# Calculate RPM given the force of a torsion spring

## Homework Statement

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.

## Homework Equations

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
= 0.667Nm
Torque = Ia
a = Torque/I = 0.667/0.0019302
Am i lacking a length of time this torque is applied for? I struggle to se where to go from here[/B]

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## Homework Statement

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.

## Homework Equations

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
= 0.667Nm
Torque = Ia
a = Torque/I = 0.667/0.0019302
Am i lacking a length of time this torque is applied for? I struggle to se where to go from here[/B]
I di dont check your numbers but the approach looks good . Note that $\alpha$ is in rad/s^2. To get to your question, I am puzzled too. It is not possible to calculate an RPM without more information. For example the amount of time it was applied, as you said (and then we would need to know if the direction of the force changes as the wheel rotates, and if so how it does). There is really no other information provided?

This is actualy the analysis of a project i've made, so yes there is no other information i can add to it. I'm thinking i'll time the flywheels motion after activation and just count the revolutions... i was just hoping there was a theory based way to calculate it! thanks for the reply

haruspex