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Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M = 1kg and radius R = 1m about an axis perpendicular to the hoop’s plane at an edge. (Express your answer in units of kg*m^2).

Respuesta :

Answer:

2 kg m^2

Explanation:

M = 1 kg, R = 1 m

The moment of inertia of the hoop about its axis perpendicular to its plane is

I = M R^2

The moment of inertia of the hoop about its edge perpendicular to it splane is given by the use of parallel axis theorem

I' = I + M x (distance between two axes)^2

I' = I + M R^2

I' = M R^2 + M R^2

I' = 2 M R^2

I' = 2 x 1 x 1 x 1 = 2 kg m^2

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