Concept Card: Equations of Rotational Motion & Comparison Between Linear and Rotational Motion
1. Fundamental Angular Kinematic Variables:
When an extended rigid body rotates about a fixed axis, the position of every constituent particle is uniquely determined by a single angular coordinate $\theta$. The fundamental kinematic quantities of rotational motion are defined analogously to linear mechanics:
- Angular Position / Displacement ($\theta$): Measured in radians ($\text{rad}$). The angular displacement $\Delta\theta = \theta_2 – \theta_1$ represents the angle swept by any radial line fixed within the rotating body.
- Instantaneous Angular Velocity ($\vec{\omega}$): Defined as the time rate of change of angular position:
$$\vec{\omega} = \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} \hat{k} = \frac{d\theta}{dt} \hat{k} \quad (\text{SI Unit: rad/s})$$
Average angular velocity over a time interval $\Delta t$:
$$\omega_{\text{avg}} = \frac{\Delta\theta}{\Delta t} = \frac{\theta_f – \theta_i}{t_f – t_i}$$ - Instantaneous Angular Acceleration ($\vec{\alpha}$): Defined as the time rate of change of angular velocity:
$$\vec{\alpha} = \frac{d\vec{\omega}}{dt} = \frac{d^2\theta}{dt^2} \hat{k} = \omega \frac{d\omega}{d\theta} \hat{k} \quad (\text{SI Unit: rad/s}^2)$$
Average angular acceleration:
$$\alpha_{\text{avg}} = \frac{\Delta\omega}{\Delta t} = \frac{\omega_f – \omega_i}{t_f – t_i}$$
2. Rigorous Derivation of the Equations of Rotational Motion (Constant $\alpha$)
When the angular acceleration is uniform ($\alpha = \text{constant}$), the kinematic differential equations integrate directly into the standard equations of rotational motion:
Equation 1: Angular Velocity-Time Relation
By definition, $\frac{d\omega}{dt} = \alpha$. Separating variables and integrating from $t = 0$ (where $\omega = \omega_0$) to time $t$ (where $\omega = \omega$):
$$\int_{\omega_0}^\omega d\omega = \int_0^t \alpha dt \implies [\omega]_{\omega_0}^\omega = \alpha [t]_0^t$$
$$\mathbf{\omega = \omega_0 + \alpha t}$$
Equation 2: Angular Displacement-Time Relation
Substitute $\omega = \frac{d\theta}{dt} = \omega_0 + \alpha t$:
$$\int_0^\theta d\theta = \int_0^t (\omega_0 + \alpha t) dt$$
$$\mathbf{\theta = \omega_0 t + \frac{1}{2} \alpha t^2}$$
Equation 3: Angular Velocity-Displacement Relation
Using the chain rule form $\alpha = \omega \frac{d\omega}{d\theta}$:
$$\omega d\omega = \alpha d\theta \implies \int_{\omega_0}^\omega \omega d\omega = \int_0^\theta \alpha d\theta$$
$$\left[ \frac{\omega^2}{2} \right]_{\omega_0}^\omega = \alpha [\theta]_0^\theta \implies \frac{\omega^2 – \omega_0^2}{2} = \alpha \theta$$
$$\mathbf{\omega^2 = \omega_0^2 + 2 \alpha \theta}$$
Equation 4: Average Angular Velocity Form
$$\mathbf{\theta = \left(\frac{\omega_0 + \omega}{2}\right) t}$$
Equation 5: Angular Displacement in the $n$-th Second
The angle swept between $t = (n – 1)$ and $t = n$:
$$\theta_n = \theta(n) – \theta(n – 1) = \left[ \omega_0 n + \frac{1}{2}\alpha n^2 \right] – \left[ \omega_0 (n – 1) + \frac{1}{2}\alpha (n – 1)^2 \right]$$
$$\mathbf{\theta_n = \omega_0 + \frac{\alpha}{2} (2n – 1)}$$
3. Handling Variable Angular Acceleration (Calculus Framework)
If $\alpha$ is non-constant, standard algebraic formulas fail and direct integration is strictly mandatory:
- Time-Dependent Acceleration $\alpha = f(t)$:
$$\omega(t) = \omega_0 + \int_0^t f(t’) dt’, \quad \theta(t) = \theta_0 + \int_0^t \omega(t’) dt’$$ - Displacement-Dependent Acceleration $\alpha = f(\theta)$:
