Mechanics Comprehensive Revision: Work-Energy-Power, Rotational Motion & Gravitation | JEE Physics Class 11

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Concept Card: Advanced Mechanics Revision – Energy, Rotation & Gravity

1. Architectural Integration of Advanced Mechanics:
The second half of classical mechanics for JEE Main and JEE Advanced synthesizes three core physical domains:

  1. Work, Energy & Power: Energy conservation theorems, conservative force fields, equilibrium stability, and constrained vertical circular motion.
  2. Rotational Dynamics & Rigid Bodies: Moment of inertia theorems, angular momentum conservation, and rolling without slipping.
  3. Gravitational Fields & Orbital Mechanics: Inverse-square field distributions, central potential energy wells, escape dynamics, and planetary orbits.

2. Module 1: Work, Energy, Power & Vertical Circular Motion

(A) Work-Energy Theorem & Conservative Potential Wells:

  • Universal Work-Energy Theorem:
    $$W_{\text{net}} = W_{\text{conservative}} + W_{\text{non-conservative}} + W_{\text{pseudo}} = \Delta K$$
  • Potential Energy Gradient: For a conservative field $\vec{F} = -\vec{\nabla} U$:
    $$F_x = -\frac{\partial U}{\partial x}, \quad F_y = -\frac{\partial U}{\partial y}, \quad F_z = -\frac{\partial U}{\partial z}$$
  • Equilibrium Classification:
    • Equilibrium condition: $\frac{dU}{dx} = 0$ (Zero net force).
    • Stable Equilibrium: $\frac{d^2 U}{dx^2} > 0$ (Local minimum in $U$, restoring force, small oscillations with $\omega = \sqrt{\frac{U”(x_0)}{m}}$).
    • Unstable Equilibrium: $\frac{d^2 U}{dx^2} < 0$ (Local maximum in $U$, destabilizing force).
    • Neutral Equilibrium: $\frac{d^2 U}{dx^2} = 0$.

(B) Vertical Circular Motion (String vs. Light Rod):

  • String of Length $L$:
    • Lowest point threshold for complete circle: $\mathbf{u_{\min} = \sqrt{5 g L}}$.
    • Tension at lowest point: $T_{\text{bottom}} = m \left(\frac{u^2}{L} + g\right) \ge 6 m g$.
    • Tension difference between bottom and top: $\mathbf{T_{\text{bottom}} – T_{\text{top}} = 6 m g}$ (always constant).
    • Oscillation condition (no slack, $\theta \le 90^\circ$): $u \le \sqrt{2 g L}$.
    • String slacks in upper half ($90^\circ < \theta < 180^\circ$): $\sqrt{2 g L} < u < \sqrt{5 g L}$. Slack occurs when $T = 0 \implies v^2 = -g L \cos\theta$.
  • Light Rigid Rod of Length $L$:
    Since a rod can support compressive stress without going slack, the minimum velocity at the top point only needs to be $v_{\text{top}} = 0$:
    $$\mathbf{u_{\min, \text{rod}} = \sqrt{4 g L} = 2 \sqrt{g L}}$$

3. Module 2: System of Particles, Rotation & Pure Rolling

(A) Angular Momentum Conservation:

  • Torque and Angular Momentum:
    $$\vec{\tau}_{\text{ext}, O} = \frac{d\vec{L}_O}{dt}$$
    If the net external torque about reference point $O$ vanishes ($\vec{\tau}_{\text{ext}, O} = \vec{0}$), then $\vec{L}_O$ is strictly conserved.
  • Angular momentum of a rigid body in planar motion about an origin $O$:
    $$\mathbf{\vec{L}_O = \vec{r}_{\text{cm}} \times (M \vec{v}_{\text{cm}}) + I_{\text{cm}} \vec{\omega}}$$

(B) Moment of Inertia Theorems:

  • Parallel Axis Theorem: $I = I_{\text{cm}} + M d^2$ (Valid for any arbitrary 3D rigid body, where axis passes through CM).
  • Perpendicular Axis Theorem: $I_z = I_x + I_y$ (Strictly restricted to planar laminas in the $xy$-plane).

