Concept Card: Rotational Kinematics – Angular Displacement, Velocity & Acceleration
1. Physical Framework & Definitions:
A rigid body executes pure rotational motion about a fixed axis if every particle moves in a circular path lying in a plane perpendicular to the axis, with its center situated on the axis of rotation.
- Angular Position ($\theta$): The angle swept by a reference line fixed in the body relative to a spatial axis. Unit: Radian ($\text{rad}$), dimensionless ($[M^0 L^0 T^0]$). Fundamental conversion: $1\text{ rev} = 2\pi\text{ rad} = 360^\circ$.
- Finite vs. Infinitesimal Angular Displacement:
- Finite Angular Displacement ($\Delta\theta$): Is a SCALAR (or pseudo-scalar), NOT a vector, because finite 3D rotations do not obey vector addition commutativity ($\theta_1 + \theta_2 \ne \theta_2 + \theta_1$).
- Infinitesimal Angular Displacement ($d\vec{\theta}$): IS a true axial vector because infinitesimal rotations commute strictly ($d\vec{\theta}_1 + d\vec{\theta}_2 = d\vec{\theta}_2 + d\vec{\theta}_1$). Direction is given by the Right-Hand Thumb Rule along the rotation axis.
- Angular Velocity ($\vec{\omega}$): Time rate of change of angular displacement:
$$\mathbf{\vec{\omega} = \frac{d\vec{\theta}}{dt} = \lim_{\Delta t \to 0} \frac{\Delta\vec{\theta}}{\Delta t}}$$
An axial vector pointing along the axis of rotation. From rotational frequency $N$ in revolutions per minute ($\text{rpm}$): $\omega = \frac{2\pi N}{60}\text{ rad/s}$.
Rigid Body Invariant: At any given instant, all particles of a rotating rigid body have the identical angular velocity $\vec{\omega}$. - Angular Acceleration ($\vec{\alpha}$): Time rate of change of angular velocity:
$$\mathbf{\vec{\alpha} = \frac{d\vec{\omega}}{dt} = \frac{d^2\theta}{dt^2} = \omega \frac{d\omega}{d\theta}}$$
Units: $\text{rad/s}^2$. Points parallel to $\vec{\omega}$ if rotation is speeding up ($d\omega/dt > 0$); points antiparallel to $\vec{\omega}$ if slowing down ($d\omega/dt < 0$).
2. Vector Relations Connecting Linear & Angular Variables
For a particle located at position vector $\vec{r}$ relative to an origin on the rotation axis:
- Linear Velocity Vector:
$$\mathbf{\vec{v} = \vec{\omega} \times \vec{r}}$$
Magnitude: $v = \omega r_{\perp} = \omega R$, where $R$ is the perpendicular distance from the axis. - Linear Acceleration Decomposition:
$$\vec{a} = \frac{d\vec{v}}{dt} = (\vec{\alpha} \times \vec{r}) + (\vec{\omega} \times \vec{v}) = \vec{a}_t + \vec{a}_c$$- Tangential Acceleration ($\vec{a}_t = \vec{\alpha} \times \vec{r}$): Magnitude $a_t = \alpha R = \frac{dv}{dt}$. Directed along the tangent to the circular path; responsible for changing the speed.
- Centripetal Acceleration ($\vec{a}_c = \vec{\omega} \times \vec{v} = -\omega^2 \vec{R}$): Magnitude $a_c = \omega^2 R = \frac{v^2}{R}$. Directed radially inward toward the rotation axis; responsible for changing the direction of motion.
