Calculating the Magnitude of Torque to Stop a Spinning Rod

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To calculate the torque required to stop a spinning rod, the initial angular velocity (ωi) is 2.1 rad/s, and the final angular velocity (ωf) is 0 rad/s after 5.0 seconds. Using the equation ωf = ωi + α*t, the angular acceleration (α) can be determined as α = (ωf - ωi) / t, resulting in α = -0.42 rad/s². The moment of inertia (I) for a rod spinning about its center is I = (1/12) * m * L², which equals 0.167 kg*m² for the given rod. Finally, applying the torque formula Torque = Iα provides the required torque values, leading to the conclusion that the correct answer is B.) 0.13 N*m.
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A 2.0 kg, 1.0 m long rigistic plastic rod is spinning about its center of mas at 2.1 rad/s. What is the magnitude of torque that will bring the rod to a halt in 5.0 s?

A.) 0.07 N*m
B.) 0.13 N*m
C.) 0.21 N*m

If you could help me set up the proper equation to use, I would greatly appreciate this.
 
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If ωi is initial angular velocity. ωf is the final angular velocity and α is the angular acceleration then
ωf = ωi + α*t.
From the given values find α.
Torque = Ια.
 


Thanks Brotha!:bugeye:
 
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