Conservation of Angular Momentum

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The angular momentum of a particle about a point O is conserved if net torque on the particle about this point is zero i.e., $\vec{L}_O=\text{constant}$ if $\vec\tau_O=\vec{0}$.

The angular momentum of a system of particles about a point O is conserved if external torque on the system about this point is zero. Since $\vec{L}$ and $\vec{\tau}$ are vector quantities, a component of the angular momentum is conserved if the same component of external torque is zero.

The conservation of angular momentum is related to the rotational symmetry (isotropy of space).

Solved Problems from IIT JEE

Problems from IIT JEE 2003

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

  1. centre of the circle.
  2. on the circumference of the circle.
  3. inside the circle.
  4. outside the circle.

Solution: In uniform circular motion, the force on the particle passes through centre of the circle so its torque about this point is zero and angular momentum remains conserved.

Problems from IIT JEE 1985

According to Kepler's second law, the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of_________

Solution: Kepler's second law is a consequence of the conservation of angular momentum.

Problems from IIT JEE 2014

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 m/s with respect to the ground. The rotational speed of the platform in rad/s after the balls leave the platform is_______

A horizontal circular platform of radius 0.5 m

Solution: Consider the balls and the platform together as a system. There is no external torque on the system about its centre. Hence, angular momentum of the system about its centre is conserved. Initial and final angular momentum of the system are \begin{align} L_i&=0,\nonumber\\ L_f&=mvr+mvr+I\omega\\ &=2mvr+\tfrac{1}{2}MR^2\omega.\nonumber \end{align} The conservation of angular momentum, $L_i=L_f$, gives \begin{align} \omega&=-\frac{4mvr}{MR^2}\\ &=-\frac{4(0.05)(9)(0.25)}{0.45(0.5)^2}\\ &=-4\,\mathrm{rad/s}.\nonumber \end{align}

Problems from IIT JEE 2015

A ring of mass $m$ and radius $r$ is rotating with angular speed $\omega$ about a fixed vertical axis passing through its centre O with two point masses, each of mass $m/8$ 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 $8\omega/9$ and one of the masses is at a distance of $3r/5$ from O. At this instant the distance of the other mass from O is

A ring of mass m and radius r
  1. $2r/3$
  2. $r/3$
  3. $3r/5$
  4. $4r/5$

Solution: Consider the ring and the two point masses together as a system. There is no external torque on the system about the fixed point O. Thus, angular momentum of the system about the point O is conserved.

Initially, two point masses (each of the mass $m/8$) were located at O and the ring of mass $m$ and radius $r$ was rotating with angular speed $\omega$. Initial value of the angular momentum of the system about O is \begin{align} L_i&=I_\text{ring}\omega+\tfrac{m}{8} (0)^2 \omega+\tfrac{m}{8} (0)^2 \omega \\ &=mr^2 \omega. \nonumber \end{align} Finally, angular speed of the system is reduced to $8\omega/9$ and one of the point mass is at a distance $3r/5$ from O. Let another point mass be at a distance $x$ from O. Final value of the angular momentum of the system about O is \begin{align} L_f&=I_\text{ring}\tfrac{8\omega}{9}+\tfrac{m}{8} \left(\tfrac{3r}{5}\right)^{\!2} \tfrac{8\omega}{9}+\tfrac{m}{8}\, x^2 \tfrac{8\omega}{9}\nonumber\\ &=m\left(\tfrac{209}{25}r^2+x^2\right) \left(\tfrac{\omega}{9}\right). \nonumber \end{align} Apply conservation of the angular momentum, $L_i=L_f$, to get $x=4r/5$.

Problems from IIT JEE 1983

A thin circular ring of mass $M$ and radius $R$ is rotating about its axis 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

  1. $\frac{\omega M}{M+m}$
  2. $\frac{\omega (M-2m)}{M+2m}$
  3. $\frac{\omega M}{M+2m}$
  4. $\frac{\omega (M+2m)}{M}$

Solution: Consider the ring and the two objects together as a system. The objects are gently attached to the ring.

A thin circular ring of mass M and radius R

The angular momentum of the system is conserved because there is no external torque on the system. Initially, angular momentum of the system about O is \begin{align} L_i=I_i\omega_i=MR^2\,\omega, \end{align} where $I_i$ is the moment of inertia of the ring about its axis of rotation. Finally, angular momentum of the system about O is \begin{align} L_f=I_f\omega_f=\left(MR^2+2mR^2\right)\omega_f. \end{align} Conservation of angular momentum, $L_i=L_f$, gives $\omega_f=\omega M/(M+2m)$.