- Total Net Linear Acceleration: Since $\vec{a}_t \perp \vec{a}_c$:
$$\mathbf{a_{\text{net}} = \sqrt{a_t^2 + a_c^2} = \sqrt{(\alpha R)^2 + (\omega^2 R)^2} = R\sqrt{\alpha^2 + \omega^4}}$$ - Angle $\beta$ of total acceleration with the velocity vector (tangent):
$$\tan\beta = \frac{a_c}{a_t} = \frac{\omega^2 R}{\alpha R} = \frac{\omega^2}{\alpha}$$
3. Kinematics for Constant Angular Acceleration ($\alpha = \text{constant}$)
- $\omega = \omega_0 + \alpha t$
- $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$
- $\omega^2 = \omega_0^2 + 2\alpha \theta$
- $\theta = \left(\frac{\omega_0 + \omega}{2}\right)t$
- Angular displacement in the $n$-th second: $\theta_n = \omega_0 + \frac{\alpha}{2}(2n – 1)$
4. Relative Angular Velocity
For two moving particles $A$ and $B$, the relative angular velocity of $B$ as viewed from $A$ is:
$$\omega_{B/A} = \frac{(v_{\text{rel}})_{\perp}}{r_{AB}} = \frac{|\vec{v}_{B/A} \times \hat{r}_{AB}|}{|\vec{r}_B – \vec{r}_A|}$$
where $(v_{\text{rel}})_{\perp}$ is the relative velocity component perpendicular to the line of sight connecting $A$ and $B$.
5. Common JEE Pitfalls & Traps
- Trap 1 (Finite Angular Displacement): Finite rotations do not commute under addition ($\vec{\theta}_1 + \vec{\theta}_2 \ne \vec{\theta}_2 + \vec{\theta}_1$). Only infinitesimal angular displacement $d\vec{\theta}$ is a valid vector.
- Trap 2 (Centripetal vs. Tangential Acceleration): Centripetal acceleration exists whenever $\omega \ne 0$, even if $\alpha = 0$ (uniform circular motion). Tangential acceleration exists only if $\alpha \ne 0$.
- Trap 3 (Angle of Acceleration Reference): Pay strict attention to whether the angle of $\vec{a}_{\text{net}}$ is requested relative to the radial inward vector ($\tan\phi = a_t/a_c = \alpha/\omega^2$) or relative to the tangential velocity vector ($\tan\beta = a_c/a_t = \omega^2/\alpha$).
Solved Examples
Example 1 (Direct Conceptual Application – Constant $\alpha$, Revolutions & Rim Acceleration):
The angular speed of a motor flywheel increases uniformly from $N_1 = 120.0\text{ rpm}$ to $N_2 = 480.0\text{ rpm}$ in an elapsed time of $t = 16.0\text{ s}$.
(a) Determine the initial and final angular speeds $\omega_1$ and $\omega_2$ in $\text{rad/s}$.
(b) Calculate the constant angular acceleration $\alpha$ of the flywheel.
(c) Find the total angular displacement $\theta$ and the number of complete revolutions $n$ made in this interval.
(d) For a particle on the rim at radius $R = 0.50\text{ m}$, calculate the tangential acceleration, centripetal acceleration, and net acceleration at $t = 16.0\text{ s}$.
Solution:
(a) Angular velocities:
$$\omega_1 = \frac{2\pi(120.0)}{60} = 4\pi\text{ rad/s} \approx 12.566\text{ rad/s}$$
$$\omega_2 = \frac{2\pi(480.0)}{60} = 16\pi\text{ rad/s} \approx 50.265\text{ rad/s}$$
(b) Angular acceleration:
$$\alpha = \frac{\omega_2 – \omega_1}{t} = \frac{16\pi – 4\pi}{16.0} = \frac{12\pi}{16.0} = \frac{3\pi}{4}\text{ rad/s}^2 \approx 2.356\text{ rad/s}^2$$
(c) Total angular displacement:
$$\theta = \left(\frac{\omega_1 + \omega_2}{2}\right)t = (10\pi)(16.0) = 160\pi\text{ radians}$$
Revolutions: $n = \frac{160\pi}{2\pi} = 80.0\text{ revolutions}$.
(d) Linear accelerations on the rim ($R = 0.50\text{ m}$) at $t = 16.0\text{ s}$:
– Tangential acceleration: $a_t = \alpha R = \left(\frac{3\pi}{4}\right)(0.50) = \frac{3\pi}{8}\text{ m/s}^2 \approx 1.178\text{ m/s}^2$.
