Concept Card: Conservation of Angular Momentum & Fundamental Applications
1. Foundational Principle:
From the rotational formulation of Newton’s Second Law for a system of particles or a rigid body:
$$\mathbf{\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}}$$
If the net external torque acting on the system about a given reference point $O$ (fixed in an inertial frame or coinciding with the centre of mass) is zero:
$$\vec{\tau}_{\text{ext}} = \vec{0} \implies \frac{d\vec{L}}{dt} = \vec{0} \implies \mathbf{\vec{L} = \text{constant (conserved)}}$$
This is the Law of Conservation of Angular Momentum: The total angular momentum vector of an isolated system (or any system experiencing zero net external torque) remains constant in both magnitude and direction over time.
2. Directional / Axis-Specific Conservation
Torque and angular momentum are three-dimensional vector quantities. Even if the net vector torque $\vec{\tau}_{\text{ext}} \neq \vec{0}$, if the torque component along a specific fixed coordinate axis (e.g., the $z$-axis) vanishes:
$$\tau_z = 0 \implies \frac{dL_z}{dt} = 0 \implies \mathbf{L_z = \text{constant}}$$
Crucial Insight for JEE: A system may experience non-zero external forces (such as normal reaction and gravity) and non-zero torques along horizontal directions ($x$ and $y$), yet its angular momentum along the vertical axis ($z$) remains strictly conserved if $\tau_z = 0$.
3. Fixed-Axis Systems with Variable Moment of Inertia
For a body or system rotating about a fixed axis with moment of inertia changing from $I_1$ to $I_2$ via internal redistribution of mass (e.g., a rotating figure skater drawing in arms, a student walking on a turntable):
$$\mathbf{I_1 \omega_1 = I_2 \omega_2 \implies \omega_2 = \left(\frac{I_1}{I_2}\right)\omega_1}$$
- Kinetic Energy Variation: Rotational kinetic energy expressed in terms of angular momentum $L$ is:
$$K = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$$ - If mass is drawn inward ($I_2 < I_1$), then $\omega_2 > \omega_1$ and:
$$K_2 = \frac{L^2}{2I_2} > \frac{L^2}{2I_1} = K_1$$
Where does the extra kinetic energy come from? It is supplied by the positive internal work done by internal forces (muscular effort or radial tension) pulling the mass inward against centrifugal tendencies: $\Delta K = W_{\text{internal}}$.
4. Angular Impulse and Collisions with Pivoted vs. Free Bodies
When a particle collides with an extended rigid body over an infinitesimal collision time interval $\Delta t \to 0$:
- Angular Impulse: $\vec{J}_{\theta} = \int \vec{\tau}_{\text{ext}}\,dt = \Delta \vec{L}$.
- Pivoted / Hinged Bodies: The hinge exerts huge, impulsive reaction forces ($\vec{F}_{\text{hinge}}$), so linear momentum is NOT conserved. However, about the pivot/hinge axis $P$, the lever arm of $\vec{F}_{\text{hinge}}$ is zero ($\vec{\tau}_{\text{hinge}} = \vec{0}$). Therefore:
$$\mathbf{\vec{L}_{P,\,\text{initial}} = \vec{L}_{P,\,\text{final}}}$$ - Free Bodies (Unconstrained on a Smooth Plane): No external horizontal forces act on the system. Therefore, both linear momentum of the system AND angular momentum about the centre of mass (CM) are simultaneously conserved:
$$\vec{P}_{\text{initial}} = \vec{P}_{\text{final}} \quad \text{and} \quad \vec{L}_{\text{cm},\,\text{initial}} = \vec{L}_{\text{cm},\,\text{final}}$$
5. Central Force Fields & Kepler’s Second Law
In any central force field where the force is directed along the radial line $\vec{r}$ (e.g., gravitational, electrostatic):
$$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times (f(r)\hat{r}) = \vec{0}$$
- Angular momentum $\vec{L} = \vec{r} \times m\vec{v}$ about the force center is strictly constant.
- Areal Velocity: The area swept by the radius vector in time $dt$ is $dA = \frac{1}{2}|\vec{r} \times d\vec{r}| = \frac{1}{2}|\vec{r} \times \vec{v}|dt = \frac{L}{2m}dt$. Hence:
$$\mathbf{\frac{dA}{dt} = \frac{L}{2m} = \text{constant}}$$
This constitutes Kepler’s Second Law of Planetary Motion.
