Rotational Kinetic Energy — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Rotational Kinetic Energy MCQs with step-by-step solutions (20 questions). Part of Rotational Dynamics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Rotational Kinetic Energy · easy · numerical
A body of moment of inertia $4\text{ kg}\cdot\text{m}^2$ rotates at $3\text{ rad/s}$. Its rotational kinetic energy is:
A. $36\text{ J}$
B. $12\text{ J}$
C. $18\text{ J}$ ✓ Correct
D. $6\text{ J}$
Solution: $K_{rot} = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}(4)(3)^2 = \dfrac{1}{2}(4)(9) = 18\text{ J}$.
Q2 — Rotational Kinetic Energy · medium · numerical
A disc of mass $4\text{ kg}$ and radius $0.5\text{ m}$ spins about its own axis at $10\text{ rad/s}$. Its rotational kinetic energy is:
A. $100\text{ J}$
B. $50\text{ J}$
C. $12.5\text{ J}$
D. $25\text{ J}$ ✓ Correct
Solution: $I = \dfrac{1}{2}MR^2 = \dfrac{1}{2}(4)(0.25) = 0.5\text{ kg}\cdot\text{m}^2$. Then $K = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}(0.5)(100) = 25\text{ J}$.
Q3 — Rotational Kinetic Energy · easy · theory
The work done by a constant torque $\tau$ in rotating a body through an angle $\theta$ is:
A. $\dfrac{1}{2}\tau\theta^2$
B. $\tau\theta^2$
C. $\tau\theta$ ✓ Correct
D. $\dfrac{\tau}{\theta}$
Solution: The rotational analogue of $W = Fs$ is $W = \tau\theta$, with $\theta$ in radian.
Q4 — Rotational Kinetic Energy · hard · numerical
A flywheel in the form of a uniform disc of mass $20\text{ kg}$ and radius $0.5\text{ m}$ starts from rest under a torque of $10\text{ N}\cdot\text{m}$. The work done by the torque in $5\text{ s}$ is:
A. $500\text{ J}$ ✓ Correct
B. $100\text{ J}$
C. $200\text{ J}$
D. $1000\text{ J}$
Solution: $I = \dfrac{1}{2}(20)(0.5)^2 = 2.5\text{ kg}\cdot\text{m}^2$, so $\alpha = \dfrac{\tau}{I} = \dfrac{10}{2.5} = 4\text{ rad/s}^2$. Then $\theta = \dfrac{1}{2}\alpha t^2 = \dfrac{1}{2}(4)(25) = 50\text{ rad}$ and $W = \tau\theta = 10 \times 50 = 500\text{ J}$.
Q5 — Rotational Kinetic Energy · easy · numerical
The power delivered by a torque of $20\text{ N}\cdot\text{m}$ acting on a body rotating at $5\text{ rad/s}$ is:
A. $50\text{ W}$
B. $25\text{ W}$
C. $4\text{ W}$
D. $100\text{ W}$ ✓ Correct
Solution: Rotational power is $P = \tau\omega = 20 \times 5 = 100\text{ W}$, the analogue of $P = Fv$.
Q6 — Rotational Kinetic Energy · easy · theory
If the angular velocity of a rotating body is doubled while its moment of inertia stays the same, its rotational kinetic energy becomes:
A. Half
B. Unchanged
C. Double
D. Four times ✓ Correct
Solution: $K = \dfrac{1}{2}I\omega^2 \propto \omega^2$, so doubling $\omega$ multiplies the kinetic energy by $4$.
Q7 — Rotational Kinetic Energy · medium · theory
The rotational kinetic energy of a body in terms of its angular momentum $L$ and moment of inertia $I$ is:
A. $\dfrac{2I}{L^2}$
B. $\dfrac{L}{2I}$
C. $\dfrac{L^2}{2I}$ ✓ Correct
D. $\dfrac{IL^2}{2}$
Solution: Substituting $\omega = \dfrac{L}{I}$ in $K = \dfrac{1}{2}I\omega^2$ gives $K = \dfrac{1}{2}I\dfrac{L^2}{I^2} = \dfrac{L^2}{2I}$.
Q8 — Rotational Kinetic Energy · medium · theory
If the moment of inertia of a body is doubled while its angular momentum remains constant, its rotational kinetic energy:
A. Doubles
B. Remains unchanged
C. Becomes four times
D. Becomes half ✓ Correct
Solution: With $L$ fixed, $K = \dfrac{L^2}{2I} \propto \dfrac{1}{I}$. Doubling $I$ therefore halves the kinetic energy.
Q9 — Rotational Kinetic Energy · medium · numerical
A constant torque of $20\text{ N}\cdot\text{m}$ acts on a flywheel of moment of inertia $5\text{ kg}\cdot\text{m}^2$ initially at rest. The kinetic energy acquired in $4\text{ seconds}$ is:
A. $320\text{ J}$
B. $640\text{ J}$ ✓ Correct
C. $1280\text{ J}$
D. $160\text{ J}$
Solution: $\alpha = \dfrac{\tau}{I} = \dfrac{20}{5} = 4\text{ rad/s}^2$, so $\omega = \alpha t = 16\text{ rad/s}$. Then $K = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}(5)(256) = 640\text{ J}$.