Problems from IIT JEE 1986

A thin uniform circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$. Another disc of the same dimensions but of mass $M/4$ is placed gently on the first disc coaxially. The angular velocity of the system now is $2\omega/\sqrt{5}$.

Solution: Consider the two discs of masses $M$ and $M/4$ and radius $R$ together as a system. There is no external torque about the axis of rotation when one disc is placed over the other coaxially. Thus, angular momentum of the system about the axis of rotation is conserved. Initial and final values of angular momenta about the axis of rotation are given by \begin{align} L_i&=I_i\omega_i \\ &=\tfrac{1}{2}MR^2\omega,\nonumber \\ L_f &=I_f\omega_f \\ &=\left(\tfrac{1}{2}MR^2+\tfrac{1}{2}\tfrac{M}{4}R^2\right)\omega_f \\ &=\tfrac{5}{8}MR^2\omega_f, \nonumber \end{align} where $I_i$ and $I_f$ are the initial and the final values of the moment of inertia about the axis of rotation. The conservation of angular momentum, $L_i=L_f$, gives $\omega_f=\frac{4}{5}\omega$.

Problems from IIT JEE 1988

A smooth uniform rod of length $L$ and mass $M$ has two identical beads of negligible size, each of mass $m$, which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with an angular velocity $\omega_0$ about an axis perpendicular to the rod and passing through the mid-point of the rod (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is____________

A smooth uniform rod of length L and mass M

Solution: Consider the rod and the two beads together as a system. There are no external torque on the system about the fixed point O (because there are no external forces). Hence, angular momentum of the system about the fixed point O is conserved. Initially, two beads of mass $m$ are at the centre of the rod of mass $M$, length $L$, and angular velocity $\omega_0$.

A smooth uniform rod of length L and mass M

Initial angular momentum of the system about the axis of rotation is \begin{align} L_0&=I_0 \omega_0 \\ &=\left[ML^2/12+m(0)^2+m(0)^2\right]\omega_0 \nonumber\\ &=ML^2\omega_0/12. \nonumber \end{align} Finally, the two beads are at the extreme ends of the rod rotating with an angular velocity $\omega$. Thus, the final angular momentum of the system about the axis of rotation is \begin{align} L&=I \omega \\ &=\left[ML^2/12+m(L/2)^2+m(L/2)^2\right]\omega\nonumber\\ &=\left[M/12+m/2\right]L^2\omega. \nonumber \end{align} Apply conservation of angular momentum about the fixed point O, $L_0=L$, to get $\omega=\frac{M\omega_0}{M+6m}$.

Problems from IIT JEE 1999

A smooth sphere A is moving on a frictionless horizontal surface with an angular speed $\omega$ and centre of mass velocity $v$. It collides elastically and head on with an identical sphere B at rest. After the collision their angular speeds are $\omega_A$ and $\omega_B$ respectively. Then,

  1. $\omega_A < \omega_B$
  2. $\omega_A = \omega_B$
  3. $\omega_A = \omega$
  4. $\omega_B =\omega$

Solution: The frictional force between the spheres is zero. The collision forces on both the sphere pass through their respective centre of masses. Thus, the torque about centre of mass is zero for each sphere. Hence, angular momentum about the centre of mass is conserved for each sphere. Initial and final angular momentum of the two spheres are \begin{align} L_\text{i,A} & = I\omega, \\ L_\text{i,B} & = 0,\nonumber\\ L_\text{f,A} & = I\omega_\text{A}, \\ L_\text{f,B} & = I\omega_\text{B}.\nonumber \end{align} The conservation of angular momentum, $L_\text{i,A}=L_\text{f,A}$ gives $\omega_\text{A}=\omega$ and $L_\text{i,B}=L_\text{f,B}$ gives $\omega_\text{B}=0$.

I encourage you to use conservation of linear momentum and conservation of energy to show that the spheres exchange their linear velocities.

Related Topics

References

  1. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
  2. 300 Solved Problems on Rotational Mechanics by Jitender Singh and Shraddhesh Chaturvedi