$$\int_{\omega_0}^\omega \omega d\omega = \int_{\theta_0}^\theta f(\theta) d\theta \implies \frac{1}{2}(\omega^2 – \omega_0^2) = \int_{\theta_0}^\theta f(\theta) d\theta$$ - Velocity-Dependent Deceleration $\alpha = f(\omega)$ (e.g., Viscous / Aerodynamic Drag):
$$t = \int_{\omega_0}^\omega \frac{d\omega’}{f(\omega’)}, \quad \theta = \int_{\omega_0}^\omega \frac{\omega’ d\omega’}{f(\omega’)}$$
4. Exhaustive Comparison: Linear vs. Rotational Dynamics
The mathematical architecture of rotational mechanics mirrors translational mechanics through an exact one-to-one mapping of physical quantities:
| Translational Parameter | Rotational Parameter | Bridging Transformation |
|---|---|---|
| Linear Displacement $x$ or $\vec{s}$ | Angular Displacement $\theta$ or $\vec{\theta}$ | $s = r \theta$ (tangential arc length) |
| Linear Velocity $v = \frac{dx}{dt}$ | Angular Velocity $\omega = \frac{d\theta}{dt}$ | $\vec{v} = \vec{\omega} \times \vec{r}$ ($v = r \omega$) |
| Linear Acceleration $a = \frac{dv}{dt}$ | Angular Acceleration $\alpha = \frac{d\omega}{dt}$ | $a_t = r \alpha$ (tangential acceleration) |
| Inertia / Mass $M$ (intrinsic scalar) | Moment of Inertia $I = \int r^2 dm$ (axis-dependent) | $I = M k^2$ ($k$: radius of gyration) |
| Force $\vec{F} = \frac{d\vec{p}}{dt} = M \vec{a}$ | Torque $\vec{\tau} = \frac{d\vec{L}}{dt} = I \vec{\alpha}$ | $\vec{\tau} = \vec{r} \times \vec{F}$ |
| Linear Momentum $\vec{p} = M \vec{v}$ | Angular Momentum $\vec{L} = I \vec{\omega}$ | $\vec{L} = \vec{r} \times \vec{p}$ |
| Linear Kinetic Energy $K = \frac{1}{2} M v^2 = \frac{p^2}{2M}$ | Rotational Kinetic Energy $K = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$ | $K_{\text{rot}} = \frac{1}{2} L \omega$ |
| Work Done $W = \int \vec{F} \cdot d\vec{r}$ | Work Done $W = \int \vec{\tau} \cdot d\vec{\theta}$ | $dW = \tau d\theta$ |
| Power $P = \vec{F} \cdot \vec{v}$ | Power $P = \vec{\tau} \cdot \vec{\omega}$ | $P = \tau \omega$ |
| Linear Impulse $\vec{J} = \int \vec{F} dt = \Delta \vec{p}$ | Angular Impulse $\vec{J}_\theta = \int \vec{\tau} dt = \Delta \vec{L}$ | $\vec{J}_\theta = \vec{r} \times \vec{J}$ |
| Conservation: If $\sum \vec{F}_{\text{ext}} = 0 \implies \vec{p} = \text{const}$ | Conservation: If $\sum \vec{\tau}_{\text{ext}} = 0 \implies \vec{L} = \text{const}$ | Rotational Noether Symmetry |
5. Vector Nature of Angular Variables & Core Traps
- Trap 1 (Infinitesimal vs. Finite Displacements):
- Infinitesimal angular displacement $d\vec{\theta}$: A true vector! It commutes under addition: $d\vec{\theta}_1 + d\vec{\theta}_2 = d\vec{\theta}_2 + d\vec{\theta}_1$.
- Finite angular displacement $\Delta\theta$: NOT a vector! Finite rotations about different axes do not commute: rotating a book $90^\circ$ about the $x$-axis then $90^\circ$ about the $y$-axis yields a completely different physical orientation than rotating about $y$ first, then $x$ ($\Delta\theta_1 + \Delta\theta_2 \ne \Delta\theta_2 + \Delta\theta_1$).
- Trap 2 (Unit Conversion Traps – rpm vs. rad/s):
$$1\text{ revolution} = 2\pi\text{ radians}$$
$$1\text{ rpm (revolution per minute)} = \frac{2\pi\text{ rad}}{60\text{ s}} = \mathbf{\frac{\pi}{30}\text{ rad/s}}$$
Never substitute frequencies in $\text{rpm}$ directly into kinematic formulas; always convert to $\text{rad/s}$ first! - Trap 3 (Variable Acceleration vs. Constant $\alpha$ Formulas): In situations where $\tau$ depends on angle (e.g., a simple/compound pendulum where $\tau = -M g \frac{L}{2} \sin\theta$), $\alpha = -\frac{3g}{2L}\sin\theta$ is not constant! Applying $\omega^2 = 2\alpha\theta$ is a fatal error. One must use energy conservation or $\omega d\omega = \alpha d\theta$.
