A solid cylinder with a mass of 266 kg and radius 0.060 m is rotating with an angular speed of 84.0 rad/s about an axis passing through its center and perpendicular to its end faces. The rotation of the cylinder is slowed down by a factor of 2 by applying a tangential frictional force to it for 5.30 s. What is the magnitude, in N, of the friction force applied to the cylinder?
oldschool
Favorite Answer
Torque = I*α = ½*266*0.06² *α = 0.4788*α = F*0.06
α = (84-42)/5.3 = 7.92453
0.4788*7.92453/0.06 = F = 63.2N
electron1
Let’s use the following equation to determine the friction force.
Force * time = moment of inertia * change of angular velocity
The following equation is used to calculate the moment of inertia of a solid cylinder.
I = ½ * m * r2 = ½ * 266 * 0.06^2 = 0.4788
Change of angular velocity = 84 – 42 = 42 rad/s
Force * 5.30 = 0.4788 * 42
F = 20.1096 ÷ 5.30
The force is approximately 3.79 N. I hope this is helpful for you.