Rigid Body Rotation: Pure Rotation About Fixed Axis & Rotational Kinetic Energy

Concept Card: Rigid Body Rotation – Pure Rotation About a Fixed Axis and Rotational Kinetic Energy

1. Kinematics of Pure Rotation About a Fixed Axis:
A rigid body is said to execute pure rotation about a given fixed axis if every constituent particle of the body moves along a circular path whose center lies on that rotation axis, and whose plane is strictly perpendicular to the rotation axis.

Universal Angular Kinematic Variables: At any given instant, every particle in the rigid body undergoes the identical angular displacement ($\theta$), rotates with the same instantaneous angular velocity ($\vec{\omega}$), and experiences the same instantaneous angular acceleration ($\vec{\alpha}$).
Vector Characteristics: Both $\vec{\omega}$ and $\vec{\alpha}$ are axial vectors directed along the fixed axis of rotation according to the right-hand screw rule:
$$\vec{\omega} = \frac{d\theta}{dt} \hat{k}, \quad \vec{\alpha} = \frac{d\vec{\omega}}{dt} = \frac{d^2\theta}{dt^2} \hat{k}$$
Particle-Level Linear Kinematics: For a particle located at a perpendicular distance $r_i$ from the rotation axis:
- Linear velocity (tangential): $v_i = \omega r_i$, or in full vector notation:
  $$\vec{v}_i = \vec{\omega} \times \vec{r}_i$$
- Tangential acceleration (responsible for changing speed):
  $$a_{t,i} = \alpha r_i$$
- Centripetal / Radial acceleration (responsible for changing direction toward the axis):
  $$a_{c,i} = \frac{v_i^2}{r_i} = \omega^2 r_i$$
- Total linear acceleration of particle $i$:
  $$\vec{a}_i = \vec{a}_{t,i} + \vec{a}_{c,i} \implies a_i = \sqrt{a_{t,i}^2 + a_{c,i}^2} = r_i \sqrt{\alpha^2 + \omega^4}$$
- The angle $\phi$ made by the total acceleration vector with the inward radial direction is given by:
  $$\tan\phi = \frac{a_t}{a_c} = \frac{\alpha r_i}{\omega^2 r_i} = \frac{\alpha}{\omega^2}$$

2. Rigorous Derivation of Rotational Kinetic Energy

Consider an extended rigid body rotating with angular velocity $\omega$ about a fixed axis $Z$. Let the body consist of $N$ discrete particles of masses $m_1, m_2, \dots, m_N$ situated at perpendicular distances $r_1, r_2, \dots, r_N$ from the axis.

The linear kinetic energy of the $i$-th particle is:
$$K_i = \frac{1}{2} m_i v_i^2 = \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2} m_i r_i^2 \omega^2$$
The total kinetic energy of the rotating rigid body is the scalar sum of the kinetic energies of all its individual particles:
$$K_{\text{rot}} = \sum_{i=1}^N K_i = \sum_{i=1}^N \frac{1}{2} m_i r_i^2 \omega^2 = \frac{1}{2} \left( \sum_{i=1}^N m_i r_i^2 \right) \omega^2$$
For a continuous mass distribution, the summation transitions into a definite volume integral:
$$I = \int r^2 dm$$
where $I$ is the Moment of Inertia of the rigid body about the specified axis of rotation.

Governing Expression for Rotational Kinetic Energy:
$$\mathbf{K_{\text{rot}} = \frac{1}{2} I \omega^2}$$
Expressed in terms of the angular momentum $L = I \omega$ about the fixed axis:
$$\mathbf{K_{\text{rot}} = \frac{L^2}{2I} = \frac{1}{2} L \omega}$$

3. Structural Analogy Between Linear and Rotational Dynamics

The mathematical architecture of pure rotational mechanics mirrors translational mechanics via direct one-to-one correspondence:

Translational Parameter	Rotational Parameter	Connecting Relationship
Displacement $x$	Angular displacement $\theta$	$s = r \theta$
Linear velocity $v = \frac{dx}{dt}$	Angular velocity $\omega = \frac{d\theta}{dt}$	$v = r \omega$
Mass (Inertia) $M$	Moment of Inertia $I$	$I = \int r^2 dm$
Force $F = M a$	Torque $\tau = I \alpha$	$\vec{\tau} = \vec{r} \times \vec{F}$
Linear Momentum $p = M v$	Angular Momentum $L = I \omega$	$\vec{L} = \vec{r} \times \vec{p}$
Kinetic Energy $K_{\text{trans}} = \frac{1}{2} M v^2 = \frac{p^2}{2M}$	Rotational Energy $K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$	$K_{\text{rot}} = \frac{1}{2} L \omega$
Work Done $W = \int F dx$	Work Done $W = \int \tau d\theta$	$dW = \vec{\tau} \cdot d\vec{\theta}$
Power $P = \vec{F} \cdot \vec{v}$	Power $P = \vec{\tau} \cdot \vec{\omega}$	$P = \tau \omega$