6. Common JEE Pitfalls & Traps
- Trap 1 (Linear vs. Angular Conservation): In collisions involving a fixed hinge, students frequently attempt to apply conservation of linear momentum. Remember: Hinge forces are external impulses; linear momentum is destroyed, but angular momentum about the hinge point is preserved!
- Trap 2 (Choice of Reference Point for Rolling Obstacle Collisions): When a sphere or cylinder strikes a step or curb, choose the corner of the step as the reference origin. Normal and friction impulses pass directly through the corner, making $\vec{\tau}_{\text{corner}} = \vec{0}$, so $\vec{L}$ about that corner is conserved.
- Trap 3 (Kinetic Energy in Inelastic Angular Collisions): In plastic collisions (where a bullet embeds into a rod), mechanical energy is never conserved ($K_f < K_i$). Only angular momentum about the pivot is conserved!
Solved Examples
Example 1 (Direct Conceptual Application – Person Walking on a Rotating Turntable):
A circular horizontal platform of mass $M = 80.0\text{ kg}$ and radius $R = 2.0\text{ m}$ is free to rotate without friction about a vertical axis through its center ($I_{\text{disc}} = \frac{1}{2}M R^2$). A person of mass $m = 40.0\text{ kg}$ stands at the outer rim while the platform rotates at an initial angular speed $\omega_0 = 1.0\text{ rad/s}$. The person then walks slowly from the rim to the exact center of the platform.
(a) Find the initial moment of inertia $I_1$ and final moment of inertia $I_2$ of the system.
(b) Determine the final angular velocity $\omega_f$ of the platform.
(c) Calculate the initial and final kinetic energies of the system and explain the physical origin of the difference.
Solution:
(a) Platform inertia: $I_{\text{disc}} = \frac{1}{2}M R^2 = \frac{1}{2}(80.0)(2.0)^2 = 160.0\text{ kg}\cdot\text{m}^2$.
– Initially, the person is at the rim ($r_1 = R = 2.0\text{ m}$):
$$I_{\text{person}, 1} = m R^2 = (40.0)(4.0) = 160.0\text{ kg}\cdot\text{m}^2$$
$$I_1 = I_{\text{disc}} + I_{\text{person}, 1} = 160.0 + 160.0 = \mathbf{320.0\text{ kg}\cdot\text{m}^2}$$
– Finally, the person stands at the center ($r_2 = 0$):
$$I_{\text{person}, 2} = m(0)^2 = 0 \implies I_2 = I_{\text{disc}} = \mathbf{160.0\text{ kg}\cdot\text{m}^2}$$
(b) Since the vertical axle exerts no vertical torque ($\tau_z = 0$), angular momentum about the vertical axis is conserved:
$$I_1 \omega_0 = I_2 \omega_f \implies \omega_f = \left(\frac{I_1}{I_2}\right)\omega_0 = \left(\frac{320.0}{160.0}\right)(1.0) = \mathbf{2.0\text{ rad/s}}$$
(c) Initial kinetic energy:
$$K_1 = \frac{1}{2}I_1 \omega_0^2 = \frac{1}{2}(320.0)(1.0)^2 = 160.0\text{ J}$$
Final kinetic energy:
$$K_2 = \frac{1}{2}I_2 \omega_f^2 = \frac{1}{2}(160.0)(2.0)^2 = 320.0\text{ J}$$
Change in kinetic energy: $\Delta K = K_2 – K_1 = 320.0 – 160.0 = \mathbf{+160.0\text{ J}}$.
Concluding takeaway: The $160.0\text{ J}$ increase in rotational kinetic energy is directly provided by the internal muscular work performed by the person pulling their body inward against the outward centrifugal inertial force.
Example 2 (Mathematical Derivation – Bullet Striking a Pivoted Thin Rod):
A uniform thin rod of mass $M$ and length $L$ hangs vertically from a frictionless horizontal pivot at its upper end $O$ ($I_O = \frac{1}{3}M L^2$). A bullet of mass $m$ travelling horizontally with velocity $v_0$ strikes the bottom tip of the rod and embeds itself into the rod instantaneously.
(a) Derive the angular velocity $\omega$ of the rod-bullet system immediately after collision.
(b) Determine the fractional loss in mechanical kinetic energy during the collision.
(c) Find the minimum speed $v_0$ required for the rod to swing through a complete vertical arc of $180^\circ$ to reach the upright position.
