Circular Motion

Practice Questions:

Question 81: 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)

1. $0.1 \text{ m/s}^2$

2. $0.15 \text{ m/s}^2$

3. $0.18 \text{ m/s}^2$

4. $0.2 \text{ m/s}^2$

View Answer

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 82: moderate

A stone is tied to a string of length ‘$\ell$’ and is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed ‘$u$’. The magnitude of the change in velocity as it reaches a position where the string is horizontal ($g$ being acceleration due to gravity) is:

(2004)

1. $\sqrt{u^2 - g\ell}$

2. $u - \sqrt{u^2 - 2g\ell}$

3. $\sqrt{2g\ell}$

4. $\sqrt{2(u^2 - g\ell)}$

View Answer

Velocity at horizontal position is $v_2 = \sqrt{u^2 - 2g\ell}$. Since velocities are perpendicular, magnitude of change in velocity is $\Delta v = \sqrt{u^2 + v_2^2} = \sqrt{2(u^2 - g\ell)}$.

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Question 83: 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)

1. $104\pi$

2. $2\pi$

3. $4\pi$

4. $12\pi$

View Answer

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 84: 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)

1. $28\pi\text{ rad/s}^2$

2. $120\pi\text{ rad/s}^2$

3. $1\text{ rad/s}^2$

4. $2\pi\text{ rad/s}^2$

View Answer

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 85: 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)

1. $10$

2. $12$

3. $4$

4. $6$

View Answer

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 86: 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)

1. $-5\hat{i} + 2\hat{j} + 3\hat{k}$

2. $-5\hat{i} + 2\hat{j} - 3\hat{k}$

3. $-5\hat{i} - 2\hat{j} + 3\hat{k}$

4. $-5\hat{i} - 2\hat{j} - 3\hat{k}$

View Answer

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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