Itotal = Ipulley + Iflywheel
Ipulley = mr^2
Iflywheel = 1/2 mr^2
Itotal = mr^2 + 1/2 mr^2
Itotal = 1.5 mr^2
Force net * radiuspulley = 1.5 mr^2 * angacceleration
Force net * radiuspulley = 1.5 * 80 * 0.230^2 * 1.67
Force net * radiuspulley = 58.904
Force net = 58.904 / 0.230
Force net = 256.3 N
Lower force = Upper force - Force net
Lower force = 135 - 256.3
Lower force = -121.3 N
An electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel. The flywheel is a solid disk with a mass of 80.0kg and a dimaeter of 1.25m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of 0.230m. If the tension in the upper (taut) segment of the belt is 135 N and the flywheel has a clockwise angular acceleration of 1.67 rad/s^2, find the tension in the lower (slack) segment of the belt.
Sol: This problem seems to involve torque; I think solid disc Icm (moment of intertia) is 1/2MR^2; there are other unknown I'm sure but if someone can walk me through this problem (step by step), I'd be most appreciative. I have a test tomorrow. I tried and tried, but finally gave up as I must move on to other practice problems.
force net= upper force - lower force
But force net* radiuspulley= Itotal*angacceleration.
Now I total is the sum of the flywheel and the pulley.
Solve for lower force.
1 answer
1 answer