– Centripetal acceleration: $a_c = \omega_2^2 R = (16\pi)^2(0.50) = 128\pi^2\text{ m/s}^2 \approx 1263.31\text{ m/s}^2$.
– Total acceleration: $a_{\text{net}} = \sqrt{a_t^2 + a_c^2} = \sqrt{(1.178)^2 + (1263.31)^2} \approx 1263.31\text{ m/s}^2$.
Example 2 (Mathematical Manipulation – Variable Angular Acceleration Integration):
A body rotates about a fixed axis such that its angular position as a function of time is given by $\theta(t) = a t – b t^2 + c t^3$, where $a = 6.0\text{ rad/s}, b = 3.0\text{ rad/s}^2,$ and $c = 1.0\text{ rad/s}^3$.
(a) Derive expressions for the instantaneous angular velocity $\omega(t)$ and angular acceleration $\alpha(t)$.
(b) Find the time $t_1$ at which the angular acceleration vanishes, and determine the angular velocity at this instant.
(c) Calculate the average angular velocity and average angular acceleration over the interval from $t = 0$ to $t = 2.0\text{ s}$.
Solution:
(a) Differentiating position:
$$\omega(t) = \frac{d\theta}{dt} = 6.0 – 6.0 t + 3.0 t^2\text{ rad/s}$$
$$\alpha(t) = \frac{d\omega}{dt} = -6.0 + 6.0 t\text{ rad/s}^2$$
(b) Setting $\alpha(t_1) = 0 \implies -6.0 + 6.0 t_1 = 0 \implies t_1 = 1.0\text{ s}$.
$$\omega(1.0) = 6.0 – 6.0(1.0) + 3.0(1.0)^2 = 3.0\text{ rad/s}$$
(c) Over $t \in [0, 2.0\text{ s}]$:
$\theta(0) = 0, \quad \theta(2.0) = 6.0(2.0) – 3.0(4.0) + 1.0(8.0) = 8.0\text{ rad}$.
$$\omega_{\text{avg}} = \frac{8.0 – 0}{2.0} = 4.0\text{ rad/s}$$
$\omega(0) = 6.0\text{ rad/s}, \quad \omega(2.0) = 6.0 – 6.0(2.0) + 3.0(4.0) = 6.0\text{ rad/s}$.
$$\alpha_{\text{avg}} = \frac{6.0 – 6.0}{2.0} = 0\text{ rad/s}^2$$
Example 3 (Standard JEE Advanced Scenario – Net Acceleration Angle & Power Braking):
A circular disc starts rotating from rest with constant angular acceleration $\alpha_0 = 2.0\text{ rad/s}^2$.
(a) Find the elapsed time $t^*$ at which the magnitude of centripetal acceleration equals that of tangential acceleration for any point on the disc.
(b) After accelerating for $t = 5.0\text{ s}$, a braking mechanism introduces a speed-dependent retardational angular acceleration $\alpha_{\text{brake}} = -k\sqrt{\omega}$. If the disc comes to rest in $t_{\text{stop}} = 10.0\text{ s}$, determine the constant $k$.
(c) Find the total number of revolutions completed during the braking phase.
Solution:
(a) Setting $a_c = a_t \implies \omega^2 R = \alpha_0 R \implies \omega^2 = \alpha_0$.
Since $\omega = \alpha_0 t \implies (\alpha_0 t^*)^2 = \alpha_0 \implies (t^*)^2 = \frac{1}{\alpha_0} = \frac{1}{2.0} \implies t^* = \frac{1}{\sqrt{2}} \approx 0.707\text{ s}$.