Solved Examples
Example 1 (Direct Conceptual Application – Uniform Angular Acceleration & Revolution Counts):
A laboratory centrifuge rotor starts from an initial angular speed of $1200\text{ rpm}$ and accelerates uniformly to its maximum operational speed of $4800\text{ rpm}$ in a time duration of $t = 12.0\text{ s}$.
(a) Determine the initial and final angular speeds in $\text{rad/s}$.
(b) Calculate the angular acceleration $\alpha$.
(c) Find the total angular displacement $\theta$ and the total number of complete revolutions $N$ made during this $12.0\text{ s}$ acceleration phase.
(d) Calculate the angular displacement swept during the 10th second of motion.
(e) For a test tube sample located at radius $r = 0.15\text{ m}$ from the rotation axis, calculate its tangential acceleration ($a_t$) and centripetal acceleration ($a_c$) at $t = 12.0\text{ s}$.
Solution:
(a) Conversion to Radians per Second:
$$\omega_0 = 1200\text{ rpm} = 1200 \times \frac{2\pi\text{ rad}}{60\text{ s}} = \mathbf{40\pi\text{ rad/s}} \approx 125.66\text{ rad/s}$$
$$\omega = 4800\text{ rpm} = 4800 \times \frac{2\pi\text{ rad}}{60\text{ s}} = \mathbf{160\pi\text{ rad/s}} \approx 502.65\text{ rad/s}$$
(b) Constant Angular Acceleration:
$$\alpha = \frac{\omega – \omega_0}{t} = \frac{160\pi – 40\pi}{12.0} = \frac{120\pi}{12.0} = \mathbf{10\pi\text{ rad/s}^2} \approx 31.416\text{ rad/s}^2$$
(c) Total Angular Displacement and Revolution Count:
Using the average velocity relation:
$$\theta = \left(\frac{\omega_0 + \omega}{2}\right) t = \left(\frac{40\pi + 160\pi}{2}\right) (12.0) = (100\pi)(12.0) = \mathbf{1200\pi\text{ radians}}$$
Number of complete revolutions $N$:
$$N = \frac{\theta}{2\pi} = \frac{1200\pi\text{ rad}}{2\pi\text{ rad/rev}} = \mathbf{600\text{ revolutions}}$$
(d) Angular Displacement in the 10th Second ($n = 10$):
$$\theta_{10} = \omega_0 + \frac{\alpha}{2} (2n – 1) = 40\pi + \frac{10\pi}{2} (2 \times 10 – 1) = 40\pi + 5\pi(19) = 40\pi + 95\pi = \mathbf{135\pi\text{ rad}} \approx 424.12\text{ rad}$$
Revolutions in the 10th second: $N_{10} = \frac{135\pi}{2\pi} = 67.5\text{ revolutions}$.
(e) Linear Acceleration Components at $r = 0.15\text{ m}$ at $t = 12.0\text{ s}$:
Tangential acceleration (constant throughout):
$$a_t = \alpha r = (10\pi\text{ rad/s}^2)(0.15\text{ m}) = \mathbf{1.5\pi\text{ m/s}^2} \approx 4.712\text{ m/s}^2$$
Centripetal acceleration at operating speed ($\omega = 160\pi\text{ rad/s}$):
$$a_c = \omega^2 r = (160\pi)^2 (0.15) = (25600\pi^2)(0.15) = \mathbf{3840\pi^2\text{ m/s}^2} \approx 37902\text{ m/s}^2 \approx 3867 g!$$
Takeaway: In high-speed rotational devices, centripetal acceleration ($a_c \propto \omega^2$) completely dwarfs tangential acceleration ($a_t \propto \alpha$).
Example 2 (Mathematical Formulation – Variable Angular Acceleration & Damped Motion):
A heavy flywheel starts from rest ($\omega_0 = 0$) at $t = 0$. For the time interval $0 \le t \le 3.0\text{ s}$, its angular acceleration is driven according to:
$$\alpha(t) = 6.0 t – 2.0 t^2 \quad (\text{in rad/s}^2)$$
(a) Derive the expressions for $\omega(t)$ and $\theta(t)$.
(b) Determine the maximum angular speed $\omega_{\max}$ attained during this interval.
(c) Calculate the total angle $\theta_1$ swept in the first $3.0\text{ s}$.