(C) Pure Rolling Dynamics on an Incline ($\theta$):

  • Condition for rolling without slipping: $v_{\text{cm}} = R \omega$ and $a_{\text{cm}} = R \alpha$.
  • Linear acceleration down the plane:
    $$\mathbf{a_{\text{cm}} = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}}$$
  • Frictional force up the plane:
    $$\mathbf{f = \frac{M g \sin\theta}{1 + \frac{R^2}{k^2}}}$$
  • Minimum coefficient of static friction to avoid slipping:
    $$\mathbf{\mu_{\min} = \frac{\tan\theta}{1 + \frac{R^2}{k^2}}}$$
  • Energy partition:
    $$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{k^2 / R^2}{1 + k^2 / R^2} = \frac{k^2}{R^2 + k^2}, \quad \frac{K_{\text{trans}}}{K_{\text{total}}} = \frac{1}{1 + k^2 / R^2}$$

4. Module 3: Gravitational Fields, Potential & Satellite Dynamics

(A) Solid Sphere Field & Potential:

  • Exterior ($r \ge R$): $E(r) = -\frac{G M}{r^2}$, $V(r) = -\frac{G M}{r}$.
  • Interior ($r \le R$): $E(r) = -\frac{G M r}{R^3}$,
    $$\mathbf{V(r) = -\frac{G M}{2 R^3} (3 R^2 – r^2)}$$
  • At the Center ($r = 0$): $\vec{E} = \vec{0}$, $\mathbf{V(0) = -\frac{3 G M}{2 R} = 1.5 V_{\text{surface}}}$.

(B) Satellite Mechanics & Escape Thresholds:

  • Orbital speed: $v_o = \sqrt{\frac{G M}{r}}$.
  • Escape speed: $v_e = \sqrt{\frac{2 G M}{r}} = \sqrt{2} v_o$.
  • Energy relationship for circular orbit:
    $$\mathbf{E = -K = \frac{U}{2}}, \quad \text{Binding Energy } BE = -E = K = \frac{G M m}{2 r}$$
  • Work required to raise an orbit from $r_1$ to $r_2$:
    $$W = \Delta E = \frac{G M m}{2}\left(\frac{1}{r_1} – \frac{1}{r_2}\right)$$

5. Critical Insights & Exam Traps in JEE:

  • Trap 1 (Work Done by Static Friction): In pure rolling on a stationary incline/floor, the point of contact is instantaneously at rest ($v_{\text{contact}} = 0$). Hence, static friction does zero work on the rolling body ($W_{f_s} = 0$), and mechanical energy is conserved! However, on a moving plank, static friction does non-zero work.
  • Trap 2 (Direction of Friction in Rolling): When a horizontal force $F$ is applied to a rolling body at height $h$ above its center of mass:
    • If $h = I_{\text{cm}} / (M R)$, friction $f = 0$. (For a solid cylinder, $h = R/2$).
    • If $h > I_{\text{cm}} / (M R)$, friction acts forward.
    • If $h < I_{\text{cm}} / (M R)$, friction acts backward.
  • Trap 3 (Central Potential of a Solid Sphere): Do not set $V = 0$ at the center. Even though the gravitational field $\vec{E} = \vec{0}$ at $r = 0$, the potential reaches its deepest negative well ($V_c = -1.5 G M / R$).

Solved Examples

Example 1 (Work-Energy Theorem & Critical Vertical Circular Motion):
A small bob of mass $m$ attached to a light inextensible string of length $L$ is projected horizontally from the lowest point with an initial speed $u = \sqrt{3 g L}$.
(a) Determine the angular displacement $\theta$ from the downward vertical at which the string becomes slack ($T = 0$).
(b) Find the speed of the bob at the instant the string slacks.
(c) Calculate the maximum height reached by the bob above the lowest point in its subsequent projectile motion.

Solution:
(a) Angular Displacement for Slackness ($T = 0$):
By conservation of mechanical energy at angle $\theta$ from downward vertical:
$$\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + m g L(1 – \cos\theta)$$
$$3 g L = v^2 + 2 g L(1 – \cos\theta) \implies v^2 = g L(1 + 2\cos\theta)$$
The radial equation of motion gives:
$$T – m g \cos(\pi – \theta) = \frac{m v^2}{L} \implies T + m g \cos\theta = \frac{m v^2}{L}$$
At the slack condition $T = 0$:
$$m g \cos\theta = -\frac{m v^2}{L} \implies v^2 = -g L \cos\theta$$
Equating the two expressions for $v^2$:
$$-g L \cos\theta = g L(1 + 2\cos\theta) \implies -\cos\theta = 1 + 2\cos\theta \implies 3\cos\theta = -1$$
$$\mathbf{\cos\theta = -\frac{1}{3} \implies \theta = \arccos\left(-\frac{1}{3}\right) \approx 109.47^\circ}$$

