If the ice on the polar caps of the Earth melts, the duration of day will
As polar ice melts and water flows towards the equator, mass is distributed further from the rotational axis, increasing the moment of inertia \(I\). Due to conservation of angular momentum, the angular velocity \(\omega\) decreases, which increases the duration of the day.
A particle of mass \(m\) is moving on a circle of radius \(R\) with kinetic energy \(K\). Then angular momentum of particle about centre of circle will be:
Kinetic energy \(K = \frac{p^2}{2m}\) gives momentum \(p = \sqrt{2mK}\) . Angular momentum is given by \(L = pR = \sqrt{2mK} R\).
Assertion: In the absence of external torque kinetic energy of a system remains conserved.
Reason: In the absence of external torque angular momentum of a system remains conserved.
No external torque means angular momentum is conserved (Reason is true). Kinetic energy may not be conserved as internal forces can change it (Assertion is false).
Assertion: A body is in translational equilibrium if the net force on it is zero.
Reason: A body is in rotational equilibrium if the net torque on it about any one point becomes zero.
Translational equilibrium requires net force to be zero. Rotational equilibrium requires net torque to be zero. Both statements are true but independent definitions.
A particle is moving along a straight line parallel to x-axis with constant velocity. Its angular momentum about the origin:
The angular momentum is \(L = m v d\), where \(d\) is the constant perpendicular distance of the line of motion from the origin. Since \(m\), \(v\), and \(d\) are all constant, \(L\) remains constant.
If the rotational kinetic energy of a body increased by 300% then determine the percentage increase in its angular momentum:
Since rotational kinetic energy \(K = \frac{L^2}{2I}\), we have \(L \propto \sqrt{K}\). An increase of 300% means \(K_f = 4K_i\), so \(L_f = 2L_i\). The percentage increase in angular momentum is 100%.
A disc of mass \(2\text{ kg}\) and radius \(0.2\text{ m}\) is rotating with angular velocity \(30\text{ rad/sec}\). If a mass of \(0.25\text{ kg}\) is put gently on periphery of disc then angular velocity of disc is :
By conservation of angular momentum, \(I_1 \omega_1 = I_2 \omega_2\). Here, \(I_1 = \frac{1}{2} M R^2\) and \(I_2 = \frac{1}{2} M R^2 + m R^2\). Substituting the values: \(1 \times 30 = (1 + 0.25) \omega_2\), which gives \(\omega_2 = 24\text{ rad/sec}\).
From a circular ring of mass \(M\) and radius \(R\) an arc corresponding to a \(90^\circ\) sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is \(K\) times \(MR^2\). Then the value of \(K\) is
The remaining mass of the ring after removing a quarter (\(90^\circ\) out of \(360^\circ\)) is \(M' = \frac{3}{4}M\). Since all parts of the remaining ring are still at a perpendicular distance \(R\) from the center axis, the moment of inertia is \(I = M' R^2 = \frac{3}{4} M R^2\). Thus, \(K = \frac{3}{4}\).
Consider the following statements \(A\) and \(B\), and identify the correct answer:
**Statement A:** In a perfectly rigid body, the net positive work done by external torques increases the rotational kinetic energy of the body.
**Statement B:** Angular acceleration of a rotating body having fixed axis of rotation is inversely proportional to the moment of inertia of the body for a given torque.
Statement A is correct because the rotational work-energy theorem states that work done equals change in rotational kinetic energy. Statement B is correct since \(alpha = frac{tau}{I}\), meaning angular acceleration is inversely proportional to the moment of inertia.
A uniform circular disc of mass \(2\text{ kg}\) and radius \(100\text{ cm}\) (hinged at centre) is subjected to a constant torque of \(4\text{ Nm}\). If the disc was initially at rest, then its angular speed after \(4\text{ s}\) will be
Moment of inertia \(I = \frac{1}{2} M R^2 = \frac{1}{2} \times 2 \times 1^2 = 1\text{ kg m}^2\). Angular acceleration \(\alpha = \frac{\tau}{I} = \frac{4}{1} = 4\text{ rad/s}^2\). Angular speed \(\omega = \alpha t = 4 \times 4 = 16\text{ rad/s}\).