Question 1:
easyA circular ring of wire of mass M and radius R is making n revolutions/sec about an axis passing through a point on its rim and perpendicular to its plane. The kinetic energy of rotation of the ring is given by
Rotational Kinetic Energy = ½ I ω²= ½(2MR²)(2πn)²= 4π²MR²n²
Question 2:
easyWhat is moment of inertia in terms of angular momentum (L) and kinetic energy (K)
\[ K = \frac{1}{2}I \omega^{2} \]
and
L = I ω ⇒  Substituting we get,
\[ I= \frac{L²}{2K} \]
Question 3:
easyIf moment of inertia of a spinning object drops to \(\left(\frac{1}{4}\right)^{\text{th}}\) of its initial value, the ratio of new rotational kinetic energy to initial rotational kinetic energy will be (Assume net external torque about the axis of rotation is zero)
Since external torque is zero, angular momentum \(L = I\omega\) is conserved. Rotational kinetic energy is \(K = \frac{L^2}{2I}\). If \(I' = I/4\), then \(K' = 4K\), so \(K' : K = 4 : 1\).
Question 4:
easyAssertion (A): Kinetic energy of a rigid body can be greater than \( \frac{1}{2}mv^2 \), where \( m \) is mass of rigid body & \( v \) is speed of centre of mass of body.
Reason (R): Kinetic energy of a particle (point mass) cannot be greater than \( \frac{1}{2}mv^2 \), where \( m \) is mass of particle & \( v \) is speed of particle.
The total kinetic energy of a rigid body is \( K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \), where the second term is rotational KE. For a particle, \( K = \frac{1}{2}mv^2 \) only. Therefore, a rigid body's KE can be greater than \( \frac{1}{2}mv^2 \).
Both A and R are true, and R explains A by highlighting the difference in KE components.
Question 5:
easyIf moment of inertia of a spinning object drops to \( \left(\frac{1}{4}\right)^{\text{th}} \) of its initial value, the ratio of new rotational kinetic energy to initial rotational kinetic energy will be (Assume net external torque about the axis of rotation is zero)
Under zero external torque, angular momentum is conserved: \( I_1 \omega_1 = I_2 \omega_2 \). If \( I_2 = I_1/4 \), then \( \omega_2 = 4\omega_1 \). The ratio of kinetic energy is \( \frac{K_2}{K_1} = \frac{\frac{1}{2}I_2\omega_2^2}{\frac{1}{2}I_1\omega_1^2} = \frac{1}{4} \times 16 = 4 \implies 4 : 1 \).
Question 6:
moderateThree 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)
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 $W_C > W_B > W_A$.
Question 7:
moderateA 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)
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 $\frac{1/5}{1} = 1:5$.
Question 8:
easyA 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)
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 $K = \frac{1}{2} mr^2 \omega^2$.
Question 9:
moderateA 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)
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 $3\text{ J}$.
Question 10:
easyA 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)
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 $5 : 7$.