4. Work-Energy Theorem and Conservation of Energy in Pure Rotation

Rotational Work-Energy Theorem: The net work done by all external torques acting on a rigid body rotating about a fixed axis is equal to the net change in its rotational kinetic energy:
$$W_{\text{net}} = \int_{\theta_i}^{\theta_f} \tau_{\text{ext}} d\theta = \int_{\omega_i}^{\omega_f} (I \alpha) d\theta = \int_{\omega_i}^{\omega_f} I \left(\omega \frac{d\omega}{d\theta}\right) d\theta = I \int_{\omega_i}^{\omega_f} \omega d\omega = \mathbf{\frac{1}{2} I \omega_f^2 – \frac{1}{2} I \omega_i^2}$$
Conservation of Mechanical Energy: In the presence of strictly conservative forces (such as gravity or ideal springs):
$$K_{\text{rot}, 1} + U_1 = K_{\text{rot}, 2} + U_2 = \text{constant}$$
For extended swinging bodies (compound pendulums), the gravitational potential energy is evaluated strictly by treating the entire mass $M$ as concentrated at the centre of mass:
$$U = M g y_{cm}$$

5. Common JEE Pitfalls & Strategic Traps

Trap 1 (Double-Counting Kinetic Energy): In pure rotation about a fixed axis $A$ (which is NOT the center of mass), the entire kinetic energy is given simply by:
$$K = \frac{1}{2} I_A \omega^2$$
By the Parallel Axis Theorem ($I_A = I_{cm} + M d^2$):
$$K = \frac{1}{2} (I_{cm} + M d^2) \omega^2 = \frac{1}{2} I_{cm} \omega^2 + \frac{1}{2} M (d \omega)^2 = \frac{1}{2} I_{cm} \omega^2 + \frac{1}{2} M v_{cm}^2$$
Warning: Students frequently commit the catastrophic error of writing $K = \frac{1}{2} I_A \omega^2 + \frac{1}{2} M v_{cm}^2$, effectively double-counting the translational contribution of the centre of mass!
Trap 2 (Zero Angular Acceleration $\ne$ Zero Linear Acceleration): Even if a body spins at a perfectly constant angular velocity ($\alpha = 0$), every particle (except those exactly on the axis) undergoes a non-zero centripetal acceleration $a_c = \omega^2 r$. Do not confuse $\alpha = 0$ with $\vec{a}_i = 0$!
Trap 3 (Work of Internal Torques): For an ideal rigid body, the internal forces between particles act along their lines of centers and their relative distances are rigidly fixed ($dr_{ij} = 0$). Consequently, the work done by all internal forces/torques in a rigid body is identically zero.

Solved Examples

Example 1 (Direct Conceptual Application – Flywheel Dynamics, Power & Work-Energy):
A heavy uniform circular flywheel of mass $M = 10.0\text{ kg}$ and radius $R = 0.40\text{ m}$ is mounted on a frictionless horizontal axle passing through its center. Starting from rest at $t = 0$, a constant driving torque $\tau$ accelerates the flywheel to an operating angular speed of $600\text{ rpm}$ in a time interval of $t = 10.0\text{ s}$.
(a) Determine the moment of inertia $I$, the operating angular velocity $\omega$, and the constant angular acceleration $\alpha$.
(b) Calculate the required constant torque $\tau$.
(c) Calculate the rotational kinetic energy $K_{\text{rot}}$ acquired by the flywheel at $600\text{ rpm}$.
(d) Calculate the total angular displacement $\theta$ and verify the Rotational Work-Energy Theorem.

Solution:
(a) Moment of Inertia and Kinematic Quantities:
For a uniform circular disc / flywheel about its central symmetry axis:
$$I = \frac{1}{2} M R^2 = \frac{1}{2} (10.0\text{ kg}) (0.40\text{ m})^2 = \frac{1}{2} (10.0)(0.16) = \mathbf{0.80\text{ kg}\cdot\text{m}^2}$$
Operating angular speed:
$$\omega = 600\text{ rpm} = 600 \times \frac{2\pi\text{ rad}}{60\text{ s}} = \mathbf{20\pi\text{ rad/s}} \approx 62.83\text{ rad/s}$$
Angular acceleration from rest ($\omega_0 = 0$):
$$\alpha = \frac{\omega – \omega_0}{t} = \frac{20\pi\text{ rad/s}}{10.0\text{ s}} = \mathbf{2\pi\text{ rad/s}^2} \approx 6.283\text{ rad/s}^2$$

(b) Driving Torque $\tau$:
$$\tau = I \alpha = (0.80\text{ kg}\cdot\text{m}^2)(2\pi\text{ rad/s}^2) = \mathbf{1.6\pi\text{ N}\cdot\text{m}} \approx 5.027\text{ N}\cdot\text{m}$$

