Rotational Dynamics Detailed Study Notes

Introduction to Rotational Dynamics

• Subject Overview: This chapter focuses on the mechanics of rotating bodies, including derivations of kinetic energy, moment of inertia, torque, and angular momentum.
• Key Contents:
  • Equations of angular motion.
  • Kinetic Energy ($K.E.$) of rotation.
  • Moment of Inertia ($MOI$).
  • Calculation of $MOI$ for various rigid bodies (rods, spheres, discs).
  • Torque ($\tau$) and angular acceleration ($\alpha$).
  • Work and Power in rotational systems.
  • Angular Momentum ($L$) and its relationship with $I$.
  • Conservation of Angular Momentum ($I\omega = \text{constant}$).

Rigid Bodies and Types of Motion

• Rigid Body Definition: A solid body in which particles are compactly arranged such that the inter-particle distance is small and fixed ($r = \text{constant}$). External forces do not disturb the relative positions of these particles, meaning the shape remains unaltered under stress. Practically, most solids are treated as rigid bodies.
• Types of Motion:
  • Translational Motion: The entire mass moves bodily from one point to another. Every particle in the mass undergoes the same linear displacement at any given time. Example: A bus moving along a straight road.
  • Rotational Motion: The body rotates about a fixed axis ($YY'$). Particles in the body generate concentric circles. While all particles have the same angular velocity ($\omega$), their linear velocities ($v$) differ based on their distance from the axis. Example: A wheel rotating about its axle, or Earth rotating on its axis.
  • Rolling Motion: A combination of translational and rotational motion. Example: A ball rolling on the ground.

Equations of Angular Motion and Linear Relations

• Angular Displacement ($\theta$):
  • Defined as the angle traversed by a body rotating about an axis.
  • Relation to linear displacement ($S$): $S = r \theta$, where $r$ is the radius.
• Angular Velocity ($\omega$):
  • Rate of change of angular displacement: $\omega = \frac{d\theta}{dt}$.
  • Unit: $\text{rad/s}$.
  • Relation to linear velocity ($v$): $v = \frac{dS}{dt} = \frac{d(r\theta)}{dt} = r \frac{d\theta}{dt} \rightarrow v = r\omega$.
• Angular Acceleration ($\alpha$):
  • Rate of change of angular velocity: $\alpha = \frac{d\omega}{dt}$.
  • Unit: $\text{rad/s}^2$.
  • Relation to linear acceleration ($a$): $a = \frac{dv}{dt} = \frac{d(r\omega)}{dt} = r \frac{d\omega}{dt} \rightarrow a = r\alpha$.
• Comparison of Equations of Motion:
  • Linear: $v = u + at$ | Rotational: $\omega = \omega_0 + \alpha t$
  • Linear: $S = ut + \frac{1}{2}at^2$ | Rotational: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$
  • Linear: $v^2 = u^2 + 2aS$ | Rotational: $\omega^2 = \omega_0^2 + 2\alpha\theta$

Moment of Inertia (MOI)

• Conceptual Definition: Moment of Inertia ($I$) is the rotational analogue of mass ($m$). It measures the "laziness" or resistance of an object to rotational motion.
  • High $MOI$: Harder to rotate or stop from rotating.
  • Low $MOI$: Easier to rotate.
• Mathematical Definition: The $MOI$ of a body about a given axis is the sum of the products of the mass of each particle and the square of its distance from the axis of rotation.
  • Formula: $I = \sum_{i=1}^n m_i r_i^2$
  • For a continuous body: $I = \int r^2 dm$
• Radius of Gyration ($k$):
  • The distance from the axis of rotation to a point where the entire mass of the body is assumed to be concentrated such that the $MOI$ remains the same.
  • Formula: $I = Mk^2 \rightarrow k = \sqrt{\frac{I}{M}}$.
  • Expression for $k$ in terms of particle radii: $k = \sqrt{\frac{r_1^2 + r_2^2 + ... + r_n^2}{n}}$.
• MOI of a Thin Uniform Rod:
  • Case A: Through Centre (Perpendicular to length):
    • Consider mass $M$ and length $l$. Mass per unit length $\lambda = \frac{M}{l}$.
    • Element mass $dm = \frac{M}{l} dx$ at distance $x$.
    • $I = \int_{-l/2}^{l/2} x^2 \frac{M}{l} dx = \frac{M}{l} \left[ \frac{x^3}{3} \right]_{-l/2}^{l/2} = \frac{M}{3l} \left( \frac{l^3}{8} + \frac{l^3}{8} \right) = \frac{Ml^2}{12}$.
  • Case B: Through One End:
    • Integration limits change from $0$ to $l$.
    • $I = \int_{0}^{l} x^2 \frac{M}{l} dx = \frac{M}{l} \left[ \frac{x^3}{3} \right]_{0}^{l} = \frac{Ml^2}{3}$.
  • Comparison: Since $\frac{Ml^2}{12} < \frac{Ml^2}{3}$, it is easier to rotate a rod from the center than from the end.

