Question

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 0.709 m and 1.11 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 6.11 rad/s in 9.71 s?

Answer 1

Use the rotational equivalent of the linear equation F = m*a => T = I*alpha where I is the moment of inertia of the rod and alpha is the angular acceleration in rad/s^2
I = m*L^2/3 = 1.03*0.709^2/3 = 0.214.
Use the rotational equivalent of the linear equation V = Vi + a*t
=> ?(t) = ?i + alpha*t = 6.23 = 0 + alpha*9.65 => alpha = 6.23/9.65 = 0.646 rad/s^2
T = I*alpha = 0.214*0.646 = 0.14N-m = 0.709*F => F = 0.14/0.709 = 0.175N <------------
If this helps you, please vote best answer.