- #1
sierra52
- 10
- 4
I am trying to work out the angular acceleration of a small flywheel (with axle running through) when an attached piston (by means of a pin) with an outstroke force acts on it. I assumed I would need to work out the torque using the formula torque = force x radius, and find out the flywheel’s and axle’s moment of inertia using I = 1/2 mr^2 , since the flywheel is shaped as a hollow cylinder, in order to calculate the angular acceleration.
Flywheel’s outer diameter = 30mm
Flywheel’s inner radius = 10mm
Axle diameter = 6mm
Mass of flywheel = 0.075 kg
Mass of axle = 0.00888 kg
Distance from centre of wheel to pin = 11.25 mm
Diameter of piston/cylinder bore = 10mm
Cylinder air pressure = 85 PSI = 0.586 MPaThis is what I did but I got an unrealistic answer. Ignoring frictional resistance
Working out the piston outstroke force
F=pa = (0.586x10^6) x π(5x10^-3)^2 = 46.0 N
Working out the torque
T=Fr = 46(11.25x10^-3) = 0.518 Nm
Working out the moment of inertia for the flywheel
I = 1/2*mr^2 = (1/2) (0.075) (15x10^-3)^2 = 8.4375x10^-6
Working out the moment of inertia for the axle
I = 1/2*mr^2 = (1/2) (0.00888) (3x10^-3)^2 = 39.96x10^-9
Therefore, total moment of inertia is 8.4375x10^-6 + 39.96x10^-9 = 8.48x10^-6
Using torque = moment of inertia x angular acceleration,
Angular acceleration = 0.518/(8.48x10^-6) = 61103 rads^-2
This value seems ridiculous. I must have gone wrong somewhere?
Flywheel’s outer diameter = 30mm
Flywheel’s inner radius = 10mm
Axle diameter = 6mm
Mass of flywheel = 0.075 kg
Mass of axle = 0.00888 kg
Distance from centre of wheel to pin = 11.25 mm
Diameter of piston/cylinder bore = 10mm
Cylinder air pressure = 85 PSI = 0.586 MPaThis is what I did but I got an unrealistic answer. Ignoring frictional resistance
Working out the piston outstroke force
F=pa = (0.586x10^6) x π(5x10^-3)^2 = 46.0 N
Working out the torque
T=Fr = 46(11.25x10^-3) = 0.518 Nm
Working out the moment of inertia for the flywheel
I = 1/2*mr^2 = (1/2) (0.075) (15x10^-3)^2 = 8.4375x10^-6
Working out the moment of inertia for the axle
I = 1/2*mr^2 = (1/2) (0.00888) (3x10^-3)^2 = 39.96x10^-9
Therefore, total moment of inertia is 8.4375x10^-6 + 39.96x10^-9 = 8.48x10^-6
Using torque = moment of inertia x angular acceleration,
Angular acceleration = 0.518/(8.48x10^-6) = 61103 rads^-2
This value seems ridiculous. I must have gone wrong somewhere?