(b) Speed at start of braking: $\omega_{\text{start}} = \alpha_0 t = 2.0(5.0) = 10.0\text{ rad/s}$.
$$\frac{d\omega}{\sqrt{\omega}} = -k\,dt \implies \left[ 2\sqrt{\omega} \right]_{10.0}^0 = -10.0k \implies -2\sqrt{10.0} = -10.0k \implies k = \frac{\sqrt{10}}{5} \approx 0.632\text{ rad}^{1/2}/\text{s}^{3/2}$$
(c) Angular displacement during braking:
$$\omega \frac{d\omega}{d\theta} = -k\sqrt{\omega} \implies \omega^{1/2}\,d\omega = -k\,d\theta$$
$$\int_{10.0}^0 \omega^{1/2}\,d\omega = -k\theta \implies -\frac{2}{3}(10.0)^{3/2} = -k\theta \implies \theta = \frac{20\sqrt{10}}{3\left(\frac{2\sqrt{10}}{10}\right)} = \frac{100}{3}\text{ rad}$$
Number of revolutions: $n = \frac{100/3}{2\pi} = \frac{50}{3\pi} \approx 5.31\text{ revolutions}$.
Example 4 (Edge Case – Relative Angular Velocity of Moving Particles):
Two particles $A$ and $B$ move along concentric circular paths of radii $r_A = 2.0\text{ m}$ and $r_B = 6.0\text{ m}$ centered at origin $O$. Both move counter-clockwise with uniform speeds $v_A = 4.0\text{ m/s}$ and $v_B = 6.0\text{ m/s}$ respectively.
(a) Determine their individual angular velocities $\omega_A$ and $\omega_B$ about the center $O$.
(b) When $O, A,$ and $B$ lie along the same radial line, find the relative angular velocity $\omega_{B/A}$ of $B$ as observed from $A$.
(c) When the position vectors $\vec{OA}$ and $\vec{OB}$ are mutually perpendicular ($\vec{OA} \perp \vec{OB}$), calculate the relative angular velocity $\omega_{B/A}$.
Solution:
(a) About center $O$: $\omega_A = \frac{v_A}{r_A} = \frac{4.0}{2.0} = 2.0\text{ rad/s}, \quad \omega_B = \frac{v_B}{r_B} = \frac{6.0}{6.0} = 1.0\text{ rad/s}$.
(b) When collinear: Relative velocity is perpendicular to the line of sight: $v_{\text{rel},\perp} = v_B – v_A = 6.0 – 4.0 = 2.0\text{ m/s}$. Distance is $r_{AB} = 6.0 – 2.0 = 4.0\text{ m}$.
$$\omega_{B/A} = \frac{v_{\text{rel},\perp}}{r_{AB}} = \frac{2.0}{4.0} = 0.50\text{ rad/s}$$
(c) When $\vec{OA} \perp \vec{OB}$:
Place $A$ at $(2.0, 0)$ with $\vec{v}_A = 4.0\hat{j}\text{ m/s}$, and $B$ at $(0, 6.0)$ with $\vec{v}_B = -6.0\hat{i}\text{ m/s}$.
$\vec{r}_{B/A} = -2.0\hat{i} + 6.0\hat{j}\text{ m} \implies r_{AB}^2 = (-2)^2 + 6^2 = 40.0\text{ m}^2$.
$\vec{v}_{B/A} = -6.0\hat{i} – 4.0\hat{j}\text{ m/s}$.
$\vec{r}_{B/A} \times \vec{v}_{B/A} = [(-2)(-4) – (6)(-6)]\hat{k} = (8 + 36)\hat{k} = 44.0\hat{k}\text{ m}^2/\text{s}$.
$$\omega_{B/A} = \frac{|\vec{r}_{B/A} \times \vec{v}_{B/A}|}{r_{AB}^2} = \frac{44.0}{40.0} = 1.10\text{ rad/s}$$
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
Which of the following rotational quantities is NOT a true vector?
