NEET Kinematics of Circular Motion MCQs & PYQs

Kinematics of circular motion is important for preparation of NEET, JEE Mains, JEE Advance , CUET and other exams in which Physics is asked

Practice NEET Kinematics of Circular Motion Questions

Question 21:

moderate

A particle of mass $10 \text{ g}$ moves along a circle of radius $6.4 text{ cm}$ with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8 \times 10^{-4} \text{ J}$ by the end of the second revolution after the beginning of the motion?

(2016-I)

Work done by tangential force $W = (ma_t)s = \Delta K$, where distance $s = 4\pi r = 4 \times \pi \times 0.064 \text{ m}$. Substituting values gives $a_t = 0.1 \text{ m/s}^2$.

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Question 22:

easy

The angular speed of a fly wheel moving with uniform angular acceleration changes from $1200\text{ rpm}$ to $3120\text{ rpm}$ in $16\text{ seconds}$. The angular acceleration in $\text{rad/s}^2$ is:

(2022)

Concept: Definition of angular acceleration. Formula: $\alpha = \frac{\omega_2 - \omega_1}{t}$. Solution: Converting rpm to rad/s and substituting values gives $\alpha = 4\pi\text{ rad/s}^2$.

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Question 23:

easy

The angular speed of the wheel of a vehicle is increased from $360\text{ rpm}$ to $1200\text{ rpm}$ in $14\text{ second}$. Its angular acceleration is.

(2020-Covid)

Concept: Kinematic equation for angular motion. Formula: $\alpha = \frac{\Delta \omega}{t}$. Solution: $\omega_1 = 12\pi$, $\omega_2 = 40\pi$, yielding $\alpha = \frac{28\pi}{14} = 2\pi\text{ rad/s}^2$.

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Question 24:

easy

A wheel has angular acceleration of $3.0\text{ rad/sec}^2$ and an initial angular speed of $2.00\text{ rad/sec}$. In a time of $2\text{ sec}$ it has rotated through an angle (in radian) of:

(2007)

Concept: Rotational kinematics equation. Formula: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$. Solution: $\theta = (2.0)(2) + \frac{1}{2}(3.0)(2)^2 = 4 + 6 = 10\text{ radians}$.

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Question 25:

moderate

For a body angular velocity $\vec{\omega} = \hat{i} – 2\hat{j} + 3\hat{k}$ and radius vector is $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ then its velocity is:

(1999)

Velocity is given by $\vec{v} = \vec{\omega} \times \vec{r}$. Evaluating the cross-product determinant yields $-5\hat{i} + 2\hat{j} + 3\hat{k}$.

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