A constant torque of `31.4 N-m` is exterted on a pivoted wheel. If the angular acceleration of the wheel is `4 pi rad//s^2`, then the moment of inertia will be.

Options

A. `2.5 kg-m^2`
B. `5.8 kg-m^2`
C. `4.5 kg-m^2`
D. `5.6 kg-m^2`

Answer

A

Explanation

(a) The moment of inertia (I) of a body about an axis is equal ot the torque (tau) required to produce unit angular acceleration (prop) in the body about that axis is
`tau = I prop`
Given. `prop = 4 pi rad//s^2, tau = 31.4 Nm`
`rArr I = (tau)/(prop) = (31.4)/(4 xx 3.14) = 2.5 kg-m^2`.