(A) Infinitesimal angular displacement
(B) Finite angular displacement
(C) Instantaneous angular velocity
(D) Instantaneous angular acceleration
Problem 2 (JEE Main – Single Correct):
A wheel initially at rest accelerates uniformly with angular acceleration $\alpha = 4.0\text{ rad/s}^2$. The angular displacement turned through by the wheel during the 3rd second of its motion is:
(A) $10.0\text{ rad}$
(B) $12.0\text{ rad}$
(C) $18.0\text{ rad}$
(D) $20.0\text{ rad}$
Problem 3 (JEE Main – Single Correct):
A particle moves in a circle of radius $R = 0.50\text{ m}$ with uniform angular velocity $\omega = 10.0\text{ rad/s}$. The magnitude of its centripetal acceleration is:
(A) $5.0\text{ m/s}^2$
(B) $10.0\text{ m/s}^2$
(C) $50.0\text{ m/s}^2$
(D) $100.0\text{ m/s}^2$
Problem 4 (JEE Main – Single Correct):
An electric motor rotating at $1800\text{ rpm}$ slows down uniformly to $600\text{ rpm}$ while completing $80$ revolutions. The time taken to decelerate is:
(A) $2.0\text{ s}$
(B) $4.0\text{ s}$
(C) $6.0\text{ s}$
(D) $8.0\text{ s}$
Problem 5 (JEE Main – Single Correct):
The angular position of a rotating disc is given by $\theta(t) = 2.0 t^3 – 6.0 t^2 + 5.0\text{ radians}$. The disc comes to momentary rest at time:
(A) $t = 1.0\text{ s}$
(B) $t = 2.0\text{ s}$
(C) $t = 3.0\text{ s}$
(D) $t = 4.0\text{ s}$
Problem 6 (JEE Advanced – One or More Correct):
For a rigid body rotating about a fixed axis with angular velocity $\vec{\omega}$ and angular acceleration $\vec{\alpha}$:
(A) Every particle of the rigid body has the identical angular velocity $\vec{\omega}$ and angular acceleration $\vec{\alpha}$.
(B) The linear speed of any particle is directly proportional to its perpendicular distance from the axis of rotation.
(C) The tangential acceleration of every particle is zero if the angular acceleration is zero.
(D) The centripetal acceleration of every particle is zero if the angular acceleration is zero.
Problem 7 (JEE Advanced – One or More Correct):
A wheel starts from rest with time-dependent angular acceleration $\alpha(t) = \alpha_0 \left(1 – \frac{t}{T}\right)$ for $0 \le t \le T$, and $\alpha = 0$ for $t > T$:
(A) The maximum angular velocity occurs at time $t = T$.
(B) The maximum angular velocity attained is $\omega_{\text{max}} = \frac{1}{2}\alpha_0 T$.
(C) The total angular displacement in time $T$ is $\theta = \frac{1}{3}\alpha_0 T^2$.
(D) For all $t > T$, the centripetal acceleration of a point on the rim remains constant.
Problem 8 (JEE Advanced – One or More Correct):
A point on the rim of a rotating wheel of radius $R$ has linear acceleration $\vec{a}$. If the angle between $\vec{a}$ and the linear velocity $\vec{v}$ is $\phi$:
(A) $\tan\phi = \frac{a_c}{a_t} = \frac{\omega^2}{\alpha}$
(B) If $\phi = 45^\circ$, then $\omega^2 = \alpha$
(C) If the wheel rotates with uniform angular speed, then $\phi = 90^\circ$
(D) The rate of change of linear speed is $\frac{dv}{dt} = a\cos\phi = a_t$
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A wheel starts from rest with constant angular acceleration $\alpha$. If it turns through an angle $\theta_1$ in the first $2.0\text{ s}$ and an additional angle $\theta_2$ in the next $2.0\text{ s}$, calculate the ratio $\frac{\theta_2}{\theta_1}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A particle moves in a circle of radius $R = 2.0\text{ m}$ starting from rest with constant angular acceleration $\alpha = 0.25\text{ rad/s}^2$. Calculate the time $t$ (in seconds) when the total linear acceleration vector makes an angle of $45^\circ$ with its linear velocity vector.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (A) | 3. (C) | 4. (B) | 5. (B) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 3 | 10. 2
Solution 1:
Finite angular displacements do not commute under vector addition in 3D ($\theta_1 + \theta_2 \ne \theta_2 + \theta_1$), making finite rotation a scalar. Infinitesimal angular displacement $d\vec{\theta}$, angular velocity $\vec{\omega}$, and angular acceleration $\vec{\alpha}$ are true axial vectors.