(b) Speed at the Instant of Slackness:
$$v^2 = -g L\left(-\frac{1}{3}\right) = \frac{g L}{3} \implies \mathbf{v = \sqrt{\frac{g L}{3}}}$$

(c) Maximum Height Above Lowest Point:
Height of the slack point above the lowest point:
$$h_1 = L(1 – \cos\theta) = L\left(1 – \left(-\frac{1}{3}\right)\right) = \frac{4}{3} L$$
At this point, the string becomes slack and the bob executes free-fall projectile motion with speed $v = \sqrt{g L/3}$ launched at an angle $\phi$ above the horizontal, where $\sin\phi = -\cos\theta = 1/3$.
Additional maximum vertical rise in projectile motion:
$$h_2 = \frac{v^2 \sin^2\phi}{2 g} = \frac{\left(\frac{g L}{3}\right) \left(\frac{1}{3}\right)^2}{2 g} = \frac{g L / 27}{2 g} = \frac{L}{54}$$
Total maximum height above the lowest point:
$$\mathbf{H_{\max} = h_1 + h_2 = \frac{4}{3} L + \frac{1}{54} L = \frac{72 + 1}{54} L = \frac{73}{54} L \approx 1.3519 L}$$

Example 2 (Pure Rolling Incline Dynamics & Critical Friction Coefficient):
A uniform solid sphere of mass $M$ and radius $R$ rolls without slipping down a rough incline of inclination $\theta = 30^\circ$ ($g = 10\text{ m/s}^2$).
(a) Determine the linear acceleration $a_{\text{cm}}$ of the sphere.
(b) Determine the friction force $f$ acting on the sphere.
(c) Find the minimum coefficient of static friction $\mu_{\min}$ required to ensure rolling without slipping.

Solution:
For a solid sphere, radius of gyration is $k^2 = \frac{2}{5} R^2$, so $\frac{k^2}{R^2} = \frac{2}{5}$.
(a) Acceleration of Center of Mass:
$$a_{\text{cm}} = \frac{g \sin\theta}{1 + k^2/R^2} = \frac{g \sin 30^\circ}{1 + 2/5} = \frac{g / 2}{7/5} = \mathbf{\frac{5}{14} g} = \frac{5}{14}(10) \approx \mathbf{3.57\text{ m/s}^2}$$

(b) Frictional Force:
$$f = \frac{M g \sin\theta}{1 + R^2/k^2} = \frac{M g (1/2)}{1 + 5/2} = \frac{M g / 2}{7/2} = \mathbf{\frac{1}{7} M g}$$

(c) Minimum Coefficient of Static Friction:
For pure rolling: $f \le \mu_s N = \mu_s M g \cos\theta$:
$$\frac{1}{7} M g \le \mu_s M g \cos 30^\circ \implies \mu_s \ge \frac{1}{7 \cos 30^\circ} = \frac{1}{7 (\sqrt{3}/2)} = \mathbf{\frac{2}{7 \sqrt{3}} \approx 0.1650}$$

Example 3 (Ballistic Impact & Angular Momentum Conservation – JEE Advanced):
A uniform thin rod of mass $M = 3.0\text{ kg}$ and length $L = 1.0\text{ m}$ is suspended vertically from a frictionless pivot at its top end $O$. A small clay ball of mass $m = 1.0\text{ kg}$ travelling horizontally with velocity $v_0 = 12.0\text{ m/s}$ strikes the bottom tip of the rod and sticks to it ($g = 10\text{ m/s}^2$).
(a) Determine the angular velocity $\omega$ of the rod-ball system immediately after the collision.
(b) Find the maximum angular deflection $\theta_{\max}$ through which the system swings before momentarily coming to rest.