(c) Rotational Kinetic Energy:
$$K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} (0.80\text{ kg}\cdot\text{m}^2)(20\pi\text{ rad/s})^2 = 0.40 \times 400\pi^2 = \mathbf{160\pi^2\text{ Joules}} \approx 1579.14\text{ J}$$

(d) Work-Energy Theorem Verification:
Total angular displacement under constant $\alpha$:
$$\theta = \frac{\omega_0 + \omega}{2} t = \left(\frac{0 + 20\pi}{2}\right) (10.0) = 100\pi\text{ radians}$$
Work done by the driving torque:
$$W = \tau \cdot \theta = (1.6\pi\text{ N}\cdot\text{m})(100\pi\text{ rad}) = \mathbf{160\pi^2\text{ Joules}}$$
Notice that $W = \Delta K_{\text{rot}} = 160\pi^2\text{ J}$, verifying the Rotational Work-Energy Theorem.

Takeaway: When torque is constant, rotational work is simply $\tau \theta$, directly equal to $\frac{1}{2} I \omega_f^2$.

Example 2 (Mathematical Formulation – Compound Pendulum: Rod Released from Horizontal):
A uniform thin rigid rod of mass $M = 3.0\text{ kg}$ and length $L = 2.0\text{ m}$ is freely pivoted at its upper end $O$ to a horizontal frictionless axle. The rod is held horizontally and released from rest at $t = 0$. (Take $g = 9.8\text{ m/s}^2$).
(a) Calculate the instantaneous angular acceleration $\alpha_0$ and the linear acceleration of the tip at the instant of release.
(b) Using the conservation of mechanical energy, determine the angular speed $\omega$ of the rod as it passes through the vertical position.
(c) Determine the linear speed of the centre of mass ($v_{cm}$) and the bottom tip ($v_{\text{tip}}$) at the lowest position.
(d) Calculate the upward reaction force exerted by the pivot on the rod as it passes the vertical position.

Solution:
(a) Instant of Release (Horizontal Orientation):
Moment of inertia about the end pivot $O$:
$$I_O = \frac{1}{3} M L^2 = \frac{1}{3} (3.0\text{ kg}) (2.0\text{ m})^2 = 4.0\text{ kg}\cdot\text{m}^2$$
At the horizontal position, the weight $M g$ acts at the centre of mass (distance $L/2 = 1.0\text{ m}$ from $O$):
$$\tau_O = M g \left(\frac{L}{2}\right) = (3.0)(9.8)(1.0) = 29.4\text{ N}\cdot\text{m}$$
Instantaneous angular acceleration:
$$\alpha_0 = \frac{\tau_O}{I_O} = \frac{M g (L/2)}{\frac{1}{3} M L^2} = \frac{3g}{2L} = \frac{3(9.8)}{2(2.0)} = \frac{29.4}{4.0} = \mathbf{7.35\text{ rad/s}^2}$$
Downward tangential acceleration of the free tip:
$$a_{\text{tip}} = \alpha_0 L = \left(\frac{3g}{2L}\right) L = \frac{3}{2} g = 1.5 \times 9.8 = \mathbf{14.7\text{ m/s}^2} > g!$$
Remarkable Result: The free tip accelerates downward faster than free fall ($a_{\text{tip}} > g$) due to internal stresses within the rod!

(b) Angular Velocity at the Vertical Position:
Between horizontal and vertical positions, the centre of mass descends by a vertical height $h = L/2 = 1.0\text{ m}$.
Loss in gravitational potential energy:
$$\Delta U = -M g \left(\frac{L}{2}\right)$$
Gain in rotational kinetic energy:
$$\Delta K_{\text{rot}} = \frac{1}{2} I_O \omega^2 = \frac{1}{2} \left(\frac{1}{3} M L^2\right) \omega^2 = \frac{1}{6} M L^2 \omega^2$$
By conservation of mechanical energy ($\Delta K + \Delta U = 0$):
$$\frac{1}{6} M L^2 \omega^2 = M g \frac{L}{2} \implies \omega^2 = \frac{3g}{L}$$
$$\omega = \sqrt{\frac{3g}{L}} = \sqrt{\frac{3(9.8)}{2.0}} = \sqrt{14.7} \approx \mathbf{3.834\text{ rad/s}}$$

(c) Linear Velocities at the Lowest Position:
$$v_{cm} = \omega \left(\frac{L}{2}\right) = \sqrt{\frac{3g}{L}} \frac{L}{2} = \frac{1}{2}\sqrt{3gL} = \frac{1}{2}\sqrt{3(9.8)(2.0)} = \frac{1}{2}\sqrt{58.8} \approx \mathbf{3.834\text{ m/s}}$$
$$v_{\text{tip}} = \omega L = \sqrt{3gL} = \sqrt{58.8} \approx \mathbf{7.668\text{ m/s}}$$