Solution:
(a) During the collision, the pivot at $O$ exerts an impulsive reaction force, so linear momentum is not conserved. However, the torque of the pivot reaction about $O$ is zero.
Initial angular momentum about pivot $O$:
$$L_i = m v_0 L$$
Total moment of inertia of the combined system about $O$ after embedding:
$$I_{\text{sys}} = I_{\text{rod}} + I_{\text{bullet}} = \frac{1}{3}M L^2 + m L^2 = \left(\frac{M + 3m}{3}\right)L^2$$
Equating $L_i = L_f = I_{\text{sys}}\omega$:
$$m v_0 L = \left(\frac{M + 3m}{3}\right)L^2 \omega \implies \mathbf{\omega = \frac{3m v_0}{(M + 3m)L}}$$
(b) Initial kinetic energy: $K_i = \frac{1}{2}m v_0^2$.
Kinetic energy immediately after impact:
$$K_f = \frac{1}{2}I_{\text{sys}}\omega^2 = \frac{L_i^2}{2I_{\text{sys}}} = \frac{(m v_0 L)^2}{2 \left(\frac{M + 3m}{3}\right)L^2} = \left(\frac{3m}{M + 3m}\right) K_i$$
Fractional loss in kinetic energy:
$$\mathbf{\frac{\Delta K_{\text{loss}}}{K_i} = 1 – \frac{K_f}{K_i} = 1 – \frac{3m}{M + 3m} = \frac{M}{M + 3m}}$$
(c) For the rod to reach the vertical inverted position, mechanical energy after collision is conserved.
Height gained by the rod’s center of mass: $\Delta h_{\text{rod}} = 2(L/2) = L$.
Height gained by the bullet: $\Delta h_{\text{bullet}} = 2L$.
Total potential energy gained at top:
$$\Delta U = M g L + m g (2L) = (M + 2m)g L$$
Setting $K_f \ge \Delta U$:
$$\frac{3m^2 v_0^2}{2(M + 3m)} \ge (M + 2m)g L \implies \mathbf{v_0 \ge \sqrt{\frac{2(M + 3m)(M + 2m)g L}{3m^2}}}$$
Example 3 (Standard JEE Advanced Scenario – Free Rod Struck by a Particle on a Smooth Horizontal Table):
A uniform rod $AB$ of mass $M = 2.0\text{ kg}$ and length $L = 1.2\text{ m}$ lies at rest on a frictionless horizontal table. A small putty ball of mass $m = 1.0\text{ kg}$ moving horizontally with velocity $v_0 = 12.0\text{ m/s}$ perpendicular to the rod strikes end $A$ (distance $L/2 = 0.60\text{ m}$ from the rod’s geometric center) and sticks to it.
(a) Find the velocity $v_{\text{cm}}$ of the centre of mass of the combined (rod + ball) system after impact.
(b) Locate the position of the new centre of mass $C’$ and calculate the moment of inertia $I_{C’}$ about $C’$.
(c) Determine the angular velocity $\omega$ of the system about $C’$ immediately after the collision.
(d) Find the instantaneous velocity of end $B$ immediately after impact.
Solution:
(a) No external horizontal forces act on the smooth table $\implies$ Linear momentum is conserved:
$$(M + m) v_{\text{cm}} = m v_0 \implies v_{\text{cm}} = \frac{m v_0}{M + m} = \frac{(1.0)(12.0)}{2.0 + 1.0} = \mathbf{4.0\text{ m/s}}$$
(b) Center of mass position: Measured from the rod’s geometric center $C$ towards end $A$:
$$y_{\text{cm}} = \frac{m(L/2) + M(0)}{M + m} = \frac{(1.0)(0.60)}{3.0} = 0.20\text{ m}$$
Thus, the new centre of mass $C’$ is $0.20\text{ m}$ from the rod center, and $d_{\text{ball}} = 0.60 – 0.20 = 0.40\text{ m}$ from the ball.
