Calculate RPM given the force of a torsion spring

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SUMMARY

The discussion focuses on calculating the RPM of a flywheel with a given inertia of 0.0019302 kg/m² when a torsion spring exerts a force of 10N. The torque produced is calculated as 0.667 Nm, leading to an angular acceleration of 345.5 rad/s². However, participants agree that additional information, such as the duration of torque application and the energy stored in the spring, is necessary to determine the RPM accurately. The analysis suggests that the flywheel behaves like a simple harmonic motion (SHM) oscillator, complicating the calculation due to variable torque.

PREREQUISITES
  • Understanding of torque calculation and its relation to angular acceleration
  • Familiarity with flywheel inertia and its significance in rotational dynamics
  • Knowledge of simple harmonic motion (SHM) principles
  • Basic concepts of energy storage in torsion springs
NEXT STEPS
  • Research the calculation of energy stored in torsion springs
  • Learn about the dynamics of simple harmonic motion (SHM) in mechanical systems
  • Explore methods for measuring angular velocity and RPM in experimental setups
  • Investigate the effects of friction and drag on rotational motion
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Mechanical engineers, physics students, and hobbyists involved in rotational dynamics and energy storage systems will benefit from this discussion.

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.