Solution:
(a) Angular Velocity Immediately After Impact:
Because the hinge at $O$ exerts an external impulsive reaction force during impact, linear momentum is not conserved. However, the external torque about pivot $O$ vanishes ($\vec{\tau}_{\text{ext}, O} = \vec{0}$). Hence, angular momentum about pivot $O$ is strictly conserved.
Initial angular momentum about $O$:
$$L_i = m v_0 L = (1.0)(12.0)(1.0) = 12.0\text{ kg}\cdot\text{m}^2/\text{s}$$
Moment of inertia of the composite system about pivot $O$:
$$I_{\text{sys}} = I_{\text{rod}} + I_{\text{ball}} = \frac{1}{3} M L^2 + m L^2 = \frac{1}{3}(3.0)(1.0)^2 + (1.0)(1.0)^2 = 1.0 + 1.0 = 2.0\text{ kg}\cdot\text{m}^2$$
By conservation of angular momentum:
$$L_i = I_{\text{sys}} \omega \implies 12.0 = 2.0 \omega \implies \mathbf{\omega = 6.0\text{ rad/s}}$$

(b) Maximum Angular Deflection $\theta_{\max}$:
After the collision, only conservative gravity acts. Mechanical energy is conserved:
Initial rotational kinetic energy:
$$K_i = \frac{1}{2} I_{\text{sys}} \omega^2 = \frac{1}{2}(2.0)(6.0)^2 = 36.0\text{ J}$$
Potential energy gain at maximum swing angle $\theta_{\max}$:
$$\Delta U = M g \left(\frac{L}{2}\right)(1 – \cos\theta_{\max}) + m g L (1 – \cos\theta_{\max}) = \left[ M g \frac{L}{2} + m g L \right] (1 – \cos\theta_{\max})$$
$$\Delta U = \left[ (3.0)(10)(0.5) + (1.0)(10)(1.0) \right] (1 – \cos\theta_{\max}) = [15.0 + 10.0](1 – \cos\theta_{\max}) = 25.0 (1 – \cos\theta_{\max})$$
Equating $K_i = \Delta U$:
$$36.0 = 25.0 (1 – \cos\theta_{\max}) \implies 1 – \cos\theta_{\max} = \frac{36.0}{25.0} = 1.44$$
$$\mathbf{\cos\theta_{\max} = 1 – 1.44 = -0.44 \implies \theta_{\max} = \arccos(-0.44) \approx 116.1^\circ}$$
The system swings past the horizontal ($90^\circ$) to $116.1^\circ$.

Example 4 (Gravitational Tunnel Escape & Multi-Concept Integration):
A particle of mass $m$ is located at the center of the Earth ($r = 0$).
(a) Find the minimum speed $u_1$ required to launch the particle through a frictionless radial tunnel to just reach the Earth’s surface ($r = R$).
(b) Find the minimum speed $u_2$ required to launch the particle from the center so that it permanently escapes to infinity.
(c) Express both launch speeds in terms of the surface escape speed $v_e = \sqrt{2 g_0 R}$.

Solution:
Gravitational potential at Earth’s center: $V_{\text{center}} = -\frac{3 G M}{2 R} = -1.5 g_0 R$.
Gravitational potential at Earth’s surface: $V_{\text{surface}} = -\frac{G M}{R} = -g_0 R$.
(a) Launch Speed to Reach the Surface ($r = R$):
$$\frac{1}{2} m u_1^2 + m V_{\text{center}} = 0 + m V_{\text{surface}}$$
$$\frac{1}{2} u_1^2 = V_{\text{surface}} – V_{\text{center}} = -g_0 R – (-1.5 g_0 R) = 0.5 g_0 R$$
$$u_1^2 = g_0 R \implies \mathbf{u_1 = \sqrt{g_0 R} = \frac{v_e}{\sqrt{2}} \approx 7.92\text{ km/s}}$$

(b) Launch Speed to Escape to Infinity ($r \to \infty$):
$$\frac{1}{2} m u_2^2 + m V_{\text{center}} = 0 + 0$$
$$\frac{1}{2} u_2^2 = -V_{\text{center}} = 1.5 g_0 R = \frac{3}{2} g_0 R$$
$$u_2^2 = 3 g_0 R \implies \mathbf{u_2 = \sqrt{3 g_0 R} = \sqrt{1.5} v_e = \sqrt{\frac{3}{2}} v_e \approx 13.72\text{ km/s}}$$

Practice Worksheet: Advanced Mechanics Revision

Section A: Single Correct Questions (Q1 to Q5)

Q1. The potential energy of a particle of mass $m$ moving along the $x$-axis is given by $U(x) = a x^2 – b x$, where $a$ and $b$ are positive constants. The stable equilibrium position and the angular frequency $\omega$ of small oscillations about it are:
(A) $x_0 = \frac{b}{2a}, \quad \omega = \sqrt{\frac{2a}{m}}$
(B) $x_0 = \frac{b}{a}, \quad \omega = \sqrt{\frac{a}{m}}$
(C) $x_0 = \frac{b}{2a}, \quad \omega = \sqrt{\frac{a}{2m}}$
(D) $x_0 = 0, \quad \omega = \sqrt{\frac{2a}{m}}$