By the parallel axis theorem, the moment of inertia of the rod about $C’$ is:
$$I_{\text{rod}, C’} = I_{\text{rod}, C} + M(y_{\text{cm}})^2 = \frac{1}{12}M L^2 + M(0.20)^2 = \frac{1}{12}(2.0)(1.2)^2 + 2.0(0.04) = 0.24 + 0.08 = 0.32\text{ kg}\cdot\text{m}^2$$
Moment of inertia of the stuck ball about $C’$:
$$I_{\text{ball}, C’} = m(d_{\text{ball}})^2 = 1.0(0.40)^2 = 0.16\text{ kg}\cdot\text{m}^2$$
Total moment of inertia about $C’$:
$$I_{C’} = 0.32 + 0.16 = \mathbf{0.48\text{ kg}\cdot\text{m}^2}$$
(c) Initial angular momentum about the system’s centre of mass $C’$:
The line of motion of the ball passes at a perpendicular distance $d_{\text{ball}} = 0.40\text{ m}$ from $C’$:
$$L_{C’, i} = m v_0 d_{\text{ball}} = (1.0)(12.0)(0.40) = 4.80\text{ J}\cdot\text{s}$$
By conservation of angular momentum about $C’$:
$$L_{C’, i} = I_{C’}\omega \implies \mathbf{\omega = \frac{4.80}{0.48} = 10.0\text{ rad/s}}$$
(d) End $B$ is located at distance $r_B = 0.60 + 0.20 = 0.80\text{ m}$ from $C’$ on the opposite side.
Velocity of end $B$:
$$v_B = v_{\text{cm}} – \omega r_B = 4.0 – (10.0)(0.80) = 4.0 – 8.0 = \mathbf{-4.0\text{ m/s}}$$
End $B$ moves backwards at $4.0\text{ m/s}$ in the opposite direction to $v_0$.
Example 4 (Edge Case – Planetary Orbit & Kepler’s Second Law):
A satellite of mass $m$ is in an elliptical orbit around Earth of mass $M_E$. The perigee distance is $r_p = 7.0 \times 10^6\text{ m}$ with orbital speed $v_p = 9.0\text{ km/s}$. The apogee distance is $r_a = 21.0 \times 10^6\text{ m}$.
(a) Calculate the orbital speed $v_a$ at apogee.
(b) Explain why angular momentum is conserved about the center of Earth but NOT about the empty focus of the orbital ellipse.
(c) Find the ratio of the areal velocity at perigee to that at apogee.
Solution:
(a) At both perigee and apogee, the velocity vector is strictly perpendicular to the position vector from the center of Earth ($\vec{r} \perp \vec{v}$).
The gravitational force is a central force directed toward the Earth’s center, producing zero external torque: $\vec{\tau}_{\text{Earth}} = \vec{0}$.
$$L = m r_p v_p = m r_a v_a \implies v_a = \left(\frac{r_p}{r_a}\right)v_p = \left(\frac{7.0 \times 10^6}{21.0 \times 10^6}\right)(9.0\text{ km/s}) = \mathbf{3.0\text{ km/s}}$$
(b) Earth is located at one focus of the ellipse. Gravitational force $\vec{F}_g$ always acts along the line joining the satellite to the Earth’s center, so $\vec{r}_{\text{Earth}} \times \vec{F}_g = \vec{0}$.
However, with respect to the empty focus $F’$, the position vector $\vec{r}_{F’}$ does not lie along the line of gravitational action. Hence $\vec{\tau}_{F’} = \vec{r}_{F’} \times \vec{F}_g \neq \vec{0}$. This non-zero torque causes the angular momentum about the empty focus to vary periodically.
(c) The areal velocity is related to angular momentum by $\frac{dA}{dt} = \frac{L}{2m}$. Since $L$ is constant at every point in the orbit:
$$\mathbf{\frac{(dA/dt)_{\text{perigee}}}{(dA/dt)_{\\text{apogee}}} = 1.0}$$
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A figure skater spinning on frictionless ice with outstretched arms has a moment of inertia $I_0$ and angular speed $\omega_0$. When she pulls her arms in close to her body, her moment of inertia reduces to $I_0/3$. The ratio of her final rotational kinetic energy to her initial kinetic energy is:
(A) $1 : 3$
(B) $1 : 1$
(C) $3 : 1$
(D) $9 : 1$
Problem 2 (JEE Main – Single Correct):
A thin uniform circular ring of mass $M$ and radius $R$ rotates about its central vertical axis perpendicular to its plane with angular velocity $\omega$. Two small particles, each of mass $m$, are gently placed without impulse onto diametrically opposite points of the ring. The new angular velocity of the ring is:
(A) $\frac{M}{M + m}\omega$
(B) $\frac{M}{M + 2m}\omega$
(C) $\frac{M – 2m}{M + 2m}\omega$
(D) $\frac{M + 2m}{M}\omega$
Problem 3 (JEE Main – Single Correct):
A horizontal turntable of mass $M$ and radius $R$ rotates freely about a vertical axle passing through its center. A small insect of mass $m$ crawls radially outward from the center to the rim. During this process:
(A) The angular velocity of the turntable increases.
