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 uniform rod of mass \(m\) and length \(\ell\) is pivoted about one end and hung vertically. Another mass \(m\) hits it perpendicular to its length with a velocity \(v\) at its midpoint and sticks to it. The initial angular velocity of the rod is:
Conserving angular momentum about pivot: \(L_i = mv\frac{\ell}{2}\). Total moment of inertia is \(I = \frac{1}{3}m\ell^2 + m(\ell/2)^2 = \frac{7}{12}m\ell^2\). Equating \(I\omega = L_i\) yields \(\omega = \frac{6v}{7\ell}\).
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 ladder of length \(\ell\) and mass m is placed against a smooth vertical wall but the ground is not smooth. Coefficient of friction between the ground and the ladder is \(\mu\). The minimum angle \(\theta\) with ground at which the ladder will stay in equilibrium is :
For translational and rotational equilibrium of the ladder, taking torque about the base gives \(N_{\text{wall}} \ell \sin\theta = mg \frac{\ell}{2} \cos\theta\). With \(N_{\text{wall}} = f \le \mu mg\), we get the minimum angle for equilibrium to be \(\tan\theta = \frac{1}{2\mu}\).
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 \(\omega_2 = 24\text{ rad/sec}\).
Three point masses \(m\), \(2m\) and \(3m\) are located at the vertices of an equilateral triangle of side length \(L\). The moment of inertia of the system about an axis passing through mid-point of the side (connecting \(m\) and \(2m\)) and perpendicular to the plane of the triangle, is
The distance of masses \(m\) and \(2m\) from the midpoint of their side is \(L/2\). The third mass \(3m\) lies at a distance of \(h = \frac{\sqrt{3}}{2}L\) (the height of the triangle). The total moment of inertia is \(I = m\left(\frac{L}{2}\right)^2 + 2m\left(\frac{L}{2}\right)^2 + 3m\left(\frac{\sqrt{3}}{2}L\right)^2 = \frac{mL^2}{4} + \frac{2mL^2}{4} + \frac{9mL^2}{4} = 3mL^2\).
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.