(d) Pivot Reaction Force at Lowest Position:
At the vertical position, the rod has zero angular acceleration ($\tau_O = 0 \implies \alpha = 0$).
However, the centre of mass undergoes purely upward centripetal acceleration:
$$a_{c,cm} = \omega^2 r_{cm} = \left(\frac{3g}{L}\right) \left(\frac{L}{2}\right) = \frac{3}{2} g$$
Let $N$ be the upward vertical force exerted by the pivot on the rod. Newton’s second law for the $CM$ motion gives:
$$N – M g = M a_{c,cm} = M \left(\frac{3}{2} g\right)$$
$$N = M g + \frac{3}{2} M g = \mathbf{\frac{5}{2} M g} = 2.5 \times (3.0\text{ kg})(9.8\text{ m/s}^2) = \mathbf{73.5\text{ N}}$$

Takeaway: At the bottom of the swing, the pivot must support not only the rod’s static weight ($M g$), but also the dynamic centripetal force required to curve the center of mass trajectory ($\frac{3}{2} M g$), totaling $\frac{5}{2} M g$.

Example 3 (Multi-Concept Linkage – Standard JEE Advanced Scenario: Pulley-Mass System with Rotational Inertia):
A bucket of mass $m = 4.0\text{ kg}$ is attached to a light, flexible, inextensible cord wound around a solid cylindrical pulley of mass $M = 6.0\text{ kg}$ and radius $R = 0.20\text{ m}$. The pulley is mounted on a frictionless horizontal axle. The cord does not slip on the pulley. The bucket is released from rest and descends through a vertical distance $h = 5.0\text{ m}$. (Take $g = 9.8\text{ m/s}^2$).
(a) Apply Newton’s dynamical laws to find the downward linear acceleration $a$ of the bucket and the tension $T$ in the cord.
(b) Use the Work-Energy Theorem to determine the speed $v$ of the bucket and the angular speed $\omega$ of the pulley after descending height $h$.
(c) Calculate the fraction of potential energy converted into translational kinetic energy of the bucket vs. rotational kinetic energy of the pulley.

Solution:
(a) Dynamical Equations of Motion:
For the descending bucket of mass $m$:
$$m g – T = m a \quad \implies \quad T = m(g – a) \quad \text{— (Equation 1)}$$
For the solid cylinder pulley ($I = \frac{1}{2} M R^2$):
$$\tau = T \cdot R = I \alpha$$
Because the cord does not slip on the rim, the tangential acceleration of the rim equals the linear acceleration of the cord: $a = \alpha R \implies \alpha = a / R$.
$$T R = \left(\frac{1}{2} M R^2\right) \left(\frac{a}{R}\right) \implies T = \frac{1}{2} M a \quad \text{— (Equation 2)}$$
Equating Equation 1 and Equation 2:
$$m(g – a) = \frac{1}{2} M a \implies m g = \left(m + \frac{M}{2}\right) a$$
$$\mathbf{a = \frac{m g}{m + \frac{M}{2}} = \frac{2 m g}{2m + M}}$$
Substitute values ($m = 4.0\text{ kg}$, $M = 6.0\text{ kg}$):
$$a = \frac{4.0 \times 9.8}{4.0 + 3.0} = \frac{39.2}{7.0} = \mathbf{5.60\text{ m/s}^2}$$
From Equation 2, the cord tension is:
$$T = \frac{1}{2} M a = \frac{1}{2} (6.0)(5.60) = \mathbf{16.80\text{ N}}$$
(Notice: $T < m g = 39.2\text{ N}$, which allows the bucket to accelerate downward).

(b) Energy Formulation & Velocity:
Total initial mechanical energy: $E_i = m g h$.
When the bucket descends by $h$, its speed is $v$ and the pulley rotates at $\omega = v/R$.
Total final kinetic energy:
$$K_{\text{total}} = K_{\text{bucket, trans}} + K_{\text{pulley, rot}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$
Substitute $I = \frac{1}{2} M R^2$ and $\omega = v/R$:
$$K_{\text{pulley, rot}} = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{4} M v^2$$
$$K_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{4} M v^2 = \frac{1}{2} \left(m + \frac{M}{2}\right) v^2$$
By Conservation of Energy:
$$m g h = \frac{1}{2} \left(m + \frac{M}{2}\right) v^2 \implies v^2 = \frac{2 m g h}{m + \frac{M}{2}} = 2 a h$$
With $h = 5.0\text{ m}$ and $a = 5.60\text{ m/s}^2$:
$$v = \sqrt{2(5.60)(5.0)} = \sqrt{56.0} \approx \mathbf{7.483\text{ m/s}}$$
Angular velocity of the pulley:
$$\omega = \frac{v}{R} = \frac{7.483}{0.20} = \mathbf{37.42\text{ rad/s}}$$