(B) The angular velocity of the turntable remains constant.
(C) The angular velocity of the turntable decreases.
(D) The total mechanical kinetic energy of the system increases.
Problem 4 (JEE Main – Single Correct):
Kepler’s second law, which states that the areal velocity of a planet revolving around the Sun is constant, is a direct consequence of the conservation of:
(A) Linear momentum
(B) Angular momentum
(C) Total mechanical energy
(D) Gravitational potential energy
Problem 5 (JEE Main – Single Correct):
A horizontal circular disc is rotating with angular speed $\omega$ about a vertical frictionless axis through its center. Another identical disc, initially stationary, is dropped coaxially onto the first disc. Frictional coupling causes slipping to cease after a short interval. The fraction of initial mechanical kinetic energy dissipated as heat is:
(A) $1/4$
(B) $1/3$
(C) $1/2$
(D) $2/3$
Problem 6 (JEE Advanced – One or More Correct):
A uniform rod of mass $M$ and length $L$ lies on a smooth horizontal table. A small puck of mass $m$ moving with speed $v_0$ perpendicular to the rod collides elastically with one end of the rod. Which of the following statements is/are correct?
(A) Total linear momentum of the (rod + puck) system is conserved.
(B) Total angular momentum of the system is conserved about any arbitrary fixed point on the table.
(C) Total angular momentum of the system is conserved about the center of mass of the rod.
(D) Total mechanical kinetic energy is conserved in the collision.
Problem 7 (JEE Advanced – One or More Correct):
A person stands on a rotating platform holding heavy dumbbells in outstretched hands. When the person pulls the dumbbells inward towards the chest:
(A) The total angular momentum of the (person + platform + dumbbells) system remains constant.
(B) The angular velocity of the platform increases.
(C) The rotational kinetic energy of the system increases.
(D) The increase in kinetic energy is equal to the mechanical work performed by the person’s muscles.
Problem 8 (JEE Advanced – One or More Correct):
A planet moves in an elliptical orbit under an inverse-square attractive central gravitational force field centered at focus $O$ (the Sun):
(A) The angular momentum vector $\vec{L}$ about focus $O$ is strictly invariant in time.
(B) The speed of the planet is maximum when its distance from $O$ is minimum.
(C) The areal velocity $\frac{dA}{dt}$ is constant throughout the orbit.
(D) The linear momentum vector $\vec{p}$ is conserved throughout the orbit.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A uniform disc of mass $M = 2.0\text{ kg}$ and radius $R = 0.50\text{ m}$ rotates freely about its central fixed vertical axis at an angular speed of $\omega_1 = 30.0\text{ rad/s}$. A small wax blob of mass $m = 1.0\text{ kg}$ is dropped gently onto the disc at a distance $r = 0.50\text{ m}$ (at the outer rim) and adheres firmly. Calculate the new angular speed of the system $\omega_2$ in $\text{rad/s}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform thin rod of mass $M = 6.0\text{ kg}$ and length $L = 2.0\text{ m}$ is pivoted smoothly at its top end $O$. A horizontal bullet of mass $m = 0.50\text{ kg}$ moving at speed $v_0 = 80.0\text{ m/s}$ strikes the rod at its lower end and gets embedded. Calculate the angular velocity $\omega$ of the rod immediately after impact in $\text{rad/s}$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (B) | 3. (C) | 4. (B) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 15 | 10. 8
Solution 1:
Angular momentum is conserved: $L = I_0 \omega_0 = I_f \omega_f \implies \omega_f = \frac{I_0}{I_f}\omega_0 = 3\omega_0$.
Kinetic energy: $K = \frac{L^2}{2I}$.
$$\frac{K_f}{K_i} = \frac{I_0}{I_f} = \frac{I_0}{I_0/3} = 3$$
Thus, $K_f : K_i = 3 : 1$.
Correct Option: (C)
Solution 2:
Initial moment of inertia: $I_i = M R^2$.
When two masses $m$ are placed at opposite ends at distance $R$ from center:
$I_f = M R^2 + m R^2 + m R^2 = (M + 2m)R^2$.
