A wire of mass m and length L is bent in the form of a circular ring. The moment of inertia of the ring about its axis is
Let the radius of Ring formed is R. Then
2πR= L ⇒ R = L/2π
Moment of Interia= m (L/2π)²= mL²/ 4π²
A uniform solid sphere and a uniform hollow sphere of the same mass have the same moment of inertia about their diameters. Then the radii of solid and hollow sphere are in the ratio
We are given \(I_{\text{solid}} = I_{\text{hollow}} ⇒ \frac{2}{5}M R_s^2 = \frac{2}{3}M R_h^2\). Thus, \(\frac{R_s}{R_h} = \sqrt{\frac{5}{3}}\).
A flywheel of moment of inertia \(1\text{ kg m}^2\) and radius 1 m starts rotating due to a constant torque 3 Nm. The velocity of a point on the rim after 1 s is (in \(\text{ms}^{-1}\))
Torque \(\tau = I\alpha ⇒ 3 = 1 \times \alpha ⇒ \alpha = 3\text{ rad/s}^2\). After 1 second, angular velocity is \(\omega = \alpha t = 3\text{ rad/s}\). The linear velocity is \(v = \omega R = 3 \times 1 = 3\text{ ms}^{-1}\).
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, \(K = \frac{3}{4}\).
The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is
About central normal axis, \(I_1 = \frac{1}{2}MR^2 = MK_1^2 \Rightarrow K_1 = \frac{R}{\sqrt{2}}\). About diameter, \(I_2 = \frac{1}{4}MR^2 = MK_2^2 \Rightarrow K_2 = \frac{R}{2}\). The ratio is \(\frac{K_1}{K_2} = \sqrt{2} : 1\).
Assertion (A): Value of radius of gyration of a body depends on axis of rotation.
Reason (R): Radius of gyration is rms distance of particles of the body from the axis of rotation.
Radius of gyration is defined as \( k = \sqrt{I/M} \), where \( I \) is the moment of inertia. Since \( I \) depends on the axis of rotation, \( k \) also depends on the axis of rotation. R correctly defines \( k \) as an RMS distance from the axis, which means its value depends on that axis. Both A and R are true, and R explains A.
Assertion (A): Moment of inertia about an axis passing through center of mass is maximum.
Reason (R): Theorem of parallel axis can be applied only for two dimensional body of negligible thickness.
Assertion (A) is false because the moment of inertia about an axis passing through the center of mass is the minimum, not maximum, for parallel axes (\(I = I_{CM} + Md^2\)).
Reason (R) is false because the parallel axis theorem is applicable to any rigid body, not just 2D bodies of negligible thickness.
Assertion (A): If the moment of inertia of a non-uniform thin circular ring is same about two different axes parallel to each other and lying in the plane of ring, then both the axis can be at same distance from geometrical centre of the ring.
Reason (R): From parallel axis theorem \(I = I_{cm} + md^2\), (where terms have usual meaning). Moment of inertia of a body about two axes parallel to each other and at a same distance from centre of mass of the body is same.
Assertion (A) is true: If two axes have the same moment of inertia, they must be equidistant from the center of mass. It is possible for them to also be equidistant from the geometrical center (e.g., if the CM coincides with the geometrical center).
Reason (R) is true: The parallel axis theorem states \(I = I_{cm} + md^2\), so if two parallel axes are at the same distance \(d\) from the CM, their moments of inertia will be equal. Both statements are true, but R does not directly explain A.
Assertion (A): It will be much easier to accelerate a merry-go-round full of children if they stand close to its axis then if they all stand at the outer edge.
Reason (R): For larger moment of inertia, the angular acceleration is small for given torque.
From the relation \(\tau = I \alpha\), where \(\tau\) is torque, \(I\) is moment of inertia, and \(\alpha\) is angular acceleration. If children stand closer to the axis, \(I\) decreases. For a given \(\tau\), a smaller \(I\) leads to a larger \(\alpha\), making it easier to accelerate. So, (A) is true. Reason (R) states that for larger \(I\), \(\alpha\) is small for given \(\tau\), which is also true and explains (A).