What torque must be applied to this flywheel?

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dancer_smiley
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Here is the problem: A chainsaw is started by pulling on a rope that is attached to a solid steel flywheel. Assume that the wheel has a mass of 1.5 kg that is completely distributed along the outside edge, and it has a diameter of 10 cm. It starts at rest and must accelerate to 4.0 rev/s in 2.0 seconds in order to start the motor. What torque must be applied to the flywheel?

Equation I used: Angular Acceleration = Torque/Moment of Inertia

Here is my attempt: (1.5 kg)(.05 m)^2 (4.0 rev/s / 2.0 s) = .0075

I don't think this is the correct answer, but I have no idea how to solve this question...
 
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Do I divide my answer by 2(pi)rad to do that?
 
1 revolution = 2pir

You can use the ratio (1 rev)/(2pir) or (2pir)/( 1 rev) to convert from revolutions to radians or vice versa. You have revs and you want to convert to radians so multiply through the ratio that will cancel out the revs.
 
thepatient said:
1 revolution = 2pir

You can use the ratio (1 rev)/(2pir) or (2pir)/( 1 rev) to convert from revolutions to radians or vice versa. You have revs and you want to convert to radians so multiply through the ratio that will cancel out the revs.
One revolution is not 2πr. One revolution is 2π radians without the "r". Then one rev/s is 2π rad/s and x revolutions per second is x*2π rad/s.
 
kuruman said:
One revolution is not 2πr. One revolution is 2π radians without the "r". Then one rev/s is 2π rad/s and x revolutions per second is x*2π rad/s.


Yes. XD 2pi = 1 rev. I don't know what I was thinking.
 
I multiplied .0075*(2pi)
So that makes my answer .047123?
Is that the correct answer for the problem?
 
What units do I use? Is it meters?
 
So my answer is 4.7x10^-2 Nm? Is this correct?
 
Thank you so much for your help! :)