Tag: law of conservation of angular momentum

Questions Related to law of conservation of angular momentum

Angular momentum of a system a particles changes, when

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

  1. Force acts on a body

  2. Torque acts on a body

  3. Direction of velocity changes

  4. None of these

Show answer
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

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

  1. The angular velocity decreases

  2. The moment of inertia decreases

  3. The angular velocity stays constant

  4. The angular momentum increases

Show answer
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?

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

  1. Linear momentum

  2. K.E

  3. Angular momentum

  4. None of these

Show answer
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 ?

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

  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.

Show answer
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 

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

  1. slightly increases

  2. slightly decreases

  3. has no change

  4. none of these

Show answer
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?

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

  1. Increase

  2. Decrease

  3. Remain same

  4. none of these

Show answer
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?

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

  1. True

  2. False

Show answer
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 :

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

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

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

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

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

Show answer
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?

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

  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

Show answer
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, ${\omega}^{\prime}$?

physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

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

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

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

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

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

Show answer
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 :      $L _i = L _f$
$\therefore$   $\dfrac{1}{2}mr^2 w = \dfrac{3}{2}mr^2 w'$                     $\implies w ' =\dfrac{w}{3}$