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A constant torque of 31.4 N-m is applied to a pivoted wheel. If the angular acceleration of the wheel is 4p rad/s2, then the moment of inertia of the wheel is (a) 1.5 kg m2 (b) 2.5 kg m2 (c) 3.5 kg m2 (d) 4.5 kg m2

A constant torque of 31.4 N-m is applied to a pivoted wheel. If the angular acceleration of the wheel is 4p rad/s2, then the moment of inertia of the wheel is
(a) 1.5 kg m2 (b) 2.5 kg m2
(c) 3.5 kg m2 (d) 4.5 kg m2

Grade:12

2 Answers

Vikas TU
14149 Points
2 years ago
Torque is the rotational  equivalent of Force. Similarly, Moment of Inertia is the rotational equivalent of Mass and Angular Acceleration is the rotational equivalent of Linear Acceleration.
Just like F = ma, in Rotational Mechanics we have:
T = I * alpha
Where:
 = Torque
 = Moment of Inertia
 = Angular Acceleration
 
We need the Moment of Inertia (I).
We see that if we take  then 
We can now easily get the answer:
I = 5/2 kgm^2

Thus, The Moment of Inertia is (d) 2.5 
Vikas TU
14149 Points
2 years ago
Correctt ans is option B .
Good Luck …............................................................................................

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