Q2. A uniform solid cylinder of mass $M$ and radius $R$ rolls without slipping on a horizontal surface with linear speed $v$. The ratio of its rotational kinetic energy to its total kinetic energy is:
(A) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) $\frac{2}{3}$
(D) $\frac{2}{5}$

Q3. A body is situated at a distance $r = 2R$ from the center of Earth (where $R$ is Earth’s radius). If $v_e = \sqrt{2 g_0 R}$ is the escape velocity from the surface of Earth, the minimum speed required to project this body to escape to infinity is:
(A) $\frac{v_e}{2}$
(B) $\frac{v_e}{\sqrt{2}}$
(C) $\sqrt{2} v_e$
(D) $\frac{v_e}{4}$

Q4. A position-dependent force $F(x) = (3x^2 + 2x)\text{ N}$ acts on a particle moving along the $x$-axis. The work done by this force as the particle moves from $x = 1\text{ m}$ to $x = 3\text{ m}$ is:
(A) $34\text{ J}$
(B) $36\text{ J}$
(C) $32\text{ J}$
(D) $28\text{ J}$

Q5. From a uniform circular disc of mass $M$ and radius $R$, a circular hole of radius $R/2$ is cut out such that its boundary passes through the center of the original disc. The moment of inertia of the remaining portion about an axis perpendicular to the plane of the disc passing through the center of the original disc is:
(A) $\frac{13}{32} M R^2$
(B) $\frac{11}{32} M R^2$
(C) $\frac{15}{32} M R^2$
(D) $\frac{7}{16} M R^2$

Section B: One or More than One Correct Questions (Q6 to Q8)

Q6. A small body is suspended by a light string of length $L$ and projected horizontally from the lowest point with speed $u$. Which of the following statements is/are correct?
(A) If $u \le \sqrt{2 g L}$, the string remains taut at all times and the body oscillates.
(B) If $\sqrt{2 g L} < u < \sqrt{5 g L}$, the string goes slack in the upper half ($90^\circ < \theta < 180^\circ$) and the body enters projectile motion.
(C) If $u \ge \sqrt{5 g L}$, the body completes full circular motion without slackness.
(D) If the light string is replaced by a light rigid rod, the minimum launch speed required to complete the circle is $2\sqrt{g L}$.

Q7. A uniform solid cylinder of mass $M$ and radius $R$ is placed on a rough horizontal floor with friction coefficient $\mu$. A horizontal force $F$ is applied at a height $h$ above the center of mass. Which of the following statements is/are correct for pure rolling?
(A) The linear acceleration of the center of mass is $a_{\text{cm}} = \frac{F + f}{M}$.
(B) If $h = R/2$, the friction force acting on the cylinder is zero.
(C) If $h > R/2$, the friction force acts in the forward direction (same direction as $F$).
(D) If $h < R/2$, the friction force acts in the backward direction (opposite to $F$).

Q8. For a satellite of mass $m$ revolving in a circular orbit of radius $r$ around Earth, which of the following statements is/are correct?
(A) Total mechanical energy is negative ($E = -K = U/2$).
(B) The binding energy of the satellite is numerically equal to its kinetic energy.
(C) As the orbital radius increases, total mechanical energy increases while orbital speed decreases.
(D) If an orbiting satellite is supplied with energy equal to its kinetic energy, it escapes permanently along a parabolic path.

Section C: Numerical Value Type Questions (Q9 to Q10)

Q9. A uniform thin rod of mass $M = 3.0\text{ kg}$ and length $L = 1.0\text{ m}$ is pivoted freely at its top end. A clay ball of mass $m = 1.0\text{ kg}$ moving horizontally at speed $v_0 = 12.0\text{ m/s}$ strikes the bottom tip of the rod and sticks to it. The angular velocity $\omega$ (in $\text{rad/s}$) of the system immediately after the collision is an integer. Find the value of $\omega$.

Q10. The ratio of the escape velocity of a particle launched from the center of Earth through a tunnel to the escape velocity from the surface of Earth is $\sqrt{\frac{k}{2}}$. Determine the integer value of $k$.