(c) Energy Partition:
– Total potential energy released: $U_{\text{lost}} = m g h = (4.0)(9.8)(5.0) = 196.0\text{ J}$.
– Translational KE of bucket: $K_{\text{trans}} = \frac{1}{2} m v^2 = \frac{1}{2}(4.0)(56.0) = \mathbf{112.0\text{ J}}$.
– Rotational KE of pulley: $K_{\text{rot}} = \frac{1}{4} M v^2 = \frac{1}{4}(6.0)(56.0) = \mathbf{84.0\text{ J}}$.
– Total $K = 112.0 + 84.0 = 196.0\text{ J}$ (Exact conservation!).
– Fraction in translation: $\frac{112}{196} = \frac{m}{m + M/2} = \frac{4}{7} \approx \mathbf{57.14\%}$.
– Fraction in rotation: $\frac{84}{196} = \frac{M/2}{m + M/2} = \frac{3}{7} \approx \mathbf{42.86\%}$.

Takeaway: The rotational inertia of a pulley acts as an effective linear mass $M_{\text{eff}} = I/R^2 = M/2$ added to the system’s translational inertia.

Example 4 (Edge Case – Inelastic Clutch Engagement & Rotational Energy Dissipation):
Two coaxial flat circular discs are mounted on a frictionless shaft. Disc 1 has moment of inertia $I_1 = 0.40\text{ kg}\cdot\text{m}^2$ and rotates with initial angular speed $\omega_1 = 300.0\text{ rad/s}$. Disc 2 has moment of inertia $I_2 = 0.60\text{ kg}\cdot\text{m}^2$ and rotates coaxially in the same direction with $\omega_2 = 100.0\text{ rad/s}$. The two discs are gently pushed into face-to-face contact. Due to interfacial friction, they eventually achieve a common final angular speed $\omega_f$ without slipping.
(a) Determine the common angular velocity $\omega_f$ of the combined system.
(b) Calculate the initial total rotational kinetic energy $K_i$ and the final rotational kinetic energy $K_f$.
(c) Determine the mechanical energy dissipated as heat ($\Delta K$).
(d) Derive the general algebraic expression for $\Delta K$ and examine the special case where the discs initially spin in opposite directions.

Solution:
(a) Common Final Angular Speed $\omega_f$:
Because the mutual frictional torques between the two discs are internal action-reaction pairs, the net external torque on the two-disc system about the common axis is zero ($\sum \tau_{\text{ext}} = 0$). Hence, the total angular momentum is strictly conserved:
$$L_i = L_f$$
$$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_f$$
$$\mathbf{\omega_f = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}}$$
Substitute numerical values:
$$\omega_f = \frac{(0.40)(300.0) + (0.60)(100.0)}{0.40 + 0.60} = \frac{120.0 + 60.0}{1.00} = \mathbf{180.0\text{ rad/s}}$$

(b) Initial and Final Rotational Kinetic Energies:
$$K_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2} (0.40)(300.0)^2 + \frac{1}{2} (0.60)(100.0)^2$$
$$K_i = (0.20)(90000) + (0.30)(10000) = 18000 + 3000 = \mathbf{21000\text{ Joules}}$$
Final rotational kinetic energy:
$$K_f = \frac{1}{2} (I_1 + I_2) \omega_f^2 = \frac{1}{2} (1.00\text{ kg}\cdot\text{m}^2)(180.0\text{ rad/s})^2 = \frac{1}{2} (32400) = \mathbf{16200\text{ Joules}}$$

(c) Energy Dissipation:
$$\Delta K = K_i – K_f = 21000 – 16200 = \mathbf{4800\text{ Joules}}$$
This $4800\text{ J}$ is irreversibly converted into thermal energy at the slipping interface.

(d) General Algebraic Formula for Energy Loss:
$$\Delta K = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 – \frac{1}{2} (I_1 + I_2) \left(\frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}\right)^2$$
Expanding and factoring reveals the elegant universal formula:
$$\mathbf{\Delta K = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} (\omega_1 – \omega_2)^2}$$
Verification with our numbers:
$$\Delta K = \frac{1}{2} \frac{(0.40)(0.60)}{1.00} (300 – 100)^2 = \frac{1}{2} (0.24) (200)^2 = (0.12)(40000) = \mathbf{4800\text{ J}}$$
Opposite Rotation Case: If Disc 2 were initially spinning in the opposite direction ($\omega_2 = -100\text{ rad/s}$):
$$\omega_f’ = \frac{120 – 60}{1.00} = 60.0\text{ rad/s}$$
$$\Delta K’ = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} [\omega_1 – (-\omega_2)]^2 = 0.12 \times (300 + 100)^2 = 0.12 \times 160000 = \mathbf{19200\text{ J}}$$
Much greater energy is dissipated because the relative slipping velocity is significantly higher.