Correct Option: (B)
Solution 2:
Angular displacement in the $n$-th second from rest ($\omega_0 = 0$):
$\theta_n = \frac{\alpha}{2}(2n – 1) = \frac{4.0}{2}(2(3) – 1) = 2.0(5) = 10.0\text{ rad}$.
Correct Option: (A)
Solution 3:
$a_c = \omega^2 R = (10.0)^2(0.50) = 100.0 \times 0.50 = 50.0\text{ m/s}^2$.
Correct Option: (C)
Solution 4:
$\omega_0 = 60\pi\text{ rad/s}, \quad \omega = 20\pi\text{ rad/s}$. Total $\theta = 80 \times 2\pi = 160\pi\text{ rad}$.
$160\pi = \left(\frac{60\pi + 20\pi}{2}\right)t = 40\pi t \implies t = 4.0\text{ s}$.
Correct Option: (B)
Solution 5:
$\omega = 6.0 t^2 – 12.0 t =:
$\omega = 6.0 t^2 – 12.0 t = 0 \implies 6.0 t(t – 2.0) = 0 \implies t = 2.0\text{ s}$.
Correct Option: (B)
Solution 6:
– (A) True: Rigid body constraint guarantees equal $\vec{\omega}$ and $\vec{\alpha}$ for all particles.
– (B) True: $v = \omega R \implies v \propto R$.
– (C) True: $a_t = \alpha R = 0$ if $\alpha = 0$.
– (D) False: Centripetal acceleration $a_c = \omega^2 R$ is non-zero as long as $\omega \ne 0$.
Correct Options: (A, B, C)
Solution 7:
– (A) True: $\alpha(t) \ge 0$ for $t \le T$, so $\omega$ peaks at $t = T$.
– (B) True: $\omega_{\text{max}} = \int_0^T \alpha_0 (1 – t/T)dt = \frac{1}{2}\alpha_0 T$.
– (C) True: $\theta = \int_0^T \omega(t)dt = \alpha_0 [T^2/2 – T^2/6] = \frac{1}{3}\alpha_0 T^2$.
– (D) True: For $t > T$, $\alpha = 0 \implies \omega = \text{const} \implies a_c = \omega^2 R = \text{const}$.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: $\vec{v}$ is along tangent ($a_t$); centripetal acceleration ($a_c$) is perpendicular to $\vec{v}$. Thus $\tan\phi = a_c/a_t = \omega^2/\alpha$.
– (B) True: $\tan 45^\circ = 1 \implies \omega^2 = \alpha$.
– (C) True: Uniform rotation means $a_t = 0 \implies \vec{a} = \vec{a}_c \perp \vec{v}$, so $\phi = 90^\circ$.
– (D) True: $a_t = a\cos\phi = \frac{dv}{dt}$.
Correct Options: (A, B, C, D)
Solution 9:
$\theta_1 = \frac{1}{2}\alpha(2.0)^2 = 2.0\alpha$. Total in $4.0\text{ s}$ is $\theta_{\text{total}} = \frac{1}{2}\alpha(4.0)^2 = 8.0\alpha$.
$\theta_2 = 8.0\alpha – 2.0\alpha = 6.0\alpha \implies \frac{\theta_2}{\theta_1} = \frac{6.0\alpha}{2.0\alpha} = 3.0$.
Correct Answer: 3
Solution 10:
$\tan 45^\circ = \frac{a_c}{a_t} = 1 \implies \omega^2 R = \alpha R \implies \omega^2 = \alpha$.
Since $\omega = \alpha t \implies (\alpha t)^2 = \alpha \implies t^2 = \frac{1}{\alpha} = \frac{1}{0.25} = 4.0 \implies t = 2.0\text{ seconds}$.
Correct Answer: 2