rotational-motion

Question: If $I_1$ is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass, and $I_2$ is the moment of inertia of the ring formed by bending the rod, then

1) $I_1 : I_2 = 1 : 1$

2) $I_1 : I_2 = \pi^2 : 3$

3) $I_1 : I_2 = \pi : 4$

4) $I_1 : I_2 = 3 : 5$

Solution:

Explanation: No Explanation

moment of inertia

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