Prepizo
Learn › MH-CET · Physics › Rotational Dynamics › Angular Motion & Kinematics

Angular Motion & Kinematics — MH-CET Physics MCQs with Solutions

Free MH-CET Physics Angular Motion & Kinematics MCQs with step-by-step solutions (22 questions). Part of Rotational Dynamics. Practise online on Prepizo — no login needed.

▶ Practise Angular Motion & Kinematics online (free)

Questions with solutions

Q1 — Angular Motion & Kinematics · easy · numerical
A wheel starts from rest and rotates with a constant angular acceleration of $2\text{ rad/s}^2$. The total angle turned in the first $5\text{ seconds}$ is:
A. $100\text{ rad}$
B. $50\text{ rad}$
C. $25\text{ rad}$  ✓ Correct
D. $10\text{ rad}$
Solution: $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = 0 + \tfrac{1}{2}(2)(5)^2 = 25\text{ rad}$.
Q2 — Angular Motion & Kinematics · easy · numerical
The angular velocity of a rotating body changes from $4\text{ rad/s}$ to $12\text{ rad/s}$ in $2\text{ seconds}$. Its angular acceleration is:
A. $2\text{ rad/s}^2$
B. $4\text{ rad/s}^2$  ✓ Correct
C. $8\text{ rad/s}^2$
D. $6\text{ rad/s}^2$
Solution: $\alpha = \dfrac{\Delta\omega}{\Delta t} = \dfrac{12 - 4}{2} = 4\text{ rad/s}^2$.
Q3 — Angular Motion & Kinematics · easy · numerical
A particle moves along a circle of radius $0.5\text{ m}$ with angular velocity $10\text{ rad/s}$. Its linear (tangential) speed is:
A. $20\text{ m/s}$
B. $0.05\text{ m/s}$
C. $2\text{ m/s}$
D. $5\text{ m/s}$  ✓ Correct
Solution: $v = \omega r = 10 \times 0.5 = 5\text{ m/s}$.
Q4 — Angular Motion & Kinematics · easy · numerical
A body moves in a circle of radius $4\text{ m}$ with a constant speed of $8\text{ m/s}$. Its centripetal acceleration is:
A. $16\text{ m/s}^2$  ✓ Correct
B. $4\text{ m/s}^2$
C. $32\text{ m/s}^2$
D. $2\text{ m/s}^2$
Solution: $a_c = \dfrac{v^2}{r} = \dfrac{8^2}{4} = \dfrac{64}{4} = 16\text{ m/s}^2$.
Q5 — Angular Motion & Kinematics · easy · numerical
A ceiling fan rotates at $300\text{ rpm}$. Its angular velocity in rad/s is:
A. $600\pi$
B. $10\pi$  ✓ Correct
C. $300$
D. $5\pi$
Solution: $\omega = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 300}{60} = 10\pi \approx 31.4\text{ rad/s}$.
Q6 — Angular Motion & Kinematics · medium · numerical
A wheel rotating at $20\text{ rad/s}$ is brought uniformly to rest in $10\text{ s}$. The number of revolutions it makes before stopping is approximately:
A. $32$
B. $16$  ✓ Correct
C. $10$
D. $100$
Solution: Angular displacement $\theta = \left(\dfrac{\omega_0 + \omega}{2}\right)t = \left(\dfrac{20 + 0}{2}\right)(10) = 100\text{ rad}$. Number of revolutions $= \dfrac{100}{2\pi} \approx 15.9 \approx 16$.
Q7 — Angular Motion & Kinematics · medium · numerical
The angular displacement of a rotating body is given by $\theta = 2t^3$ radian. Its angular velocity at $t = 2\text{ s}$ is:
A. $48\text{ rad/s}$
B. $24\text{ rad/s}$  ✓ Correct
C. $16\text{ rad/s}$
D. $12\text{ rad/s}$
Solution: $\omega = \dfrac{d\theta}{dt} = 6t^2$. At $t = 2\text{ s}$, $\omega = 6(4) = 24\text{ rad/s}$.