(d) For $t > 3.0\text{ s}$, the driving torque ceases, and a viscous damping torque produces a retarding acceleration proportional to angular velocity: $\alpha = -k \omega$, where $k = 0.50\text{ s}^{-1}$. Calculate the total additional angle $\theta_2$ the flywheel rotates before coming to rest.
Solution:
(a) Integration for Velocity and Displacement:
Since $\alpha(t) = \frac{d\omega}{dt} = 6.0 t – 2.0 t^2$:
$$\omega(t) = \int_0^t (6.0 t’ – 2.0 t’^2) dt’ = \mathbf{3.0 t^2 – \frac{2}{3} t^3} \quad (\text{for } 0 \le t \le 3.0\text{ s})$$
Integrating $\omega(t) = \frac{d\theta}{dt}$ with $\theta(0) = 0$:
$$\theta(t) = \int_0^t \left(3.0 t’^2 – \frac{2}{3} t’^3\right) dt’ = \mathbf{t^3 – \frac{1}{6} t^4}$$
(b) Maximum Angular Velocity:
Maximum $\omega$ occurs when $\frac{d\omega}{dt} = \alpha(t) = 0$:
$$6.0 t – 2.0 t^2 = 0 \implies 2.0 t (3.0 – t) = 0 \implies t = 3.0\text{ s}$$
Substituting $t = 3.0\text{ s}$:
$$\omega_{\max} = \omega(3.0) = 3.0(3.0)^2 – \frac{2}{3}(3.0)^3 = 27.0 – 18.0 = \mathbf{9.0\text{ rad/s}}$$
(c) Angular Displacement in First $3.0\text{ s}$:
$$\theta_1 = \theta(3.0) = (3.0)^3 – \frac{1}{6}(3.0)^4 = 27.0 – \frac{81.0}{6} = 27.0 – 13.5 = \mathbf{13.5\text{ radians}}$$
(d) Viscous Deceleration Phase ($t > 3.0\text{ s}$):
Here $\alpha = \omega \frac{d\omega}{d\theta} = -k \omega$.
Canceling $\omega$ (for $\omega \ne 0$):
$$\frac{d\omega}{d\theta} = -k \implies d\omega = -k d\theta$$
Integrating from initial state ($\omega = 9.0\text{ rad/s}$, $\theta = 0$) to complete stop ($\omega = 0$, $\theta = \theta_2$):
$$\int_{9.0}^0 d\omega = -k \int_0^{\theta_2} d\theta$$
$$0 – 9.0 = -k \theta_2 \implies \theta_2 = \frac{9.0}{k} = \frac{9.0}{0.50\text{ s}^{-1}} = \mathbf{18.0\text{ radians}}$$
Total angular displacement from start to rest:
$$\theta_{\text{total}} = \theta_1 + \theta_2 = 13.5 + 18.0 = \mathbf{31.5\text{ radians}}$$
Takeaway: When acceleration is velocity-dependent, changing the independent variable from time $t$ to displacement $\theta$ via $\alpha = \omega \frac{d\omega}{d\theta}$ allows one-step direct integration for displacement without computing time.
Example 3 (Multi-Concept Linkage – Coupled Linear and Rotational System with Friction Torque):
A uniform solid cylinder of mass $M = 4.0\text{ kg}$ and radius $R = 0.20\text{ m}$ is mounted on a frictionless horizontal axle. A light inextensible cord wrapped around the cylinder supports a hanging block of mass $m = 2.0\text{ kg}$. In addition, a constant frictional brake torque $\tau_b = 1.0\text{ N}\cdot\text{m}$ opposes the rotation of the cylinder. (Take $g = 9.8\text{ m/s}^2$).
(a) Derive the downward linear acceleration $a$ of the block, the angular acceleration $\alpha$ of the cylinder, and the cord tension $T$.
(b) If the block is released from rest and descends a distance $h = 3.0\text{ m}$, find its speed $v$ and the angular speed $\omega$ of the cylinder.
(c) Verify the result using the Work-Energy Theorem.
Solution:
(a) Dynamical Equations:
Moment of inertia of cylinder: $I = \frac{1}{2} M R^2 = \frac{1}{2}(4.0)(0.20)^2 = 0.08\text{ kg}\cdot\text{m}^2$.
