Question

The M.I. of a ring of mass 500 gm and radius 0.5 m about a tangent perpendicular to its plane is...

Answer 1

Answer:

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Answer 2

Answer:

0.250 kg m^2.

Explanation:

mass M = 0.5 kg, Radius= R = 0.5 m.

The moment of Inertia of the ring about its axis = M R^2.

I_axis = 0.5 × 0.5^2 = 0.125 kg m^2.

The tangent to the ring, perpendicular to its plane is parallel to the axis and distance R away from the axis. We use the parallel axes theorem of the moment of Inertia.

I_tangent = I_axis + M R^2

= 2 M R^2 = 0.250 kg m^2.