Conserving angular momentum: $I_i \omega = I_f \omega’ \implies (M R^2)\omega = (M + 2m)R^2 \omega’$.
$$\omega’ = \frac{M}{M + 2m}\omega$$
Correct Option: (B)
Solution 3:
As the insect crawls from $r = 0$ to $r = R$, the system’s moment of inertia $I(r) = \frac{1}{2}M R^2 + m r^2$ increases monotonically.
Since no external torque acts about the vertical axle, $L = I(r)\omega(r) = \text{constant}$.
As $I$ increases, $\omega$ must decrease.
Correct Option: (C)
Solution 4:
The central gravitational force passes through the Sun, exerting zero torque about the Sun: $\vec{\tau} = \vec{0} \implies \vec{L} = \text{constant}$.
Since areal velocity is $\frac{dA}{dt} = \frac{L}{2m}$, constancy of areal velocity is directly due to the conservation of angular momentum.
Correct Option: (B)
Solution 5:
Each disc has moment of inertia $I$. Initial angular momentum $L = I\omega$.
After dropping, the combined moment of inertia is $I_f = 2I$.
Final angular velocity: $\omega_f = \frac{I\omega}{2I} = \frac{\omega}{2}$.
Initial kinetic energy: $K_i = \frac{1}{2}I\omega^2$.
Final kinetic energy: $K_f = \frac{1}{2}(2I)\left(\frac{\omega}{2}\right)^2 = \frac{1}{4}I\omega^2 = \frac{1}{2}K_i$.
Fractional energy dissipated: $\frac{\Delta K}{K_i} = 1 – \frac{1}{2} = \frac{1}{2}$.
Correct Option: (C)
Solution 6:
– (A) True: No external horizontal forces act on the smooth table $\implies \vec{P}$ is conserved.
– (B) True: Zero net external force and zero external torque about any fixed point on the plane $\implies \vec{L}$ is conserved about any fixed point.
– (C) True: Angular momentum about the center of mass is conserved.
– (D) True: The collision is specified as elastic $\implies$ total kinetic energy is conserved.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: Frictionless platform has zero external vertical torque $\implies L$ is conserved.
– (B) True: $I$ decreases as masses move inward $\implies \omega$ increases.
– (C) True: $K = \frac{L^2}{2I}$ increases as $I$ decreases.
– (D) True: By the work-energy theorem, $\Delta K = W_{\text{internal, muscular}}$.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: Gravitational force is central about focus $O \implies \vec{\tau}_O = \vec{0} \implies \vec{L}_O = \text{constant}$.
– (B) True: $m v r = L = \text{constant} \implies v \propto 1/r$. Minimum distance $r$ (perihelion) has maximum speed.
– (C) True: $\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$.
– (D) False: Gravitational force is a non-zero external force on the planet, continually altering the direction and magnitude of its linear momentum $\vec{p}$.
Correct Options: (A, B, C)
Solution 9:
Disc inertia: $I_{\text{disc}} = \frac{1}{2}M R^2 = \frac{1}{2}(2.0)(0.50)^2 = 0.25\text{ kg}\cdot\text{m}^2$.
Initial angular momentum: $L_i = I_{\text{disc}}\omega_1 = (0.25)(30.0) = 7.50\text{ J}\cdot\text{s}$.
Blob inertia at rim: $I_{\text{blob}} = m R^2 = (1.0)(0.50)^2 = 0.25\text{ kg}\cdot\text{m}^2$.
Total final inertia: $I_f = I_{\text{disc}} + I_{\text{blob}} = 0.25 + 0.25 = 0.50\text{ kg}\cdot\text{m}^2$.
Final angular speed: $\omega_2 = \frac{L_i}{I_f} = \frac{7.50}{0.50} = 15.0\text{ rad/s}$.
Correct Answer: 15
Solution 10:
Initial angular momentum about pivot $O$ is due solely to the bullet:
$L_i = m v_0 L = (0.50)(80.0)(2.0) = 80.0\text{ J}\cdot\text{s}$.
Total moment of inertia about pivot $O$ after embedding:
$I_{\text{sys}} = I_{\text{rod}} + I_{\text{bullet}} = \frac{1}{3}M L^2 + m L^2 = \frac{1}{3}(6.0)(2.0)^2 + (0.50)(2.0)^2 = 8.0 + 2.0 = 10.0\text{ kg}\cdot\text{m}^2$.
Conservation of angular momentum about pivot $O$:
$L_i = I_{\text{sys}}\omega \implies 80.0 = 10.0 \omega \implies \omega = 8.0\text{ rad/s}$.
Correct Answer: 8