Rotational Kinetic Energy

Practice Questions:

Question 1: moderate

Three objects, A: (a solid sphere), B: (a thin circular disk) and C: (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation

(2018)

1. $W_B > W_A > W_C$

2. $W_A > W_B > W_C$

3. $W_C > W_B > W_A$

4. $W_A > W_C > W_B$

View Answer

Work required equals rotational kinetic energy $\frac{1}{2}I\omega^2$. Comparing moments of inertia, $I_C = MR^2 > I_B = 0.5MR^2 > I_A = 0.4MR^2$, leading to $W_C > W_B > W_A$.

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Question 2: moderate

A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ($E_{sphere} / E_{cylinder}$) will be:

(2016 – II)

1. $1:4$

2. $3:1$

3. $2:3$

4. $1:5$

View Answer

Kinetic energy $E = \frac{1}{2}I\omega^2$. For sphere, $E_1 = \frac{1}{2}(\frac{2}{5}mR^2)\omega^2 = \frac{1}{5}mR^2\omega^2$. For cylinder, $E_2 = \frac{1}{2}(\frac{1}{2}mR^2)(2\omega)^2 = mR^2\omega^2$. Ratio is $\frac{1/5}{1} = 1:5$.

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Question 3: easy

A ring of mass $m$ and radius $r$ rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. Its kinetic energy is:

(1988)

1. $\frac{1}{2} mr^2 \omega^2$

2. $mr \omega^2$

3. $mr^2 \omega^2$

4. $\frac{1}{2} mr \omega^2$

View Answer

For a ring rotating about its central perpendicular axis, the moment of inertia is $I = mr^2$. The rotational kinetic energy is defined as $K = \frac{1}{2} I \omega^2$. Substituting $I$, we get $K = \frac{1}{2} mr^2 \omega^2$.

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Question 4: moderate

A disc of radius $2text{ m}$ and mass $100text{ kg}$ rolls on a horizontal floor. Its centre of mass has speed of $20text{ cm/s}$. How much work is needed to stop it?

(2019)

1. $3\text{ J}$

2. $30\text{ kJ}$

3. $2\text{ J}$

4. $1\text{ J}$

View Answer

Total kinetic energy of the rolling disc is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3}{4}mv^2$. Substituting $m = 100\text{ kg}$ and $v = 0.2\text{ m/s}$ gives $K = 3\text{ J}$. Work required to stop it is equal to its total kinetic energy, which is $3\text{ J}$.

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Question 5: easy

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy ($K_t$) as well as rotational kinetic energy ($K_r$) simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is:

(2018)

1. $10 : 7$

2. $5 : 7$

3. $7 : 10$

4. $2 : 5$

View Answer

For a solid sphere, $K_t = \frac{1}{2}mv^2$ and $K_r = \frac{1}{5}mv^2$. The total kinetic energy is $K_t + K_r = \frac{7}{10}mv^2$. The ratio $K_t : (K_t + K_r)$ evaluates to $5 : 7$.

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