Takeaway: Just like a completely inelastic linear collision where $\Delta K = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(v_1 – v_2)^2$, angular momentum is conserved during clutch engagement, but mechanical energy is inevitably dissipated.

Worksheet: 10 Practice Problems

Problem 1 (JEE Main – Single Correct):
Two rotating rigid bodies $A$ and $B$ have moments of inertia $I_A$ and $I_B$ respectively, with $I_A > I_B$. If their angular momenta about their respective axes of rotation are identical ($L_A = L_B$), then the relationship between their rotational kinetic energies $K_A$ and $K_B$ is:
(A) $K_A > K_B$
(B) $K_A < K_B$
(C) $K_A = K_B$
(D) $K_A K_B = 1$

Problem 2 (JEE Main – Single Correct):
A flywheel of moment of inertia $I = 2.0\text{ kg}\cdot\text{m}^2$ is rotating freely at an initial angular speed of $\omega_0 = 30.0\text{ rad/s}$. If a constant retarding brake torque $\tau = 6.0\text{ N}\cdot\text{m}$ is applied to the flywheel, the total number of complete revolutions made by the flywheel before coming to rest is:
(A) $\frac{150}{\pi}$
(B) $\frac{75}{\pi}$
(C) $75\pi$
(D) $150\pi$

Problem 3 (JEE Main – Single Correct):
A uniform thin rod of length $L$ and mass $M$ is pivoted smoothly at its bottom end on a horizontal floor and held vertically. If it is released from rest and falls under gravity, its angular velocity $\omega$ at the exact instant it strikes the horizontal floor is:
(A) $\sqrt{\frac{g}{L}}$
(B) $\sqrt{\frac{2g}{L}}$
(C) $\sqrt{\frac{3g}{L}}$
(D) $\sqrt{\frac{6g}{L}}$

Problem 4 (JEE Main – Single Correct):
A uniform solid cylinder of mass $M$ and radius $R$ is mounted on a frictionless horizontal axle. A light string is wrapped around the cylinder and carries a mass $m$ hanging vertically from its free end. When the system is released from rest, the downward linear acceleration $a$ of the mass $m$ is:
(A) $\frac{m g}{M + m}$
(B) $\frac{2m g}{M + 2m}$
(C) $\frac{2m g}{2M + m}$
(D) $\frac{m g}{2M + m}$

Problem 5 (JEE Main – Single Correct):
The angular speed of an electric motor rotor increases uniformly from $1200\text{ rpm}$ to $3120\text{ rpm}$ in a time duration of $16.0\text{ seconds}$. If the moment of inertia of the rotor about its rotation axis is $I = 0.20\text{ kg}\cdot\text{m}^2$, the total work done by the motor during this acceleration phase is:
(A) $460.8\pi^2\text{ J}$
(B) $921.6\pi^2\text{ J}$
(C) $1843.2\pi^2\text{ J}$
(D) $3686.4\pi^2\text{ J}$

Problem 6 (JEE Advanced – One or More Correct):
A rigid body is rotating about a fixed axis with instantaneous angular velocity $\vec{\omega}$ and angular acceleration $\vec{\alpha}$. For a particle inside the body located at a perpendicular distance $r$ from the axis, which of the following statements is/are correct?
(A) The tangential component of its linear acceleration is $a_t = \alpha r$.
(B) The centripetal component of its linear acceleration is $a_c = \omega^2 r$.
(C) The angle $\phi$ between the net linear acceleration vector and the inward radial direction satisfies $\tan\phi = \frac{\alpha}{\omega^2}$.
(D) If $\alpha = 0$, the net linear acceleration of the particle is identically zero.

Problem 7 (JEE Advanced – One or More Correct):
A uniform thin rod of mass $M$ and length $L$ hangs vertically at rest, freely pivoted about a horizontal frictionless hinge at its top end $O$. A brief horizontal strike delivers an impulse $J$ to the rod at a distance $x$ below the pivot $O$. Which of the following statements is/are correct?
(A) The instantaneous angular velocity acquired by the rod is $\omega = \frac{3 J x}{M L^2}$.
(B) The impulsive reaction force delivered by the hinge to the rod is zero if the strike occurs at $x = \frac{2}{3} L$.
(C) The point at $x = \frac{2}{3} L$ is the Centre of Percussion of the rod relative to pivot $O$.
(D) The rotational kinetic energy acquired by the rod is $K = \frac{3 J^2 x^2}{2 M L^2}$.

Problem 8 (JEE Advanced – One or More Correct):
A uniform circular thin ring and a uniform circular disc have identical mass $M$ and radius $R$. Both are rotating about their respective central perpendicular axes with identical rotational kinetic energy $K$. Which of the following statements is/are correct?
(A) The ratio of the angular velocity of the disc to that of the ring is $\sqrt{2} : 1$.
(B) The ratio of the angular momentum of the ring to that of the disc is $\sqrt{2} : 1$.
(C) If identical constant retarding torques $\tau$ are applied to both bodies, they will come to rest after sweeping through the exact same angular displacement.
(D) The time taken by the ring to come to a stop is greater than that for the disc by a factor of $\sqrt{2}$.

