A constant torque of 1500 N-m turns a wheel of M.I 300 kg m^2 about an axis passipassing through its center the angular velocity of the wheel after 3 second will be
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Answer:15 rad/sec
Explanation:
Torque = Moment of inertia✖️ angular acceleration
Angular acceleration = 1500/300 = 5 rad/sec^2
Omega F. = omega I.+Angular acceleration✖️ time
Omega F. = 0 + 5✖️3
=15 rad/sec
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