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?

Question

Raiden Friedman

Physics

29 November, 15:29

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?

+1

Answers (1)

Know the Answer?

Anya Lowe · Answer 1

(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