A torque of 1.45 N-m is applied to a thin rod, fixed to rotate about a point through its center. If the rod is 2.50 kg in mass and 1.11 m in length, find the resulting angular acceleration (in rad/s2) of the rod. (The moment of inertia for a thin rod rotating about a point through its center is /= ML2/12)

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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A torque of 1.45 N-m is applied to a thin rod, fixed to rotate about a point through its center. If the rod is 2.50 kg in mass and 1.11 m in length, find the resulting angular acceleration (in
rad/s2) of the rod. (The moment of inertia for a thin rod rotating about a point through its center is I = ML2/12)
Transcribed Image Text:A torque of 1.45 N-m is applied to a thin rod, fixed to rotate about a point through its center. If the rod is 2.50 kg in mass and 1.11 m in length, find the resulting angular acceleration (in rad/s2) of the rod. (The moment of inertia for a thin rod rotating about a point through its center is I = ML2/12)
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