– For the hanging block (downward motion):
$$m g – T = m a \implies T = m(g – a) \quad \text{— (Equation 1)}$$
– For the cylinder (rotation about fixed axis):
The cord tension produces a driving torque $T R$, while $\tau_b$ opposes:
$$\tau_{\text{net}} = T R – \tau_b = I \alpha$$
Since the cord unwinds without slipping: $a = \alpha R \implies \alpha = a / R$.
$$T R – \tau_b = I \left(\frac{a}{R}\right) \implies T = \frac{I}{R^2} a + \frac{\tau_b}{R} = \frac{1}{2} M a + \frac{\tau_b}{R} \quad \text{— (Equation 2)}$$
Equating Equation 1 and Equation 2:
$$m(g – a) = \frac{1}{2} M a + \frac{\tau_b}{R} \implies m g – \frac{\tau_b}{R} = \left(m + \frac{M}{2}\right) a$$
$$\mathbf{a = \frac{m g – \frac{\tau_b}{R}}{m + \frac{M}{2}}}$$
Substitute numerical values ($m = 2.0\text{ kg}$, $M = 4.0\text{ kg}$, $R = 0.20\text{ m}$, $\tau_b = 1.0\text{ N}\cdot\text{m}$):
$$\frac{\tau_b}{R} = \frac{1.0\text{ N}\cdot\text{m}}{0.20\text{ m}} = 5.0\text{ N}$$
$$m g = (2.0)(9.8) = 19.6\text{ N}$$
$$m + \frac{M}{2} = 2.0 + 2.0 = 4.0\text{ kg}$$
$$a = \frac{19.6 – 5.0}{4.0} = \frac{14.6}{4.0} = \mathbf{3.65\text{ m/s}^2}$$
Angular acceleration:
$$\alpha = \frac{a}{R} = \frac{3.65}{0.20} = \mathbf{18.25\text{ rad/s}^2}$$
Cord tension:
$$T = m(g – a) = 2.0(9.8 – 3.65) = 2.0(6.15) = \mathbf{12.30\text{ N}}$$
(b) Velocity After Descending $h = 3.0\text{ m}$:
$$v = \sqrt{2 a h} = \sqrt{2(3.65)(3.0)} = \sqrt{21.9} \approx \mathbf{4.68\text{ m/s}}$$
$$\omega = \frac{v}{R} = \frac{4.68}{0.20} = \mathbf{23.40\text{ rad/s}}$$
(c) Work-Energy Theorem Verification:
Total angular rotation of cylinder: $\theta = \frac{h}{R} = \frac{3.0}{0.20} = 15.0\text{ rad}$.
– Work done by gravity: $W_g = m g h = (2.0)(9.8)(3.0) = 58.8\text{ J}$.
– Work done against braking torque: $W_{\text{loss}} = \tau_b \theta = (1.0)(15.0) = 15.0\text{ J}$.
– Net work done on the system: $W_{\text{net}} = 58.8 – 15.0 = 43.8\text{ J}$.
– Total kinetic energy acquired:
$$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2}(2.0)(21.9) + \frac{1}{2}(0.08)(23.4)^2 = 21.9 + 21.9 = \mathbf{43.8\text{ J}}$$
The work-energy theorem ($W_{\text{net}} = \Delta K$) is verified.
Takeaway: Resisting torque $\tau_b$ acts as an effective retarding linear force $\tau_b / R$ opposing the gravitational pull of the suspended mass.
Example 4 (Edge Case – Non-Linear Aerodynamic Retardation $\alpha = -k\sqrt{\omega}$):
A gas turbine rotor spinning at an initial angular speed $\omega_0 = 100.0\text{ rad/s}$ is cut off from fuel and experiences drag such that its angular deceleration is governed by:
$$\alpha = \frac{d\omega}{dt} = -k \sqrt{\omega}$$
where $k = 2.0\text{ rad}^{1/2}\text{s}^{-3/2}$.
(a) Calculate the exact time $T_{\text{stop}}$ required for the rotor to come to a complete stop.
(b) Derive the functional form of $\omega(t)$ during the rundown.
(c) Calculate the total angular displacement $\theta_{\text{total}}$ swept before stopping.
(d) Contrast this finite stopping time with linear viscous damping ($\alpha = -k\omega$).
