Q.Obtain Eq. (6.36), $\omega = \omega_0 + \alpha t$, from first principles.

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1. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If a wheel starting from rest is rotating with an angular acceleration of $\pi\ \mathrm{rad\,s^{-2}}$, then the number of rotations made by the wheel in the first 6 seconds time is (A) 36 (B) 9 (C) 18 (D) 12

   ›Reveal solutionSolution

   Straightforward rotational kinematics: angle swept from rest under constant angular acceleration, converted to number of full rotations, giving 9.

   Concept and Intuition

   Rotational motion with constant angular acceleration mirrors linear motion with constant linear acceleration. Starting from rest, the angle covered grows quadratically with time, exactly like distance under constant linear acceleration.

   Step-by-Step Solution

   1. Given $\omega_0 = 0$, $\alpha = \pi\ \mathrm{rad\,s^{-2}}$, $t = 6\ \mathrm{s}$.
   2. Angle swept: $\theta = \omega_0 t + \tfrac12\alpha t^2 = 0 + \tfrac12(\pi)(6)^2 = \tfrac12(\pi)(36) = 18\pi\ \mathrm{rad}$. …

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2. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.A wheel of angular speed 600 rev/min is made to slow down at a rate of $2\ rad\,s^{-2}$. The number of revolutions made by the wheel before coming to rest is (A) 157 (B) 314 (C) 177 (D) 117

   ›Reveal solutionSolution

   This is a rotational kinematics problem: convert rev/min to rad/s, apply $\omega^2=\omega_0^2-2\alpha\theta$, then convert the angle back into revolutions.

   Concept and Intuition

   Angular deceleration problems mirror linear ones: replace $v,u,a,s$ with $\omega,\omega_0,\alpha,\theta$. The wheel decelerates uniformly, so the standard kinematic equation $\omega^2=\omega_0^2-2\alpha\theta$ applies directly, with the final angular speed zero ("comes to rest").

   Step-by-Step Solution

   1. Convert initial angular speed: $\omega_0=600\ \text{rev/min}=\dfrac{600\times2\pi}{60}\ \text{rad/s}=20\pi\ \text{rad/s}\approx62.83\ \text{rad/s}$.
   2. At rest, $\omega=0$. Using $\omega^2=\omega_0^2-2\alpha\theta$: $0=(20\pi)^2-2(2)\theta\Rightarrow\theta=\dfrac{(20\pi)^2}{4}=100\pi^2\ \text{rad}$. …

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3. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.A uniform rod of length '2L' is placed with one end in contact with the earth and is then inclined at an angle $\alpha$ to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be (A) $\sqrt{\dfrac{3g\sin\alpha}{2L}}$ (B) $\sqrt{\dfrac{2L}{3g\sin\alpha}}$ (C) $\sqrt{\dfrac{6g\sin\alpha}{L}}$ (D) $\sqrt{\dfrac{L}{g\sin\alpha}}$

   ›Reveal solutionSolution

   This tests rotational energy conservation for a rod pivoted (hinged, no slipping) at one end; the answer is $\omega=\sqrt{3g\sin\alpha/(2L)}$.

   Concept and Intuition

   Because the lower end of the rod stays in contact with the ground without slipping, that end acts as a fixed pivot (hinge). The rod's fall is then pure rotation about this end, not translation, so the natural tool is conservation of mechanical energy using the rotational kinetic energy $\frac12 I\omega^2$ about the pivot, not $\frac12 mv^2$.

   Step-by-Step Solution

   1. The rod has length $2L$, so its centre of mass (COM) lies at distance $L$ from the pivoted end.
   2. Take the moment of inertia of a uniform rod about one end: $I = \frac13 M (2L)^2 = \frac{4}{3}ML^2$.
   3. Initially the rod makes angle $\alpha$ with the horizontal, so the COM height above the ground is $h_i = L\sin\alpha$. When the rod becomes horizontal, the COM height is $h_f = 0$.
   4. Loss in gravitational PE = gain in rotational KE (no slipping ⇒ no friction losses, energy is conserved):

   $MgL\sin\alpha = \frac12 I\omega^2 = \frac12\left(\frac{4}{3}ML^2\right)\omega^2 = \frac{2}{3}ML^2\omega^2$

   5. Cancel $M$ and solve for $\omega$: …

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4. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.A disc starts rotating from rest with constant acceleration and attains angular velocity of $20\ \mathrm{rad\,s^{-1}}$ in 5 seconds. The total angular displacement during this interval is (A) $50\ \mathrm{rad}$ (B) $100\ \mathrm{rad}$ (C) $200\ \mathrm{rad}$ (D) $400\ \mathrm{rad}$

   ›Reveal solutionSolution

   A rotational-kinematics analogue of $s = ut+\tfrac12 at^{2}$. Answer: (A) $50\ \mathrm{rad}$.

