Calculate the moment of inertia of a ring about a tangent to the circle of the ring normal to its plane. Text Solution. Solution : Moment of inertia of the ring through CM is. I=MR2.

We know that the moment of inertia about an axis perpendicular to the rod and passing through its centre is ML212. Now we need to M.O.I about an axis through its edge and perpendicular to the rod. Hence, the correct option is C.

Moment of inertia is the name given to rotational inertia, the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation.

Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force).

The moment of inertia of an object is a calculated measure for a rigid body that is undergoing rotational motion around a fixed axis: that is to say, it measures how difficult it would be to change an object’s current rotational speed.

Analyzing rotational inertia Rotational inertia depends both on an object’s mass and how the mass is distributed relative to the axis of rotation. If two objects have the same shape but different mass, the heavier one will have a larger moment of inertia.

Rotational collisions conserve angular momentum When objects collide without a net external torque, the angular momentum is constant. The two objects exert equal, but opposite angular impulses upon each other to maintain the total angular momentum of the colliding system.

Firstly, in horizontal circular motion, the vertical displacement of the object, relative to the ground, remains constant, while in vertical circular motion it does not. In vertical circular motion, the object’s weight force also pulls it towards the ground.