Home › JEE Advanced › Physics › Rotational Motion › By **parallel-axis theorem**, the moment of iner…
By **parallel-axis theorem**, the moment of inertia of a thin ring of mass $M$, radius $R$, about a tangent in its plane is:
A$MR^2$
B$\tfrac{1}{2}MR^2$
C$\tfrac{3}{2}MR^2$
D$2MR^2$
Answer & Solution
Correct answer: C. $\tfrac{3}{2}MR^2$
About diameter (in plane), $I_d = MR^2/2$ (perpendicular-axis theorem: $I_z = 2I_d = MR^2$ gives $I_d = MR^2/2$). Tangent in plane: $I = MR^2/2 + MR^2 = 3MR^2/2$.
Related questions
The branch describing the motion of a rotating rigid body about a fixed axis is rotationalSix washers on a light rod illustrate the moment of inertia of a system of:The more massive an object is, the more of this it has in linear motion:Unlike linear position measured in metres, angular position is measured in units that are:Acceleration caused by the change in direction of tangential velocity is called:Several torques acting about one axis are combined by finding their:The perpendicular distance from the axis to the line of the force is called the:The SI unit of torque is the newton multiplied by the: