Calculate RPM given the force of a torsion spring

AI Thread Summary
To calculate the RPM of a flywheel driven by a torsion spring, the torque produced by the spring's force must first be determined, which in this case is calculated as 0.667 Nm. The angular acceleration is then found to be 345.5 rad/s² based on the flywheel's moment of inertia. However, to convert this acceleration into RPM, additional information is needed, such as the duration the torque is applied. The discussion highlights that the torque is not constant due to the nature of the torsion spring, suggesting that the flywheel behaves like a simple harmonic oscillator. Ultimately, measuring the flywheel's motion post-activation may provide practical insights into its RPM.
saaaaam
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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
= 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]
 
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saaaaam said:

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
= 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]
I di don't 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
 
saaaaam said:
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
Since it is a torsion spring the torque will not be constant. In principle you have an SHM oscillator. The flywheel speed will be maximised each time the torsion spring is at its relaxed position. So what you need to know is the energy initially stored in the spring.
But that is ignoring practical considerations of friction and drag.
 
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