   Concept and Intuition

   For uniformly accelerated rotational motion starting from rest, the same kinematic structure as linear motion applies with angular quantities: $\omega = \omega_0+\alpha t$ and $\theta = \omega_0 t + \tfrac12\alpha t^{2}$.

   Step-by-Step Solution

   1. Given $\omega_0=0$, $\omega=20\ \mathrm{rad\,s^{-1}}$, $t=5\ \mathrm{s}$.
   2. Find angular acceleration: $\alpha = \dfrac{\omega-\omega_0}{t} = \dfrac{20}{5}=4\ \mathrm{rad\,s^{-2}}$.
   3. Find angular displacement: $\theta = \omega_0 t + \tfrac12\alpha t^{2} = 0 + \tfrac12(4)(5^{2}) = \tfrac12(4)(25)=50\ \mathrm{rad}$.

   Common Mistakes …

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5. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.The angular speed of a rigid body rotating about a fixed axis is (8-2t) rad s$^{-1}$. The angle through which the body rotates before it comes to rest is (A) 8 rad (B) 12 rad (C) 16 rad (D) 20 rad

   ›Reveal solutionSolution

   Integrating the given $\omega(t)$ up to the moment it hits zero gives the total angle rotated: 16 rad.

   Concept and Intuition

   Angle rotated is the time-integral of angular speed, exactly analogous to distance being the integral of speed. Since $\omega$ decreases linearly with time here, the body undergoes uniform angular deceleration, and we integrate over the time it takes to reach zero angular speed.

   Step-by-Step Solution

   1. Given $\omega(t) = 8-2t$ rad/s.
   2. Find when the body comes to rest: $\omega=0 \Rightarrow 8-2t=0 \Rightarrow t=4\,\text{s}$. …

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6. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.A flywheel is rotating at a rate of 150 rev/minute. If it slows at constant retardation of $\pi\ \mathrm{rad\,s}^{-2}$, then the time required for the wheel to come to rest is (A) 2.5 s (B) 5 s (C) 4 s (D) 6 s

   ›Reveal solutionSolution

   Convert the initial angular speed to rad/s, then use $t=\omega_0/\alpha$ for constant retardation to rest.

   Concept and Intuition

   Rotational kinematics under constant angular retardation mirrors linear kinematics under constant deceleration: $\omega=\omega_0-\alpha t$, and the wheel stops when $\omega=0$.

   Step-by-Step Solution

   1. Convert $150$ rev/min to rad/s: $150$ rev/min $=\dfrac{150}{60}$ rev/s $=2.5$ rev/s. Each revolution is $2\pi$ rad, so $\omega_0=2.5\times2\pi=5\pi$ rad/s.
   2. Constant retardation: $\alpha=\pi$ rad/s$^2$ (deceleration). …

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7. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.A body rotating with uniform acceleration about its geometrical axis makes 8 rotations in the first 2 seconds. The number of rotations the body makes in the next 3 seconds is (Initially the body is at rest) (A) 50 (B) 25 (C) 42 (D) 21

   ›Reveal solutionSolution

   Using $\theta=\tfrac12\alpha t^2$ from rest, the body makes 50 total rotations by $t=5$s and 8 in the first 2s, so 42 more in the next 3s.

   Concept and Intuition

   This is the rotational analogue of a body starting from rest under uniform (linear) acceleration: $\theta(t)=\tfrac12\alpha t^2$. Since the equation is valid at any instant measured from $t=0$, the cleanest approach is to first find $\alpha$ from the given data at $t=2$s, then compute the total rotations at the later time $t=5$s, and subtract off what already happened in the first 2s.

