Moment of inertia, radius of gyration, MI of simple objects

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A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is : Answer: 3. 7/57 Key idea: moment of inertia of the remaining part = full sphere - removed sphere, using the parallel-axis theorem for the cu... Moment of inertia, radius of gyration, MI of simple objects 1 answer
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm^2. The length of the 400 g rod is nearly : Answer: 2. 8.5 cm Key idea: For a thin rod about its midpoint and perpendicular to its length, I=ML²/12 . L²=12I/M=12×2400/400=72 , so L=√(7... Moment of inertia, radius of gyration, MI of simple objects 1 answer
The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is : Answer: Dropped in the official answer key: full marks to every candidate Radius of gyration is k=√(I/M) . Solid sphere about its axis: I=2/... Moment of inertia, radius of gyration, MI of simple objects 1 answer
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 : Answer: 3. √(2) : 1 Radius of gyration is k=√(I/M) . For a disc, about the normal axis through the centre I=1/2MR² ; about a diameter I=1/4M... Moment of inertia, radius of gyration, MI of simple objects 1 answer
From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° 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 i… Answer: 1. 3/4 Key idea: moment of inertia of a ring about its axis depends only on mass and radius, since every part is at distance R. Remo... Moment of inertia, radius of gyration, MI of simple objects 1 answer
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 ω about their own symmetry axes. The amounts of work (W) required to bring th… Answer: 1. W_C>W_B>W_A Work needed to stop each body equals its rotational kinetic energy, W=1/2Iω² . Same M , R , ω , so W∝ I . Solid spher... Moment of inertia, radius of gyration, MI of simple objects 1 answer
A light rod of length l has two masses m_1 and m_2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is Answer: 4. m₁m₂/m₁+m₂l² Take the two masses at distances x₁ and x₂ from the centre of mass, with x₁+x₂=l and m₁x₁=m₂x₂ . This gives x₁=m₂ l/... Moment of inertia, radius of gyration, MI of simple objects 1 answer
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… Answer: 1. 1 : 5 Rotational kinetic energy is E=1/2 Iω² , so find each moment of inertia first. Sphere about a diameter: I_s=2/5mR² , so E_s... Moment of inertia, radius of gyration, MI of simple objects 1 answer
From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ? Answer: 2. 13 MR²/32 Key idea: remaining part = full disc minus the cut-out disc, using the parallel axis theorem for the hole. Full disc: I... Moment of inertia, radius of gyration, MI of simple objects 1 answer