I've got a flywheel of Inertia = 0.0019302kg/m^2 (found via solidworks), when a torsion spring is released, a force of 10N acts on the wheel via astring attached 0.065m above and 0.0325m to the right of the wheel's axis at an angle of 40 degrees.
What is the flywheel's RPM (or rad/s or Hz)?
The Attempt at a Solution
Torque produced = FhDv + FvDh = (10*sin(40))0.0325 + (10*cos(40))0.065
Torque = Ia
a = Torque/I = 0.667/0.0019302
= 345.5 rad/s(?)
Am i lacking a length of time this torque is applied for? I struggle to se where to go from here[/B]