Solution:
(a) Stopping Time:
Separating variables:
$$\frac{d\omega}{\sqrt{\omega}} = -k dt \implies \omega^{-1/2} d\omega = -k dt$$
Integrating from $t = 0$ ($\omega = \omega_0$) to $t = T_{\text{stop}}$ ($\omega = 0$):
$$\int_{\omega_0}^0 \omega^{-1/2} d\omega = -k \int_0^{T_{\text{stop}}} dt$$
$$\left[ 2\sqrt{\omega} \right]_{\omega_0}^0 = -k T_{\text{stop}} \implies 0 – 2\sqrt{\omega_0} = -k T_{\text{stop}}$$
$$\mathbf{T_{\text{stop}} = \frac{2\sqrt{\omega_0}}{k}} = \frac{2\sqrt{100.0}}{2.0} = \frac{2(10.0)}{2.0} = \mathbf{10.0\text{ seconds}}$$
(b) Angular Velocity Function:
Integrating up to time $t \le T_{\text{stop}}$:
$$2\sqrt{\omega(t)} – 2\sqrt{\omega_0} = -k t \implies \sqrt{\omega(t)} = \sqrt{\omega_0} – \frac{k}{2} t$$
$$\mathbf{\omega(t) = \left(10.0 – t\right)^2 \quad (\text{for } 0 \le t \le 10.0\text{ s})}$$
(c) Total Angular Displacement:
Method 1 (Time Integration):
$$\theta_{\text{total}} = \int_0^{T_{\text{stop}}} \omega(t) dt = \int_0^{10.0} (10.0 – t)^2 dt = \left[ -\frac{(10.0 – t)^3}{3} \right]_0^{10.0} = 0 – \left( -\frac{1000}{3} \right) = \mathbf{\frac{1000}{3}\text{ rad}} \approx 333.33\text{ rad}$$
Method 2 (Phase Space Integration):
Using $\alpha = \omega \frac{d\omega}{d\theta} = -k \sqrt{\omega}$:
$$\frac{\omega d\omega}{\sqrt{\omega}} = -k d\theta \implies \omega^{1/2} d\omega = -k d\theta$$
$$\int_{\omega_0}^0 \omega^{1/2} d\omega = -k \int_0^{\theta_{\text{total}}} d\theta$$
$$\left[ \frac{2}{3} \omega^{3/2} \right]_{\omega_0}^0 = -k \theta_{\text{total}} \implies -\frac{2}{3} \omega_0^{3/2} = -k \theta_{\text{total}}$$
$$\theta_{\text{total}} = \frac{2 \omega_0^{3/2}}{3 k} = \frac{2(100.0)^{3/2}}{3(2.0)} = \frac{2(1000)}{6.0} = \mathbf{\frac{1000}{3}\text{ rad}}$$
(d) Conceptual Contrast:
Under linear damping $\alpha = -k\omega$, $\omega(t) = \omega_0 e^{-kt}$, which approaches zero asymptotically as $t \to \infty$ (infinite stopping time). By contrast, power-law damping with exponent $p < 1$ (such as $\alpha \propto -\omega^{1/2}$) decelerates the body to a dead stop in a strictly finite time.
Takeaway: Sublinear velocity dependence of retarding torque allows the angular velocity to reach absolute zero in a well-defined finite duration.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A wheel starts from rest and rotates with a constant angular acceleration $\alpha = 4.0\text{ rad/s}^2$. The angular displacement swept by the wheel during the 5th second of its motion is:
(A) $10.0\text{ rad}$
(B) $18.0\text{ rad}$
(C) $20.0\text{ rad}$
(D) $36.0\text{ rad}$
Problem 2 (JEE Main – Single Correct):
A motor shaft rotating at an initial frequency of $240\text{ rpm}$ slows down uniformly to $60\text{ rpm}$ in a time duration of $6.0\text{ seconds}$ under a constant frictional torque. The number of complete revolutions made by the shaft during this time interval is:
(A) $10$
(B) $15$
(C) $20$
(D) $30$
Problem 3 (JEE Main – Single Correct):
The angular displacement of a particle moving in a circular path is given by the time function $\theta(t) = 2.0 t^3 – 5.0 t^2 + 8.0 t + 1.0$ (with $\theta$ in radians and $t$ in seconds). The angular acceleration of the particle is zero at time $t$ equal to:
(A) $\frac{5}{6}\text{ s}$
(B) $\frac{6}{5}\text{ s}$
(C) $\frac{5}{12}\text{ s}$
(D) $1.0\text{ s}$
Problem 4 (JEE Main – Single Correct):
A ceiling fan rotating at initial angular speed $\omega_0$ is switched off and comes to rest after completing $36$ complete rotations under a uniform retarding torque. How many rotations did the fan complete during the time its angular speed dropped from $\omega_0$ to $\frac{\omega_0}{2}$?
(A) $9$
(B) $18$
(C) $27$
(D) $30$
Problem 5 (JEE Main – Single Correct):
In establishing the formal mathematical correspondence between linear translational dynamics and fixed-axis rotational dynamics, which pair of variables represents the exact dynamical analogues of mass $M$ and force $F$?