Q8 — Angular Motion & Kinematics · medium · numerical
The angular velocity of a body varies as $\omega = (4t^2 + 2t)\text{ rad/s}$. Its angular acceleration at $t = 1\text{ s}$ is:
A. $6\text{ rad/s}^2$
B. $10\text{ rad/s}^2$  ✓ Correct
C. $12\text{ rad/s}^2$
D. $8\text{ rad/s}^2$
Solution: $\alpha = \dfrac{d\omega}{dt} = 8t + 2$. At $t = 1\text{ s}$, $\alpha = 8 + 2 = 10\text{ rad/s}^2$.
Q9 — Angular Motion & Kinematics · medium · numerical
The angular velocity of the Earth about its own axis is approximately:
A. $7.27 \times 10^{-5}\text{ rad/s}$  ✓ Correct
B. $1.16 \times 10^{-5}\text{ rad/s}$
C. $7.27 \times 10^{-4}\text{ rad/s}$
D. $7.27 \times 10^{-6}\text{ rad/s}$
Solution: The Earth turns once in $24\text{ hours} = 86400\text{ s}$, so $\omega = \dfrac{2\pi}{86400} \approx 7.27 \times 10^{-5}\text{ rad/s}$.
Q10 — Angular Motion & Kinematics · medium · numerical
If the angular velocity of a particle in circular motion is tripled while the radius is halved, its linear speed becomes:
A. $1.5$ times  ✓ Correct
B. $6$ times
C. $0.5$ times
D. $3$ times
Solution: $v = \omega r$, so $v' = (3\omega)\left(\dfrac{r}{2}\right) = 1.5\,\omega r = 1.5\,v$.
Q11 — Angular Motion & Kinematics · easy · theory
The tangential acceleration $a_t$ of a particle moving on a circle of radius $r$ with angular acceleration $\alpha$ is:
A. $\omega^2 r$
B. $\alpha r$  ✓ Correct
C. $\alpha^2 r$
D. $\dfrac{\alpha}{r}$
Solution: Differentiating $v = \omega r$ with $r$ constant gives $a_t = \dfrac{dv}{dt} = r\dfrac{d\omega}{dt} = \alpha r$. ($\omega^2 r$ is the centripetal, not tangential, acceleration.)
Q12 — Angular Motion & Kinematics · easy · numerical
A particle on a circle of radius $2\text{ m}$ speeds up uniformly from rest to $10\text{ m/s}$ in $5\text{ s}$. Its tangential acceleration is:
A. $0.4\text{ m/s}^2$
B. $2\text{ m/s}^2$  ✓ Correct
C. $5\text{ m/s}^2$
D. $10\text{ m/s}^2$
Solution: Tangential acceleration is the rate of change of speed: $a_t = \dfrac{10 - 0}{5} = 2\text{ m/s}^2$.
Q13 — Angular Motion & Kinematics · easy · numerical
A particle moves on a circle of radius $2\text{ m}$. At the instant its speed is $10\text{ m/s}$, its centripetal acceleration is:
A. $20\text{ m/s}^2$
B. $100\text{ m/s}^2$
C. $5\text{ m/s}^2$
D. $50\text{ m/s}^2$  ✓ Correct
Solution: $a_c = \dfrac{v^2}{r} = \dfrac{100}{2} = 50\text{ m/s}^2$, directed towards the centre.
Q14 — Angular Motion & Kinematics · medium · numerical
A flywheel speeds up uniformly from $60\text{ rpm}$ to $180\text{ rpm}$ in $10\text{ s}$. Its angular acceleration is:
A. $0.4\pi\text{ rad/s}^2$  ✓ Correct
B. $12\text{ rad/s}^2$
C. $4\pi\text{ rad/s}^2$
D. $0.2\pi\text{ rad/s}^2$
Solution: $\omega_0 = \dfrac{2\pi(60)}{60} = 2\pi$ and $\omega = \dfrac{2\pi(180)}{60} = 6\pi\text{ rad/s}$. So $\alpha = \dfrac{6\pi - 2\pi}{10} = 0.4\pi \approx 1.26\text{ rad/s}^2$.
Q15 — Angular Motion & Kinematics · medium · numerical