   Step-by-Step Solution

   1. Body starts from rest ($\omega_0=0$) with uniform angular acceleration $\alpha$: $\theta(t)=\tfrac12\alpha t^2$.
   2. Given $\theta(2)=8$ rotations: $8=\tfrac12\alpha(2)^2=2\alpha \Rightarrow \alpha=4$ rotations/s².
   3. Total rotations up to $t=2+3=5$ s: $\theta(5)=\tfrac12(4)(5)^2=\tfrac12(4)(25)=50$ rotations. …

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8. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.A fan is rotating with an angular speed 300 rpm. The fan is switched off, and it takes 80 s to come to rest. Assuming constant angular deceleration, the number of revolutions made by the fan before it comes to rest is (A) 400 (B) 200 (C) 300 (D) 314

   ›Reveal solutionSolution

   Use the average-angular-velocity trick for uniformly decelerated angular

   motion: total angle = average $\omega$ $\times$ time. This gives exactly 200

   revolutions.

   Concept and Intuition

   This is rotational motion with constant angular deceleration, exactly

   analogous to a ball thrown up and decelerating uniformly under gravity. For

   constant deceleration from $\omega_0$ to $0$, the average angular speed

   during the process is simply the arithmetic mean $\dfrac{\omega_0+0}{2}$, and

   the total angle swept is that average speed times the total time — no need to

   separately solve for the deceleration $\alpha$.

   Step-by-Step Solution

   1. Initial angular speed: $\omega_0 = \dfrac{2\pi\times300}{60} = 10\pi\ \mathrm{rad/s}$.
   2. Since deceleration is constant and final $\omega=0$, average angular speed $= \dfrac{\omega_0+0}{2} = 5\pi\ \mathrm{rad/s}$.
   3. Total angle swept in $t=80$ s: $\theta = \bar\omega\, t = 5\pi\times80 = 400\pi$ rad. …

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9. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.A wheel undergoes a constant acceleration starting form rest at t = 0. The angular velocity of the wheel is 3.14 $\dfrac{rad}{sec}$ when t = 2 s. The accelerator is abruptly ceased at t = 20 s. The number of revolutions, wheel makes in the interval t = 0 to t = 40 s is (A) 100 (B) 175 (C) 225 (D) 150

   ›Reveal solutionSolution

   This tests rotational kinematics in two phases — constant angular acceleration, then constant angular velocity — and converting total angle swept to revolutions. Answer: 150 revolutions.

   Concept and Intuition

   The motion has two distinct regimes: from $t=0$ to $t=20$ s the wheel speeds up uniformly (constant $\alpha$, found from the given data point at $t=2$ s); from $t=20$ s to $t=40$ s the acceleration stops, so the wheel simply coasts at whatever angular velocity it had reached at $t=20$ s. Total angle is the sum of the angle swept in each phase, then divide by $2\pi$ to get revolutions.

   Step-by-Step Solution

   1. Find $\alpha$ from the given data: starting from rest, $\omega = \alpha t$. At $t=2$ s, $\omega=3.14\approx\pi\ \text{rad s}^{-1}$, so $\alpha = \pi/2\ \text{rad s}^{-2}$.
   2. Phase 1 ($0\to20$ s, constant acceleration from rest): angle swept $\theta_1 = \tfrac12\alpha t^2 = \tfrac12\left(\dfrac{\pi}{2}\right)(20)^2 = \tfrac12\times\dfrac{\pi}{2}\times400 = 100\pi$ rad.
   3. Angular velocity reached at $t=20$ s: $\omega_{20} = \alpha\times20 = \dfrac{\pi}{2}\times20 = 10\pi\ \text{rad s}^{-1}$.
   4. Phase 2 ($20\to40$ s, acceleration ceased, so constant $\omega_{20}$ for $20$ s): $\theta_2 = \omega_{20}\times20 = 10\pi\times20=200\pi$ rad. …

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10. AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
    Q.A body, initially at rest, starts rotating with a uniform acceleration and covers $100\pi\ rad$ in the first 5 seconds. Its angular speed at the end of 5 seconds is ______ (A) $20\pi\ rad.s^{-1}$ (B) $30\pi\ rad.s^{-1}$ (C) $40\pi\ rad.s^{-1}$ (D) $10\pi\ rad.s^{-1}$

    ›Reveal solutionSolution

    Rotational kinematics with constant $\alpha$ starting from rest: $\theta=\frac12\alpha t^2$ gives $\alpha = 8\pi$, hence $\omega = \alpha t = 40\pi\ \mathrm{rad\,s^{-1}}$.

