^2 \times \omega= \dfrac { MR^2 \times \omega ^{'}}{ M + 2m}$,
$\Rightarrow \omega ^{'} = \dfrac {\omega M}{M+2m}$

A student is rotating on a stool at an angular velocity $'\omega'$ with their arms outstretched while holding a pair of masses. The frictional effects of the stool are negligible.
Which of the following actions would result in a change in angular momentum for the student?

  1. A clockwise torque of $50 Nm$ and a counterclockwise torque of $25 Nm$ are both applied to the students arms by fellow students

  2. The student brings the masses closer to their body

  3. The student stretches the masses further away from their body

  4. A second student steps onto the stool with the first student

  5. A clockwise torque of $50 Nm$ and a counterclockwise torque of $50 Nm$ are both applied to the students arms by fellow students


Correct Option: A
Explanation:
When no external force is acting, the total angular momentum of the system remains constant. So in this case, angular momentum will change only is an external force is applied on the system thus option (A) is correct

A solid cylinder of mass, $m$, and radius, $r$, is rotating at an angular velocity, $\omega$ when a non-rotating hoop of equal mass and radius drops onto the cylinder.
In terms of its initial angular velocity, $\omega$, what is its new angular velocity, ${\omega}^{\prime}$?

  1. ${\omega}^{\prime}=\dfrac{\omega}{3}$

  2. ${\omega}^{\prime}=\dfrac{3\omega}{4}$

  3. ${\omega}^{\prime}=\omega$

  4. ${\omega}^{\prime}=3\omega$

  5. ${\omega}^{\prime}=\dfrac{2\omega}{3}$


Correct Option: A
Explanation:

Moment of inertia of solid cylinder       $I _{cylinder}= \dfrac{1}{2}mr^2$

Thus initial angular momentum     $L _i = \dfrac{1}{2}mr^2 w$
Final moment of inertia of the system      $I' = I _{cylinder}+ I _{hoop} = \dfrac{1}{2}mr^2+ mr^2   = \dfrac{3}{2}mr^2$ 
Final angular momentum     $L _f =I'w' = \dfrac{3}{2}mr^2 w'$

Using conservation of angular momentum :      $L _i = L _f$
$\therefore$   $\dfrac{1}{2}mr^2 w = \dfrac{3}{2}mr^2 w'$                     $\implies w ' =\dfrac{w}{3}$

A bar of length l carrying a small mass m at one of its ends rotates with a uniform angular speed $\omega$ in a vertical plane about the mid-point of the bar. During the rotation, at some instant of time when the bar is horizontal, the mass is detached from the bar but the bar continues to rotate with same $\omega$. The mass moves vertically up, comes back and reches the bar at the same point. At that place, the acceleration due to gravity is g.

  1. This is possible if the quantity $\dfrac{{\omega}^2 \ell}{2\pi g}$ is an integer

  2. The total time of flight of the mass is proportional to ${\omega}^2$

  3. The total distance travelled by the mass n air is proportional to ${\omega}^2$

  4. The total distance travelled by the mass in air and its total time of flight are both independent on its mass.


Correct Option: A,C,D
Explanation:
The whole system is possible only if $\dfrac{\omega^2L}{2mg}$ is an integer, where the total distance travelled by the mass is proportional to $\omega^2$ and the distance travelled by the mass in air and its total time of flight are both independent on its mass.
The law of conservation of angular momentum states that '' When the net external torque acting on a system about a given axis is zero, the total angular momentum of the system about the axis remains constant''.
Hence, this question has multiple solutions.

A body is moving on a rough horizontal plate in a circular path being tide to a nail (at the centre) by a string, while the body is in motion the friction force of the body

  1. changes direction

  2. changes magnitude

  3. changes both magnitude and direction

  4. none of he above


Correct Option: A
Explanation:

As the motion is circular, friction opposes the centripetal acceleration, thus its direction changes regularly.

A dancer is rotating on smooth horizontal floor with an angular momentum $L$. The dancer folds her hands so that her momentof inertia decreases by $25$%. The new angular momentum is.

  1. $\dfrac {3L}{4}$

  2. $\dfrac {L}{4}$

  3. $\dfrac {L}{2}$

  4. $L$


Correct Option: D
Explanation:

The law of conservation of angular momentum states that , in an isolated system , total angular momentum remains constant . A rotating dancer is an isolated system therefore when she folds hands , her moment of inertia decreases but her angular speed increases such that angular momentum remains constant as L i.e.

                $L=I\omega$
when $I$ decreases  , $\omega$ increases .
Hence , total angular momentum would be L .

Two men of equal masses stand at opposite ends of the diameter of a turntable disc of a certain mass, moving with constant angular velocity. The two men make their way to the middle of the turntable at equal rates. In doing so will

  1. kinetic energy of rotation has increased while angular momentum remains same.