(A) Angular momentum $L$ and Torque $\tau$
(B) Moment of inertia $I$ and Torque $\tau$
(C) Moment of inertia $I$ and Angular velocity $\omega$
(D) Angular momentum $L$ and Angular acceleration $\alpha$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements regarding the vector properties of angular kinematic quantities is/are correct?
(A) Infinitesimal angular displacements $d\vec{\theta}$ obey the commutative law of vector addition ($d\vec{\theta}_1 + d\vec{\theta}_2 = d\vec{\theta}_2 + d\vec{\theta}_1$) and behave as true vectors.
(B) Finite angular displacements do not commute under vector addition and therefore cannot be treated as vector quantities.
(C) Instantaneous angular velocity $\vec{\omega}$ and angular acceleration $\vec{\alpha}$ are axial vectors (pseudovectors).
(D) For any particle on a rotating rigid body, the tangential acceleration vector is always perpendicular to the inward radial centripetal acceleration vector.
Problem 7 (JEE Advanced – One or More Correct):
A rigid body is rotating about a fixed axis with angular velocity $\omega$ and angular acceleration $\alpha$. For a particle located at a perpendicular distance $r$ from the rotation axis:
(A) The magnitude of its total linear acceleration is $a = r\sqrt{\alpha^2 + \omega^4}$.
(B) If $\alpha < 0$ and $\omega > 0$, the linear speed of the particle is decreasing.
(C) The time rate of change of the particle’s speed is $\frac{dv}{dt} = \alpha r$.
(D) The magnitude of the rate of change of the particle’s velocity vector is $\left|\frac{d\vec{v}}{dt}\right| = r\sqrt{\alpha^2 + \omega^4}$.
Problem 8 (JEE Advanced – One or More Correct):
A flywheel of moment of inertia $I$ is spinning with initial angular velocity $\omega_0$. It is subjected to a retarding torque directly proportional to its angular speed, $\tau = c \omega$, where $c$ is a positive constant:
(A) The instantaneous angular velocity is given by $\omega(t) = \omega_0 e^{-(c/I)t}$.
(B) The total angle swept by the flywheel before coming to rest is $\theta_{\text{total}} = \frac{I \omega_0}{c}$.
(C) The flywheel takes an infinite time to come to a complete stop.
(D) The initial rate of energy dissipation by the retarding torque is $P_0 = c \omega_0^2$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A particle moves in a circular path of radius $R = 2.0\text{ m}$. Its angular speed increases uniformly from $\omega_1 = 2.0\text{ rad/s}$ to $\omega_2 = 8.0\text{ rad/s}$ in a time duration of $t = 3.0\text{ s}$. Calculate the magnitude of the tangential acceleration $a_t$ (in $\text{m/s}^2$) of the particle.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A heavy industrial flywheel rotating at an operational speed of $600\text{ rpm}$ is brought to rest under a uniform angular deceleration of $\alpha = 2\pi\text{ rad/s}^2$. Calculate the total number of complete revolutions $N$ made by the flywheel before coming to a stop.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (B) | 3. (A) | 4. (C) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 4 | 10. 50
Solution 1:
Displacement in the $n$-th second under constant $\alpha$ with $\omega_0 = 0$:
$$\theta_n = \omega_0 + \frac{\alpha}{2}(2n – 1) = 0 + \frac{4.0}{2}(2 \times 5 – 1) = 2.0(9) = \mathbf{18.0\text{ rad}}$$
Correct Option: (B)
Solution 2:
Average rotational frequency:
$$n_{\text{avg}} = \frac{n_1 + n_2}{2} = \frac{240 + 60}{2} = 150\text{ rpm} = \frac{150}{60}\text{ rev/s} = 2.5\text{ rev/s}$$
Total revolutions completed in $t = 6.0\text{ s}$:
$$N = n_{\text{avg}} \times t = (2.5\text{ rev/s})(6.0\text{ s}) = \mathbf{15\text{ revolutions}}$$
Correct Option: (B)
Solution 3:
Given $\theta(t) = 2.0 t^3 – 5.0 t^2 + 8.0 t + 1.0$:
$$\omega(t) = \frac{d\theta}{dt} = 6.0 t^2 – 10.0 t + 8.0$$
$$\alpha(t) = \frac{d\omega}{dt} = 12.0 t – 10.0$$
Setting $\alpha(t) = 0$:
$$12.0 t – 10.0 = 0 \implies t = \frac{10.0}{12.0} = \mathbf{\frac{5}{6}\text{ s}}$$
Correct Option: (A)
Solution 4:
Using $\omega^2 = \omega_0^2 – 2\alpha \theta$:
– For total stop from $\omega_0$: $0 = \omega_0^2 – 2\alpha (36) \implies 2\alpha (36) = \omega_0^2$.