    Concept and Intuition

    Rotational motion with constant angular acceleration is the exact mirror of straight-line motion with constant linear acceleration — just replace $s\to\theta$, $u\to\omega_0$, $v\to\omega$, $a\to\alpha$:

    $\omega = \omega_0 + \alpha t,\qquad \theta = \omega_0 t + \tfrac12\alpha t^2,\qquad \omega^2 = \omega_0^2 + 2\alpha\theta$

    The physical intuition worth carrying away: when a body starts from rest and accelerates uniformly, its average speed over the interval is exactly half its final speed. So the final speed is always twice the "angle covered ÷ time".

    Step-by-Step Solution

    1. Data: $\omega_0 = 0$, $\theta = 100\pi$ rad, $t = 5$ s, $\alpha$ constant.
    2. Use $\theta = \omega_0 t + \tfrac12 \alpha t^2 = \tfrac12 \alpha t^2$:

    $100\pi = \tfrac12 \alpha (25) \Rightarrow \alpha = \frac{2 \times 100\pi}{25} = 8\pi\ \text{rad·s}^{-2}$

    3. Use $\omega = \omega_0 + \alpha t = 8\pi \times 5 = 40\pi\ \text{rad·s}^{-1}$. …

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11. AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQ
    Q.The angular velocity of a ceiling fan reduces to 50% after 36 rotations since it is switched off. Assuming uniform retardation, the number of rotations it further makes before coming to rest is ____ (A) 12 (B) 18 (C) 48 (D) 36

    ›Reveal solutionSolution

    Under uniform angular retardation, using $\omega^2=\omega_0^2-2\alpha\theta$ at the halfway-speed point and again at rest shows the fan needs 48 total rotations to stop, i.e. only 12 more after the first 36 — option (A).

    Concept and Intuition

    Uniform (constant) angular retardation means the fan behaves exactly like uniformly decelerated linear motion, just with angle in place of distance. The rotational analogue of $v^2=u^2-2as$ is $\omega^2=\omega_0^2-2\alpha\theta$. Since $\theta$ (or equivalently the number of rotations $N$, since $\theta=2\pi N$) appears linearly, we can work directly in "rotations" as our displacement variable and the constant $\alpha$ absorbs the $2\pi$ factor — it cancels out cleanly, so we never need to convert to radians at all.

    Step-by-Step Solution

    1. Let $\omega_0$ be the initial angular velocity and let $k$ be a positive constant such that $\omega^2 = \omega_0^2 - kN$, where $N$ is the number of rotations completed since switch-off (this is just $\omega^2=\omega_0^2-2\alpha\theta$ rewritten with $N=\theta/2\pi$ and $k=4\pi\alpha$).
    2. After $N_1=36$ rotations, $\omega = \omega_0/2$: $\left(\dfrac{\omega_0}{2}\right)^2 = \omega_0^2 - k(36)$.
    3. Simplify: $\dfrac{\omega_0^2}{4} = \omega_0^2 - 36k \;\Rightarrow\; 36k = \omega_0^2 - \dfrac{\omega_0^2}{4} = \dfrac{3\omega_0^2}{4}$, so $k = \dfrac{\omega_0^2}{48}$. …

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12. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.A wheel having moment of inertia $40\ kg.m^2$ about its axis, rotates at $50\ rpm$. The angular retardation required to stop this wheel in $90$ seconds is ________ $rad.s^{-2}$ (A) $\dfrac{\pi}{45}$ (B) $\dfrac{\pi}{30}$ (C) $\dfrac{\pi}{54}$ (D) $\dfrac{\pi}{24}$

    ›Reveal solutionSolution

    Convert the initial angular speed to rad/s, then use the kinematic relation $\omega_f = \omega_i - \alpha t$ with $\omega_f=0$ to find the constant angular retardation.

    Concept and Intuition

    Angular retardation (deceleration) uniformly brings a rotating wheel to rest, analogous to linear deceleration. We just need $\omega_i$ in proper units and the stopping time; the moment of inertia given in the problem is a distractor not needed to find $\alpha$ directly (it would only matter if we needed to find the retarding torque).

    Step-by-Step Solution

    1. Convert $\omega_i = 50\ rpm$ to rad/s: $\omega_i = 50 \times \dfrac{2\pi}{60} = \dfrac{100\pi}{60} = \dfrac{5\pi}{3}\ rad.s^{-1}$.
    2. The wheel comes to rest in $t=90\ s$, so final angular velocity $\omega_f=0$.
    3. Using $\omega_f = \omega_i - \alpha t$: $0 = \dfrac{5\pi}{3} - \alpha(90)$. …

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