A wheel starting from rest rotates with constant angular acceleration $4\text{ rad/s}^2$. The angle turned during the $3^{\text{rd}}$ second alone is:
A. $6\text{ rad}$
B. $20\text{ rad}$
C. $18\text{ rad}$
D. $10\text{ rad}$  ✓ Correct
Solution: Angle in the $n^{\text{th}}$ second: $\theta_n = \omega_0 + \dfrac{\alpha}{2}(2n - 1) = 0 + \dfrac{4}{2}(2 \times 3 - 1) = 2 \times 5 = 10\text{ rad}$.
Q16 — Angular Motion & Kinematics · easy · theory
The angular velocity $\omega$ of a body making $n$ revolutions per second is:
A. $\dfrac{n}{2\pi}$
B. $2\pi n$  ✓ Correct
C. $\pi n$
D. $\dfrac{2\pi}{n}$
Solution: One revolution corresponds to $2\pi$ radian, so $n$ revolutions per second means $\omega = 2\pi n\text{ rad/s}$.
Q17 — Angular Motion & Kinematics · easy · theory
For a particle in uniform circular motion, which of the following remains constant?
A. Its acceleration vector
B. Its velocity vector
C. The magnitude of its velocity  ✓ Correct
D. Its momentum vector
Solution: In uniform circular motion the speed is constant, but the direction of motion changes continuously. Hence velocity, acceleration and momentum are all changing vectors, while the magnitude of velocity (the speed) stays fixed.
Q18 — Angular Motion & Kinematics · easy · numerical
A wheel rotating at $10\text{ rad/s}$ is subjected to a constant angular retardation of $2\text{ rad/s}^2$. The time it takes to come to rest is:
A. $10\text{ s}$
B. $20\text{ s}$
C. $2.5\text{ s}$
D. $5\text{ s}$  ✓ Correct
Solution: $\omega = \omega_0 - \alpha t \Rightarrow 0 = 10 - 2t \Rightarrow t = 5\text{ s}$.
Q19 — Angular Motion & Kinematics · easy · numerical
A flywheel starting from rest has a constant angular acceleration of $3\text{ rad/s}^2$. Its angular velocity after $6\text{ s}$ is:
A. $0.5\text{ rad/s}$
B. $54\text{ rad/s}$
C. $9\text{ rad/s}$
D. $18\text{ rad/s}$  ✓ Correct
Solution: $\omega = \omega_0 + \alpha t = 0 + 3 \times 6 = 18\text{ rad/s}$.
Q20 — Angular Motion & Kinematics · easy · numerical
A wheel rotates at $120\text{ rpm}$. Its angular velocity is:
A. $2\pi\text{ rad/s}$
B. $120\pi\text{ rad/s}$
C. $240\pi\text{ rad/s}$
D. $4\pi\text{ rad/s}$  ✓ Correct
Solution: $\omega = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 120}{60} = 4\pi \approx 12.57\text{ rad/s}$.
Q21 — Angular Motion & Kinematics · easy · numerical
A point on the rim of a disc of radius $0.2\text{ m}$ rotating at $15\text{ rad/s}$ has a linear speed of:
A. $75\text{ m/s}$
B. $30\text{ m/s}$
C. $3\text{ m/s}$  ✓ Correct
D. $0.75\text{ m/s}$
Solution: $v = \omega r = 15 \times 0.2 = 3\text{ m/s}$.
Q22 — Angular Motion & Kinematics · medium · numerical
A wheel starting from rest turns through $100\text{ radian}$ in $10\text{ s}$ under constant angular acceleration. Its angular acceleration is:
A. $1\text{ rad/s}^2$
B. $20\text{ rad/s}^2$
C. $2\text{ rad/s}^2$  ✓ Correct
D. $10\text{ rad/s}^2$
Solution: $\theta = \dfrac{1}{2}\alpha t^2 \Rightarrow 100 = \dfrac{1}{2}\alpha(100) \Rightarrow \alpha = 2\text{ rad/s}^2$.