Moment of Inertia for Common Shapes

• Thin Circular Ring: $I = MR^2$ (Axis through centre, perpendicular to plane).
• Thin Circular Disc: $I = \frac{1}{2}MR^2$ (Axis through centre, perpendicular to plane).
• Solid Sphere: $I = \frac{2}{5}MR^2$ (Axis through diameter).
• Hollow Sphere: $I = \frac{2}{3}MR^2$ (Axis through diameter).
• Solid Cylinder: $I = \frac{1}{2}MR^2$ (Axis along the geometric center).
• Hollow Cylinder: $I = MR^2$ (Axis along the geometric center).

Kinetic Energy of Rotation

• Derivation: Consider a body with particles $m_1, m_2, ... m_n$ at distances $r_1, r_2, ... r_n$ rotating with uniform $\omega$.
  • Linear velocity of $i$-th particle: $v_i = r_i \omega$.
  • $K.E.$ of $i$-th particle: $E_i = \frac{1}{2} m_i v_i^2 = \frac{1}{2} m_i r_i^2 \omega^2$.
  • Total $K.E. = \sum E_i = \frac{1}{2} \omega^2 (\sum m_i r_i^2) = \frac{1}{2} I \omega^2$.
• Total K.E. in Rolling: $K.E_{\text{total}} = \frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2$.

Torque ($\tau$) and Angular Momentum ($L$)

• Torque (Turning Effect): The product of the force and the perpendicular distance ($r$) from the axis of rotation.
  • Vector: $\vec{\tau} = \vec{r} \times \vec{F}$.
  • Magnitude: $\tau = rF\sin(\theta)$.
  • Relation to $MOI$: $\tau = I \alpha$ (Rotational equivalent of $F = ma$).
• Angular Momentum ($L$): The moment of linear momentum about the axis of rotation.
  • For a particle: $L = mvr = mr^2 \omega$.
  • For a rigid body: $L = I \omega$.
• Relation between $\tau$ and $L$:
  • Linear variant: $F = \frac{dp}{dt}$.
  • Rotational variant: $\tau = \frac{dL}{dt}$.
  • Proof: $\frac{dL}{dt} = \frac{d(I\omega)}{dt} = I\frac{d\omega}{dt} = I\alpha = \tau$.
• Principle of Conservation of Angular Momentum:
  • If no external torque acts on a system ($\tau_{ext} = 0$), the total angular momentum remains constant.
  • $I\omega = \text{constant} \rightarrow I_1 \omega_1 = I_2 \omega_2$.

Power and Work

• Rotational Work: $W = \tau \theta$ (Linear: $W = FS$).
• Rotational Power: $P = \tau \omega$ (Linear: $P = Fv$).