Q10 — Rotational Kinetic Energy · easy · theory
A flywheel is used in machines mainly because it:
A. Increases the torque produced by the engine
B. Stores rotational kinetic energy and smooths out speed fluctuations ✓ Correct
C. Reduces the moment of inertia of the machine
D. Eliminates friction in the bearings
Solution: Its large moment of inertia lets a flywheel absorb energy during the power stroke and release it during the idle strokes, keeping the shaft speed nearly uniform.
Q11 — Rotational Kinetic Energy · medium · numerical
A wheel of moment of inertia $3\text{ kg}\cdot\text{m}^2$ slows from $10\text{ rad/s}$ to $4\text{ rad/s}$. The loss in its rotational kinetic energy is:
A. $84\text{ J}$
B. $42\text{ J}$
C. $150\text{ J}$
D. $126\text{ J}$ ✓ Correct
Solution: $\Delta K = \dfrac{1}{2}I(\omega_1^2 - \omega_2^2) = \dfrac{1}{2}(3)(100 - 16) = 1.5 \times 84 = 126\text{ J}$.
Q12 — Rotational Kinetic Energy · easy · numerical
A flywheel of moment of inertia $0.5\text{ kg}\cdot\text{m}^2$ rotates at $20\text{ rad/s}$. The energy stored in it is:
A. $100\text{ J}$ ✓ Correct
B. $200\text{ J}$
C. $50\text{ J}$
D. $10\text{ J}$
Solution: $K = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}(0.5)(400) = 100\text{ J}$.
Q13 — Rotational Kinetic Energy · easy · theory
The work-energy theorem for rotational motion states that the work done by the net torque equals:
A. The change in rotational kinetic energy ✓ Correct
B. The change in angular velocity
C. The change in angular momentum
D. The change in moment of inertia
Solution: Just as $W = \Delta\left(\dfrac{1}{2}mv^2\right)$ for translation, in rotation $W = \Delta\left(\dfrac{1}{2}I\omega^2\right)$.
Q14 — Rotational Kinetic Energy · hard · numerical
Two bodies have the same rotational kinetic energy, but the moment of inertia of the first is four times that of the second. The ratio of their angular velocities $\omega_1 : \omega_2$ is:
A. $1 : 4$
B. $4 : 1$
C. $1 : 2$ ✓ Correct
D. $2 : 1$
Solution: Equal $K = \dfrac{1}{2}I\omega^2$ means $I\omega^2$ is the same, so $\omega \propto \dfrac{1}{\sqrt{I}}$. With $I_1 = 4I_2$, $\dfrac{\omega_1}{\omega_2} = \sqrt{\dfrac{I_2}{I_1}} = \dfrac{1}{2}$.
Q15 — Rotational Kinetic Energy · medium · numerical
Two bodies have the same angular momentum, but the moment of inertia of the first is four times that of the second. The ratio of their rotational kinetic energies $K_1 : K_2$ is:
A. $2 : 1$
B. $1 : 2$
C. $1 : 4$ ✓ Correct
D. $4 : 1$
Solution: With $L$ equal, $K = \dfrac{L^2}{2I} \propto \dfrac{1}{I}$. Since $I_1 = 4I_2$, $K_1 : K_2 = 1 : 4$.
Q16 — Rotational Kinetic Energy · easy · theory
The expression $\dfrac{1}{2}I\omega^2$ is the rotational analogue of:
A. $mgh$
B. $mv$
C. $\dfrac{1}{2}mv^2$ ✓ Correct
D. $\dfrac{1}{2}kx^2$
Solution: Replacing mass $m$ by moment of inertia $I$ and linear speed $v$ by angular speed $\omega$ converts translational kinetic energy $\dfrac{1}{2}mv^2$ into $\dfrac{1}{2}I\omega^2$.
Q17 — Rotational Kinetic Energy · easy · numerical
A body of moment of inertia $8\text{ kg}\cdot\text{m}^2$ rotates at $5\text{ rad/s}$. Its rotational kinetic energy is:
A. $100\text{ J}$ ✓ Correct
B. $200\text{ J}$
C. $20\text{ J}$
D. $40\text{ J}$
Solution: $K = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}(8)(25) = 100\text{ J}$.
Q18 — Rotational Kinetic Energy · hard · numerical
A solid sphere of mass $2\text{ kg}$ and radius $0.1\text{ m}$ spins about its diameter at $10\text{ rad/s}$. Its rotational kinetic energy is:
A. $0.8\text{ J}$
B. $4.0\text{ J}$
C. $1.0\text{ J}$
D. $0.4\text{ J}$ ✓ Correct
Solution: $I = \dfrac{2}{5}(2)(0.01) = 0.008\text{ kg}\cdot\text{m}^2$, so $K = \dfrac{1}{2}(0.008)(100) = 0.4\text{ J}$.
Q19 — Rotational Kinetic Energy · easy · numerical
A constant torque of $6\text{ N}\cdot\text{m}$ turns a wheel through $10\text{ radian}$. The work done is:
A. $16\text{ J}$
B. $60\text{ J}$ ✓ Correct
C. $30\text{ J}$
D. $0.6\text{ J}$
Solution: $W = \tau\theta = 6 \times 10 = 60\text{ J}$.
Q20 — Rotational Kinetic Energy · easy · numerical
A torque of $15\text{ N}\cdot\text{m}$ acts on a body rotating at $8\text{ rad/s}$. The power delivered is:
A. $60\text{ W}$
B. $120\text{ W}$ ✓ Correct
C. $23\text{ W}$
D. $1.875\text{ W}$
Solution: $P = \tau\omega = 15 \times 8 = 120\text{ W}$.