Comprehensive Solutions and Answer Key

Answer Key

Question Correct Answer Core Physics Concept
Q1 (A) Potential Minimum Equilibrium: $x_0 = \frac{b}{2a}, \omega = \sqrt{2a/m}$
Q2 (A) Energy Partition in Pure Rolling: $K_{\text{rot}} / K_{\text{total}} = 1/3$
Q3 (B) Escape Speed Scaling: $v = \sqrt{2GM/r} = \sqrt{2GM/2R} = v_e / \sqrt{2}$
Q4 (A) Variable Force Work Integral: $W = [x^3 + x^2]_1^3 = 36 – 2 = 34\text{ J}$
Q5 (A) Negative Mass Moment of Inertia: $I = \frac{1}{2}MR^2 – \frac{3}{32}MR^2 = \frac{13}{32}MR^2$
Q6 (A, B, C, D) Vertical Circular Motion Regimes (String vs Rod)
Q7 (A, B, C, D) Height of Force Line & Friction Direction in Rolling
Q8 (A, B, C, D) Satellite Energy Balance & Escape Thresholds
Q9 6 Angular Momentum Conservation: $\omega = 12.0 / 2.0 = 6\text{ rad/s}$
Q10 3 Center vs Surface Escape Ratio: $\sqrt{3 g_0 R} / \sqrt{2 g_0 R} = \sqrt{3/2} \implies k = 3$

Detailed Explanations

Solution 1:
Given $U(x) = a x^2 – b x$.
First derivative:
$$\frac{dU}{dx} = 2 a x – b = 0 \implies \mathbf{x_0 = \frac{b}{2a}}$$
Second derivative:
$$\frac{d^2 U}{dx^2} = 2 a > 0 \quad (\text{since } a > 0)$$
Because the second derivative is positive, this equilibrium is stable.
For small displacements, restoring force $F = -k_{\text{eff}} \Delta x$ where $k_{\text{eff}} = \frac{d^2 U}{dx^2} = 2a$.
Angular frequency of small oscillations:
$$\mathbf{\omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{2a}{m}}}$$
Correct Answer: (A)

Solution 2:
For a uniform solid cylinder, moment of inertia $I = \frac{1}{2} M R^2$, and $k^2 = R^2 / 2$.
Rotational kinetic energy:
$$K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2}\left(\frac{1}{2} M R^2\right)\left(\frac{v^2}{R^2}\right) = \frac{1}{4} M v^2$$
Translational kinetic energy:
$$K_{\text{trans}} = \frac{1}{2} M v^2$$
Total kinetic energy:
$$K_{\text{total}} = \frac{1}{4} M v^2 + \frac{1}{2} M v^2 = \frac{3}{4} M v^2$$
Taking the ratio:
$$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{4} M v^2}{\frac{3}{4} M v^2} = \mathbf{\frac{1}{3}}$$
Correct Answer: (A)

Solution 3:
At distance $r = 2R$ from the center of Earth, the gravitational potential is:
$$V(2R) = -\frac{G M}{2R}$$
To escape to infinity ($E_f = 0$):
$$\frac{1}{2} m v^2 + m V(2R) = 0 \implies v^2 = -2 V(2R) = \frac{G M}{R}$$
Surface escape velocity is $v_e = \sqrt{\frac{2 G M}{R}} \implies \frac{G M}{R} = \frac{v_e^2}{2}$.
Therefore:
$$v = \sqrt{\frac{v_e^2}{2}} = \mathbf{\frac{v_e}{\sqrt{2}}}$$
Correct Answer: (B)

Solution 4:
Work done by a position-dependent 1D force:
$$W = \int_{x_1}^{x_2} F(x) dx = \int_1^3 (3x^2 + 2x) dx = \left[ x^3 + x^2 \right]_1^3$$
$$W = (3^3 + 3^2) – (1^3 + 1^2) = (27 + 9) – (1 + 1) = 36 – 2 = \mathbf{34\text{ J}}$$
Correct Answer: (A)

