Question

A 1.79 m long uniform rod with a mass of 2.03 kg is allowed to rotate. By

pushing perpendicularly on the edge of the rod, a student produces an
angular acceleration of 12.9 rad/s2. What is the minimum force the
student could be providing if (a) the rod is rotating around its center.
What is the minimum force the student could be providing if (b) the rod
is rotating around the opposite end?

Answer 1

(a) 7.81 N

(b) 15.6 N

Step-by-step explanation:

For a rod rotating around its center, I = 1/12 mL².

For a rod rotating around its end, I = 1/3 mL².

(a) Sum of the torques on the rod:

∑τ = Iα

F L/2 = (1/12 mL²) α

F = 1/6 m L α

F = 1/6 (2.03 kg) (1.79 m) (12.9 rad/s²)

F = 7.81 N

(b) Sum of the torques on the rod:

∑τ = Iα

FL = (1/3 mL²) α

F = 1/3 m L α

F = 1/3 (2.03 kg) (1.79 m) (12.9 rad/s²)

F = 15.6 N