– For dropping to $\omega_0/2$:
$$\left(\frac{\omega_0}{2}\right)^2 = \omega_0^2 – 2\alpha \theta_1 \implies \frac{\omega_0^2}{4} = \omega_0^2 – 2\alpha \theta_1 \implies 2\alpha \theta_1 = \frac{3}{4} \omega_0^2$$
Dividing the two equations:
$$\frac{2\alpha \theta_1}{2\alpha (36)} = \frac{\frac{3}{4}\omega_0^2}{\omega_0^2} \implies \frac{\theta_1}{36} = \frac{3}{4} \implies \theta_1 = 36 \times \frac{3}{4} = \mathbf{27\text{ rotations}}$$
Correct Option: (C)
Solution 5:
In linear mechanics, mass $M$ quantifies translational inertia and force $F = M a$ causes linear acceleration. In rotational mechanics, moment of inertia $I$ quantifies rotational inertia and torque $\tau = I \alpha$ causes angular acceleration. Hence, $M \leftrightarrow I$ and $F \leftrightarrow \tau$.
Correct Option: (B)
Solution 6:
– (A, B) True: Infinitesimal angular rotations commute and follow vector rules, while finite rotations do not commute and are not vectors.
– (C) True: Angular velocity and acceleration change sign under coordinate inversion (pseudovectors) and align with the axis of rotation.
– (D) True: $\vec{a}_t$ is directed tangentially while $\vec{a}_c$ is directed radially inward; they are mutually orthogonal at all times ($\vec{a}_t \cdot \vec{a}_c = 0$).
All options are correct.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: $a = \sqrt{a_t^2 + a_c^2} = \sqrt{(\alpha r)^2 + (\omega^2 r)^2} = r\sqrt{\alpha^2 + \omega^4}$.
– (B) True: Since speed is $v = \omega r$, $\frac{dv}{dt} = \alpha r$. When $\alpha < 0$, $\frac{dv}{dt} < 0$, so speed decreases.
– (C) True: The rate of change of speed is solely determined by tangential acceleration: $\frac{d|\vec{v}|}{dt} = a_t = \alpha r$.
– (D) True: $\left|\frac{d\vec{v}}{dt}\right| = |\vec{a}| = r\sqrt{\alpha^2 + \omega^4}$.
All options are correct.
Correct Options: (A, B, C, D)
Solution 8:
Given retarding torque $\tau = I \left(-\frac{d\omega}{dt}\right) = c \omega \implies \frac{d\omega}{dt} = -\frac{c}{I} \omega$.
– (A) True: Direct integration yields $\omega(t) = \omega_0 e^{-(c/I)t}$.
– (B) True: Using $\omega \frac{d\omega}{d\theta} = -\frac{c}{I} \omega \implies d\omega = -\frac{c}{I} d\theta$. Integrating: $0 – \omega_0 = -\frac{c}{I} \theta_{\text{total}} \implies \theta_{\text{total}} = \frac{I \omega_0}{c}$.
– (C) True: $\omega(t) \to 0$ only as $t \to \infty$.
– (D) True: $P_0 = \tau_0 \omega_0 = (c \omega_0)(\omega_0) = c \omega_0^2$.
All options are correct.
Correct Options: (A, B, C, D)
Solution 9:
Angular acceleration:
$$\alpha = \frac{\omega_2 – \omega_1}{t} = \frac{8.0 – 2.0}{3.0} = \frac{6.0}{3.0} = 2.0\text{ rad/s}^2$$
Tangential acceleration of the particle at radius $R = 2.0\text{ m}$:
$$a_t = \alpha R = (2.0\text{ rad/s}^2)(2.0\text{ m}) = \mathbf{4.0\text{ m/s}^2}$$
Correct Answer: 4
Solution 10:
Initial angular velocity:
$$\omega_0 = 600\text{ rpm} = 600 \times \frac{2\pi}{60} = 20\pi\text{ rad/s}$$
Using $\omega^2 = \omega_0^2 – 2\alpha \theta$ with $\omega = 0$ and $\alpha = 2\pi\text{ rad/s}^2$:
$$0 = (20\pi)^2 – 2(2\pi) \theta \implies 4\pi \theta = 400\pi^2 \implies \theta = \frac{400\pi^2}{4\pi} = 100\pi\text{ radians}$$
Total number of complete revolutions $N$:
$$N = \frac{\theta}{2\pi} = \frac{100\pi}{2\pi} = \mathbf{50}$$
Correct Answer: 50