Solution 5:
Mass of the complete disc is $M$, area $A_0 = \pi R^2$.
Mass of the removed circular hole of radius $R/2$:
$$m = M \frac{\pi (R/2)^2}{\pi R^2} = \frac{M}{4}$$
The center of the removed hole is at distance $d = R/2$ from the center of the original disc.
Moment of inertia of the complete disc about its central axis:
$$I_0 = \frac{1}{2} M R^2$$
Moment of inertia of the removed portion about the original disc’s center (by Parallel Axis Theorem):
$$I_{\text{removed}} = \frac{1}{2} m \left(\frac{R}{2}\right)^2 + m d^2 = \frac{1}{2}\left(\frac{M}{4}\right)\left(\frac{R^2}{4}\right) + \left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2$$
$$I_{\text{removed}} = \frac{M R^2}{32} + \frac{M R^2}{16} = \frac{M R^2 + 2 M R^2}{32} = \frac{3 M R^2}{32}$$
By the principle of superposition, the remaining moment of inertia is:
$$I = I_0 – I_{\text{removed}} = \frac{M R^2}{2} – \frac{3 M R^2}{32} = \frac{16 M R^2 – 3 M R^2}{32} = \mathbf{\frac{13}{32} M R^2}$$
Correct Answer: (A)

Solution 6:
– (A) True: For $u \le \sqrt{2 g L}$, the velocity becomes zero before or at $\theta = 90^\circ$ while tension remains non-negative ($T \ge 0$); hence oscillation occurs.
– (B) True: For $\sqrt{2 g L} < u < \sqrt{5 g L}$, $T$ vanishes before $v$ reaches zero in the upper half ($90^\circ < \theta < 180^\circ$); the string goes slack and projectile motion ensues. - (C) True: $u \ge \sqrt{5 g L}$ ensures $T \ge 0$ everywhere, including the topmost point where $v_{\text{top}} \ge \sqrt{g L}$. - (D) True: For a rigid rod, tension can be negative (compressive support); hence the bob only needs to reach the top with zero velocity: $u_{\min} = \sqrt{2 g (2L)} = 2\sqrt{g L}$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)

Solution 7:
Let $F$ act at height $h$ above center of mass and assume friction $f$ acts forward:
Equations of motion:
1. $F + f = M a_{\text{cm}}$
2. Torque about center of mass: $\tau_{\text{cm}} = F h – f R = I_{\text{cm}} \alpha = \left(\frac{1}{2} M R^2\right) \frac{a_{\text{cm}}}{R} = \frac{1}{2} M R a_{\text{cm}}$.
Substitute $M a_{\text{cm}} = F + f$:
$$F h – f R = \frac{1}{2} R (F + f) = \frac{1}{2} F R + \frac{1}{2} f R$$
$$F\left(h – \frac{R}{2}\right) = \frac{3}{2} f R \implies f = \frac{2 F}{3 R}\left(h – \frac{R}{2}\right)$$
– If $h = R/2$: $f = 0$. (B is True)
– If $h > R/2$: $f > 0$ (acts forward). (C is True)
– If $h < R/2$: $f < 0$ (acts backward). (D is True) - Acceleration: $a_{\text{cm}} = \frac{F + f}{M}$. (A is True)
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)

Solution 8:
– (A) True: In circular orbit, $K = \frac{G M m}{2r}$, $U = -\frac{G M m}{r}$, $E = -\frac{G M m}{2r} = -K = U/2$.
– (B) True: Binding energy $BE = -E = K$.
– (C) True: $E \propto -1/r$ (increases with $r$) while $v_o = \sqrt{G M / r}$ decreases.
– (D) True: Adding $\Delta E = K$ results in $E_{\text{total}} = -K + K = 0$, the parabolic escape condition.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)

Solution 9:
Conservation of angular momentum about pivot $O$:
$$L_i = m v_0 L = (1.0)(12.0)(1.0) = 12.0\text{ kg}\cdot\text{m}^2/\text{s}$$
$$I_{\text{sys}} = \frac{1}{3} M L^2 + m L^2 = \frac{1}{3}(3.0)(1.0)^2 + (1.0)(1.0)^2 = 2.0\text{ kg}\cdot\text{m}^2$$
$$\omega = \frac{L_i}{I_{\text{sys}}} = \frac{12.0}{2.0} = \mathbf{6\text{ rad/s}}$$
Correct Answer: 6

Solution 10:
Launch escape velocity from Earth’s center:
$$u = \sqrt{3 g_0 R}$$
Launch escape velocity from Earth’s surface:
$$v_e = \sqrt{2 g_0 R}$$
Taking the ratio:
$$\frac{u}{v_e} = \frac{\sqrt{3 g_0 R}}{\sqrt{2 g_0 R}} = \sqrt{\frac{3}{2}}$$
Comparing with $\sqrt{\frac{k}{2}}$:
$$\mathbf{k = 3}$$
Correct Answer: 3

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