  2. kinetic energy of rotation has decreased while angular momentum remains same.

  3. kinetic energy of rotation has decreased but angular momentum has increased.

  4. both, kinetic energy of rotation and angular momentum have decreased.


Correct Option: B
Explanation:

Potential energy $U = mV$
$\Rightarrow U=(50x^2+100)10^{-2}$
     $F=-\dfrac{dU}{dx}=-(100x)10^{-2}$
$\Rightarrow m\omega ^2x=-(100\times 10^{-2})x$
      $10\times 10^{-3}\omega ^2x=100\times 10^{-2}x\Rightarrow \omega ^2=100,\omega =10$
$\Rightarrow f=\dfrac{\omega }{2\pi }=\dfrac{10}{2\pi}=\dfrac{5}{\pi }$

A circular disk of moment of inertia $I _t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega _i$. Another disk of moment of inertia $I _b$ is dropped co-axially onto the rotating disk.Initially the second disk has zero angular speed.Eventually both the disks rotate with a constant angular speed $\omega _p$ .The energy lost by the initially rotating disc due to friction is 

  1. $\dfrac{1}{2} \dfrac{I^2 _b}{(I _t+I _b)}\omega^2 _1$

  2. $\dfrac{1}{2} \dfrac{I^2 _t}{(I _t+I _b)}\omega^2 _1$

  3. $\dfrac{I _b-I _t}{(I _t+I _b)}\omega^2 _1$

  4. $\dfrac{1}{2} \dfrac{I _bI _t}{(I _t+I _b)}\omega^2 _1$


Correct Option: D
Explanation:

Initial angular momentum $={ I } _{ t }{ w } _{ 1 }4$

Now another disc is dropped coaxially on to the rotating disc

Final angular momentum

$=\left( { I } _{ t }+{ I } _{ b } \right) { w } _{ 2 }$

According to the law of conservation of momentum

${ I } _{ t }{ w } _{ 1 }=\left( { I } _{ t }+{ I } _{ b } \right) { w } _{ 2 }$

or$ { w } _{ 2 }=\cfrac{ { I } _{ t } }{ \left( { I } _{ t }+{ I } _{ b } \right)  } { w } _{ 1 }$

Loss in kinetic energy

$\Delta K=\cfrac{ 1 }{ 2 } { I } _{ t }{ w } _{ 1 }^{ 2 }-\cfrac{ 1 }{ 2 } \left( { I } _{ t }+{ I } _{ b } \right) { \left[ \cfrac{ { I } _{ t } }{ \left( { I } _{ t }+{ I } _{ b } \right)  } { w } _{ 1 } \right]  }^{ 2 }\\ \quad \quad =\cfrac{ 1 }{ 2 } { w } _{ 1 }^{ 2 }\cfrac{ { I } _{ t }{ I } _{ b } }{ \left( { I } _{ t }+{ I } _{ b } \right)  } $

 

Two particles each of mass m move in opposite direction along Y-axis. One particle moves in positive direction with velocity v while the other particle moves in negative direction with speed 2v. The total angular momentum of the system with respect to origin is:

  1. Is zero

  2. Goes on increasing

  3. Goes on decreasing

  4. None of these


Correct Option: A
Explanation:

Since both are moving in same line through origin total angular momentum is zero.

The shape of the orbit a planet depends on:

  1. angular momentum

  2. total energy

  3. both angular momentum and total

  4. None of the above


Correct Option: A
Explanation:

The shape of the orbit a planet depends on angular momentum.

A man standing on a platform holds weight in his outstreached arms. The system rotates freely about a central vertical axis. If he now draws the weights inward close to his body 

  1. The angular velocity of the system will increase

  2. The angular momentum of the system will increase

  3. The kinetic energy of the system will increase

  4. He will have to expend some energy to draw the weights in


Correct Option: D
Explanation:
$\tau=0=\cfrac{dl}{dt}$
$\Rightarrow L=constant$, Angular momentum remains constant.
By drawing arms in, $I$ is decreased.
$L=I\omega$
For $L$ to remain constant, $\omega$ increases.
$KE=\cfrac{L^{2}}{2I}$, $I$ decreases.
$\Rightarrow KE \,\,increases$
The energy spent in drawing weight is converted into KE of rotation

A force $\vec { F } =\alpha \hat { i } +3\hat { j } +6\hat { k }$ is acting at a point $\vec { r } =2\hat { i } -6\hat { j } -12\hat { k }$. The value of $\alpha$ for which angular momentum about origine is conserved is

  1. $1$

  2. $-1$

  3. $2$

  4. Zero


Correct Option: B
Explanation:

For  conservation of angular momentum about origin, external torque acting about origin should be zero.

