A uniform rod is set up so that it can rotate about a perpen

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are 0.789 m and 1.27 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of 6.45 rad/s in 8.23 s?

Solution

Here ,

L = 0.789 m

m = 1.27 Kg

wf = 6.45 rad/s

time ,t = 8.23 s

let the force acting is F

Using second law of motion

F * time * r = I * w

F * (0.789/2) * 8.23 = (1/12) * 1.27 * 0.789^2 * 6.45

solving for F

F = 0.131 N

the force needed is 0.131 N