Answer:
(a) 7.81 N
(b) 15.6 N
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