$\therefore$ Torque $=r \times F$
$\Rightarrow (2\hat i-6\hat j- 12\hat k ) \times (\alpha \hat i + 3 \hat j + 6 \hat k)= (-36 +36) \hat i + (-12 \alpha -12 ) \hat j + ( 6+6\alpha ) \hat k=0$   $\Rightarrow \alpha =-1$

A point sized sphere of mass $'m'$ is suspended from a point using a string of length $'l'$. It is pulled to a side till the string is horizontal and released. As the mass passes through the portion where the string is vertical, magnitude of its angular momentum is:

  1. $ ml\sqrt { gl } $

  2. $ ml\sqrt { 2gl } $

  3. $ml\sqrt { \dfrac { gl }{ 2 } } $

  4. $ ml\sqrt { 3gl } $


Correct Option: B
Explanation:

By conservation of energy

$\dfrac{1}{2}mv^2=mgl$
$v=\sqrt{2gl}$
Angular momentum$=mvr=$$ml\sqrt{2gl}$

A disc of mass $100\ g$ and radius $10\ cm$ has a projection on its circumference. The mass of projection is negligible. A $20\ g$ bit of putty moving tangential to the disc with a velocity of $5\ m\ s^{-1}$ strikes the projection and sticks to it. The angular velocity of disc is

  1. $14.29\ rad\ s _1$

  2. $17.3\ rad\ s\ _1$

  3. $12.4\ rad\ s\ _1$

  4. $9.82\ rad\ s _1$


Correct Option: A
Explanation:

In this case, the angular momentum of  bit of putty about the axis of rotation = angular momentum of system of disc and bit of putty about the axis of rotation.


Let:

$M$ = Mass of puty

$m$ = Mass of disc

$ \therefore MvR=\left( \dfrac{m{{R}^{2}}}{2}+M{{R}^{2}} \right)\omega  $

 $ \omega =\dfrac{MvR}{\left( \dfrac{m{{R}^{2}}}{2}+M{{R}^{2}} \right)} $

 Putting all the values

 $ \omega =14.298\ m/s $

A stationary body explodes into two fragments of masses ${m} _{1}$ and ${m} _{2}$. If momentum of one fragment is $p$, the minimum energy of explosion is

  1. $\cfrac { { p }^{ 2 } }{ 2\left( { m } _{ 1 }+{ m } _{ 2 } \right) } $

  2. $\cfrac { { p }^{ 2 } }{ 2\left( \sqrt { { m } _{ 1 }{ m } _{ 2 } } \right) } $

  3. $\cfrac { { p }^{ 2 }\left( { m } _{ 1 }+{ m } _{ 2 } \right) }{ 2{ m } _{ 1 }{ m } _{ 2 } } $

  4. $\cfrac { { p }^{ 2 } }{ 2\left( { m } _{ 1 }-{ m } _{ 2 } \right) } $


Correct Option: A
Explanation:

Initially body is stationary (zero velocity)

using conservation of momentum
$\left( A \right) O=P+{ P _{ 2 } },{ P _{ 2 } }$ is momentum of mass ${m _2}$ after explosion
$\begin{array}{l} { P _{ 2 } }=-P \ Energy={ E _{ 1 } }\Rightarrow \dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 1 } } } } d{ E _{ 2 } }=\dfrac { { { { \left( { -P } \right)  }^{ 2 } } } }{ { 2{ m _{ 2 } } } }  \ net\, \, energy\Rightarrow { E _{ 1 } }+{ E _{ 2 } }=\dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 2 } } } } +\dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 2 } } } } \Rightarrow \dfrac { { { P^{ 2 } } } }{ { 2\left( { { m _{ 1 } }+{ m _{ 2 } } } \right)  } }  \end{array}$

A particle of mass $5kg$ is moving with a uniform speed $3\sqrt{2}$ in $XOY$ plane along the line $Y=X+4$. The magnitude of its angular momentum about the origin is:

  1. $40$units

  2. $60$units

  3. $0$

  4. $40\sqrt{2}$ units


Correct Option: B

A circular platform is mounted on a vertical frictionless axle. Its radius is $r=2m$ and its moment inertia is $I=200kg$ ${m}^{2}$. It is initially at rest. A $70kg$ man stands on the edge of the platform and begins to walk along the edge at speed ${v} _{0}=10{ms}^{-1}$ relative to the ground. The angular velocity of the platform is

  1. $1.2rad$ ${s}^{-1}$

  2. $0.4rad$ ${s}^{-1}$

  3. $2.0rad$ ${s}^{-1}$

  4. $7.0rad$ ${s}^{-1}$


Correct Option: D
Explanation:

As there is no external torque, thus $L$ is conserved.

Let angular velocity of the platform be $w.$
$L _i= L _f$
$0 = Iw - mv _o r$
$0 = 200 (w) - 70 (10 ) (2)$
$\implies w= 7    rad/s$

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