Rotational MotionMedium
Question
A constant torque of 31.4 Nm is exerted on a pivoed wheel. If the angular acceleration of the wheel is 4π rad/ sec2, then the moment of inertia will be:
Options
A.5.8kg - m2
B.4.5kg - m2
C.5.6kg - m2
D.2.5kg - m2
Solution
Torque = I α
[Iis moment of inertia motion the angular acceleration]
31.4 = 1 × 4p
I =
2.50 kg-m2
[Iis moment of inertia motion the angular acceleration]
31.4 = 1 × 4p
I =
2.50 kg-m2Create a free account to view solution
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