Calculate RPM given the force of a torsion spring

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Homework Help Overview

The discussion revolves around calculating the RPM of a flywheel subjected to a force from a torsion spring. The flywheel's moment of inertia and the force applied at specific distances and angles are provided, but the participants are exploring the implications of these factors on the RPM calculation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the torque produced by the applied force and its relation to angular acceleration. Questions arise regarding the necessity of additional information, such as the duration of the applied torque, to determine the RPM accurately.

Discussion Status

Some participants acknowledge the initial calculations and express uncertainty about proceeding without further information. Suggestions include timing the flywheel's motion to count revolutions as a practical approach, while also noting the complexities introduced by the non-constant torque of the torsion spring.

Contextual Notes

Participants highlight the lack of information regarding the duration of the torque application and the potential effects of friction and drag on the flywheel's motion. The discussion also touches on the energy stored in the spring as a factor in the analysis.

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.