Solved Numerical Problems

• Numerical [2081 Set D Q.No. 12c]: A wheel starts from rest and reaches $\omega = 8\,\text{rev/s}$ in $5\,\text{s}$.
  • Given: $\omega_0 = 0$, $f = 8\,\text{rev/s} \rightarrow \omega = 2\pi \times 8 = 16\pi\,\text{rad/s}$, $t = 5\,\text{s}$.
  • Calculations:
    1. $\alpha = \frac{\omega - \omega_0}{t} = \frac{16\pi}{5} = 10.05\,\text{rad/s}^2$.
    2. Angle at $t=3\,\text{s}$: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(10.05)(3)^2 = 45.21\,\text{rad}$.
    3. Revolutions at $t=3\,\text{s}$: $n = \frac{\theta}{2\pi} = \frac{45.21}{2\pi} = 7.2\,\text{rev}$.
• Numerical [2080 Set P Q.No. 12b]: Motor engine speed decreases from $900\,\text{rev/min}$ to $600\,\text{rev/min}$ in $10\,\text{s}$.
  • Given: $f_1 = 15\,\text{rev/s}$, $f_2 = 10\,\text{rev/s}$, $t = 10\,\text{s}$.
  • Calculations:
    1. $\alpha = \frac{2\pi(f_2 - f_1)}{t} = \frac{2\pi(10 - 15)}{10} = -\pi = -3.14\,\text{rad/s}^2$.
    2. Revolutions: $\theta = \omega_1 t + \frac{1}{2}\alpha t^2 = (30\pi)(10) + \frac{1}{2}(-\pi)(100) = 250\pi \approx 785\,\text{rad}$. $n = \frac{785}{2\pi} = 125\,\text{rev}$.
    3. Time to rest: $0 = \omega_2 + \alpha t_{add} \rightarrow 0 = 20\pi - \pi t_{add} \rightarrow t_{add} = 20\,\text{s}$.
• Numerical [2081 GIE Set B Q.No. 12b]: Disc ($I = 5 \times 10^{-4}\,\text{kg\,m}^2$) at $60\,\text{rpm}$. Wax ($0.01\,\text{kg}$) added at $0.06\,\text{m}$.
  • Calculation: New $I_2 = I_1 + mr^2 = 5 \times 10^{-4} + (0.01)(0.06)^2 = 5.36 \times 10^{-4}\,\text{kg\,m}^2$. By conservation: $I_1 f_1 = I_2 f_2 \rightarrow 5 \times 60 = 5.36 \times f_2 \rightarrow f_2 \approx 56\,\text{rpm}$.
• Numerical [2081 Set B/C Q.No. 1c]: Flywheel ($I=0.32\,\text{kg\,m}^2$) at $120\,\text{rad/s}$ by $50\,\text{W}$ motor.
  • Calculations:
    1. $K.E. = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.32)(120)^2 = 2304\,\text{J}$.
    2. Frictional couple (Torque): $P = \tau \omega \rightarrow 50 = \tau (120) \rightarrow \tau = 0.42\,\text{N\,m}$.

Applications and Conceptual Questions

• Helicopter Propellers: Helicopters have two propellers to balance angular momentum. If there were only one, the body of the helicopter would rotate in the opposite direction to conserve angular momentum.
• Earth's Size Shrink/Expansion: If Earth's size doubles ($R \rightarrow 2R$),
  • $I_1 = \frac{2}{5}MR^2$ while $I_2 = \frac{2}{5}M(2R)^2 = 4 I_1$.
  • By $I_1 \omega_1 = I_2 \omega_2$, the new angular velocity $\omega_2 = \frac{\omega_1}{4}$.
  • Since $T = \frac{2\pi}{\omega}$, the new time period $T_2 = 4 T_1 = 4(24) = 96\,\text{hours}$.
• K.E. of Earth: Mass $M = 6 \times 10^{24}\,\text{kg}$, radius $R = 6400\,\text{km}$. Period $T = 86400\,\text{s}$.
  • $I = \frac{2}{5} MR^2 = 9.83 \times 10^{37}\,\text{kg\,m}^2$.
  • $\omega = \frac{2\pi}{86400}\,\text{rad/s}$.
  • $K.E. = \frac{1}{2}I\omega^2 \approx 2.6 \times 10^{29}\,\text{J}$.
• Ballet Dancer: When a dancer spins at $2.4\,\text{rev/s}$ with arms outstretched and then folds them (reducing $I$ from $I$ to $0.6I$), their new spin rate is:
  • $I_1 f_1 = I_2 f_2 \rightarrow I(2.4) = 0.6I(f_2) \rightarrow f_2 = 4.0\,\text{rev/s}$.