Problem 9 (JEE Main / Advanced – Numerical Value Type):
A uniform circular disc of mass $M = 4.0\text{ kg}$ and radius $R = 0.50\text{ m}$ rotates freely about its central perpendicular axis at an initial angular speed of $\omega_0 = 20.0\text{ rad/s}$. A light tangential brake shoe is pressed radially against its rim with a normal force $N = 10.0\text{ N}$. The coefficient of kinetic friction between the shoe and the rim is $\mu = 0.40$. Calculate the time duration $t$ (in seconds) taken for the disc to come to a complete stop.

Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform rigid rod of mass $M = 6.0\text{ kg}$ and length $L = 2.0\text{ m}$ is freely hinged at one end to a fixed ceiling. The rod is held horizontally and released from rest. As the rod swings through its lowest vertical position, the upward vertical reaction force exerted by the hinge on the rod is $F_{\text{hinge}}$ in Newtons. Calculate the value of $F_{\text{hinge}}$ (take $g = 10.0\text{ m/s}^2$).

Solutions & Explanations

Answer Key Summary:
1. (B) | 2. (B) | 3. (C) | 4. (B) | 5. (B) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 5 | 10. 150

Solution 1:
Rotational kinetic energy expressed in terms of angular momentum is:
$$K = \frac{L^2}{2I}$$
Given $L_A = L_B = L$:
$$\frac{K_A}{K_B} = \frac{I_B}{I_A}$$
Since $I_A > I_B$, it follows directly that $K_A < K_B$. The body with the smaller rotational inertia possesses greater kinetic energy for a given angular momentum.
Correct Option: (B)

Solution 2:
Angular deceleration produced by the brake torque:
$$\alpha = \frac{\tau}{I} = \frac{6.0\text{ N}\cdot\text{m}}{2.0\text{ kg}\cdot\text{m}^2} = 3.0\text{ rad/s}^2$$
Using the rotational equation of motion $\omega^2 = \omega_0^2 – 2\alpha \theta$ with $\omega = 0$:
$$0 = (30.0)^2 – 2(3.0) \theta \implies 6.0 \theta = 900 \implies \theta = 150.0\text{ radians}$$
Number of complete revolutions $N$:
$$N = \frac{\theta}{2\pi} = \frac{150.0}{2\pi} = \mathbf{\frac{75}{\pi}}$$
Correct Option: (B)

Solution 3:
Moment of inertia about the end pivot: $I = \frac{1}{3} M L^2$.
As the rod falls from the vertical position to the horizontal floor, its centre of mass falls by a vertical distance $h = L/2$.
By conservation of mechanical energy:
$$\Delta U = M g \left(\frac{L}{2}\right) = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{3} M L^2\right) \omega^2 = \frac{1}{6} M L^2 \omega^2$$
$$M g \frac{L}{2} = \frac{1}{6} M L^2 \omega^2 \implies \omega^2 = \frac{3g}{L} \implies \mathbf{\omega = \sqrt{\frac{3g}{L}}}$$
Correct Option: (C)

Solution 4:
For the hanging mass $m$: $m g – T = m a$.
For the cylinder ($I = \frac{1}{2} M R^2$): $\tau = T R = I \alpha = I (a/R) \implies T = \frac{I}{R^2} a = \frac{1}{2} M a$.
Substituting $T$ into the linear equation:
$$m g – \frac{1}{2} M a = m a \implies m g = \left(m + \frac{M}{2}\right) a = \left(\frac{2m + M}{2}\right) a$$
$$\mathbf{a = \frac{2m g}{2m + M}}$$
Correct Option: (B)

Solution 5:
Initial and final angular speeds in $\text{rad/s}$:
$$\omega_1 = 1200 \times \frac{2\pi}{60} = 40\pi\text{ rad/s}$$
$$\omega_2 = 3120 \times \frac{2\pi}{60} = 104\pi\text{ rad/s}$$
By the Rotational Work-Energy Theorem:
$$W = \Delta K = \frac{1}{2} I (\omega_2^2 – \omega_1^2)$$
$$W = \frac{1}{2} (0.20) \left[(104\pi)^2 – (40\pi)^2\right] = 0.10 \pi^2 \left[10816 – 1600\right] = 0.10 \pi^2 (9216) = \mathbf{921.6\pi^2\text{ Joules}}$$
Correct Option: (B)

Solution 6:
– (A) True: By definition, $a_t = \alpha r$.
– (B) True: By definition, $a_c = \omega^2 r = v^2/r$.
– (C) True: $\tan\phi = \frac{a_t}{a_c} = \frac{\alpha r}{\omega^2 r} = \frac{\alpha}{\omega^2}$.
– (D) False: If $\alpha = 0$, $a_t = 0$, but $a_c = \omega^2 r \ne 0$. The net acceleration is $a_c$, not zero.
Correct Options: (A, B, C)

Solution 7:
– (A) True: Angular impulse about pivot $O$ is $J x = I_O \omega = (\frac{1}{3} M L^2) \omega \implies \omega = \frac{3 J x}{M L^2}$.
– (B, C) True: Linear impulse-momentum for the center of mass gives $J + J_{\text{hinge}} = M v_{cm} = M (\omega L/2) = M \left(\frac{3Jx}{2ML}\right) = \frac{3Jx}{2L}$.
Setting $J_{\text{hinge}} = 0 \implies J = \frac{3Jx}{2L} \implies x = \frac{2}{3} L$. This special point is the centre of percussion.
– (D) True: $K = \frac{1}{2} I_O \omega^2 = \frac{1}{2} \left(\frac{1}{3} M L^2\right) \left(\frac{3Jx}{M L^2}\right)^2 = \frac{3 J^2 x^2}{2 M L^2}$.
All four options are analytically correct.
Correct Options: (A, B, C, D)

Solution 8:
For ring: $I_{\text{ring}} = M R^2$. For disc: $I_{\text{disc}} = \frac{1}{2} M R^2$.
Given $K_{\text{ring}} = K_{\text{disc}} = K$.
– (A) $\frac{1}{2} I_r \omega_r^2 = \frac{1}{2} I_d \omega_d^2 \implies \frac{\omega_d}{\omega_r} = \sqrt{\frac{I_r}{I_d}} = \sqrt{\frac{M R^2}{\frac{1}{2} M R^2}} = \sqrt{2}$. (Correct)
– (B) $K = \frac{L^2}{2I} \implies L = \sqrt{2 I K}$. Thus $\frac{L_r}{L_d} = \sqrt{\frac{I_r}{I_d}} = \sqrt{2}$. (Correct)
– (C) Work-energy gives $W = \tau \theta = K \implies \theta = K/\tau$. Since both have identical $K$ and identical $\tau$, they sweep identical angular displacement $\theta$ before stopping. (Correct)
– (D) Angular impulse-momentum gives $\tau t = L \implies t = L/tau$. Since $L_r = \sqrt{2} L_d$, $t_r = \sqrt{2} t_d$. (Correct)
All statements are correct.
Correct Options: (A, B, C, D)

Solution 9:
Moment of inertia of the disc:
$$I = \frac{1}{2} M R^2 = \frac{1}{2} (4.0\text{ kg}) (0.50\text{ m})^2 = 0.50\text{ kg}\cdot\text{m}^2$$
Kinetic friction force exerted by the brake shoe on the rim:
$$f_k = \mu N = (0.40)(10.0\text{ N}) = 4.0\text{ N}$$
Retarding torque:
$$\tau = f_k \cdot R = (4.0\text{ N})(0.50\text{ m}) = 2.0\text{ N}\cdot\text{m}$$
Angular deceleration:
$$\alpha = \frac{\tau}{I} = \frac{2.0\text{ N}\cdot\text{m}}{0.50\text{ kg}\cdot\text{m}^2} = 4.0\text{ rad/s}^2$$
Time to stop from $\omega_0 = 20.0\text{ rad/s}$:
$$t = \frac{\omega_0}{\alpha} = \frac{20.0\text{ rad/s}}{4.0\text{ rad/s}^2} = \mathbf{5.0\text{ s}}$$
Correct Answer: 5

Solution 10:
Moment of inertia about the hinge: $I_O = \frac{1}{3} M L^2$.
By conservation of mechanical energy between horizontal and vertical orientations:
$$M g \left(\frac{L}{2}\right) = \frac{1}{2} I_O \omega^2 = \frac{1}{2} \left(\frac{1}{3} M L^2\right) \omega^2 \implies \omega^2 = \frac{3g}{L}$$
At the lowest position, the centre of mass (at distance $L/2$ from hinge) undergoes upward centripetal acceleration:
$$a_{c,cm} = \omega^2 \left(\frac{L}{2}\right) = \left(\frac{3g}{L}\right) \left(\frac{L}{2}\right) = \frac{3}{2} g$$
Applying Newton’s second law for the vertical motion of the rod:
$$F_{\text{hinge}} – M g = M a_{c,cm} = M \left(\frac{3}{2} g\right)$$
$$F_{\text{hinge}} = \frac{5}{2} M g = 2.5 \times (6.0\text{ kg}) \times (10.0\text{ m/s}^2) = 2.5 \times 60.0 = \mathbf{150\text{ N}}$$
Correct Answer: 150

Rigid Body Rotation: Pure Rotation About Fixed Axis & Rotational Kinetic Energy | JEE Physics Class 11