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Two flywheel problem

by curiouschris
Tags: flywheel
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curiouschris
#1
Nov23-08, 06:49 PM
P: 116
Hi all

First of all this is not a homework problem.

Say I have a flywheel of an arbitrary weight and diameter it is fixed in position with a vertical axis and spinning at oh I don't know 1000 revs per min.

Now I bring a second flywheel of the same size and weight which is not spinning at all into contact with the first.

Now my question is how do I calculate the force on the axles of either flywheel.

I am not sure this describes what I am looking for so I will use a clock face analogy to try and make it clearer.

You are looking down at the first flywheel (axle coming out of the screen) and it is spinning clockwise. twelve o'clock is at the top of the screen 6 o'clock at the bottom 3 to the right and 9 to the left.

If the second flywheel came into contact with the first flywheel at the first flywheels 3 o'clock position I imagine the axle of the first flywheel would attempt to move towards the 12 o'clock position, exactly opposite the tangent described at the 3 o'clock position.

With what force would the axle attempt to move? or more precisely how would I calculate that force?

Is that clearer or more muddy?

Please note: the only purpose of the second flywheel is to introduce friction, a rubber block would server the purpose equally well, I am only interested on the forces on the axle of the first flywheel. it just seemed easier to describe that way.

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rcgldr
#2
Nov24-08, 12:27 PM
HW Helper
P: 7,133
The linear force on the axle of the first flywheel is the same regardless of the point of application of that force. The point of application of the force only affects the torque about the axis. If the rubber block is generating 10 lbs of force eastward, then there is a eastward 10 lbs of force applied to the axle. The torque would be equal to 10 lbs times the radius of the flywheel.
curiouschris
#3
Nov24-08, 06:08 PM
P: 116
Quote Quote by Jeff Reid View Post
The linear force on the axle of the first flywheel is the same regardless of the point of application of that force. The point of application of the force only affects the torque about the axis. If the rubber block is generating 10 lbs of force eastward, then there is a eastward 10 lbs of force applied to the axle. The torque would be equal to 10 lbs times the radius of the flywheel.
So neglecting the directional easterly pressure applied by the rubber block the only force on the axle is torque (twisting force).

In other words if I configured the rubber block like the pads in a disc brake, where the flywheel was squeezed equally between two rubber blocks. Then the only force that was 'felt' by the axle was the torque applied by the rapidly slowing flywheel.

Is that correct?

Why wouldn't the axle try to move in the opposite direction to the tangent at the point of contact with the blocks?

CC

Savant13
#4
Nov24-08, 06:53 PM
P: 85
Two flywheel problem

The force exerting the torque would be friction.
curiouschris
#5
Nov24-08, 09:04 PM
P: 116
The force exerting the torque would be friction.
Of course it would.

Perhaps a drawing may help people understand what I am looking for.

The "unknown force" is the force I am looking for a formula to calculate, I assume it would be an acceleration in the direction indicated.

Please take note the "braking force" is only applied at one point on the flywheels surface, as opposed to the whole surface of the flywheel.
Attached Thumbnails
flywheel.png  
rcgldr
#6
Nov24-08, 09:11 PM
HW Helper
P: 7,133
Quote Quote by curiouschris View Post
So neglecting the directional easterly pressure applied by the rubber block the only force on the axle is torque (twisting force).
It can't be ignored. You'd need 2 rubber blocks on opposite sides of the flywheel for the linear forces to cancel and end up with just a torque about the axle.

Why wouldn't the axle try to move in the opposite direction to the tangent at the point of contact with the blocks?
Blocks or block? If there are 2 blocks on opposite sides and they generate equal and opposite forces, then there is no net force on the axle, just a torque. Otherwise there is a linear force that can't be neglected.
curiouschris
#7
Nov24-08, 11:19 PM
P: 116
Quote Quote by Jeff Reid View Post
Blocks or block?
I did state two blocks, configured as in a disc brake like on a car. so yes two blocks, but not on opposite edges but on opposite sides of the flywheel. If you have seen how a car disc brake works then you'll understand.

See the diagram I uploaded in my previous post.
m.e.t.a.
#8
Nov25-08, 10:31 AM
P: 112
The problem is analogous to spinning up the wheels of a toy car, and then placing the car on the tabletop. The car will accelerate forwards for a few moments while the rotational kinetic stored in the spinning wheels transfers to linear kinetic energy of the whole car (approaching an equilibrium velocity where no more energy transfer occurs, but that's not important here).

If you spin up the wheels again, but this time hold the car's wheels in contact with the tabletop instead of letting it accelerate, then we have a situation similar to your flywheel/brakes scenario. It seems to me that the initial force of friction (per wheel) of the wheels grinding against the tabletop must be equal and opposite to the "pulling" force on each axle (the forwards driving force exerted on the car by the spinning wheels). I only have a minute to finish writing, but at a glance it seems that this pulling force must be the same as the pulling force if you were not holding the car with your hand. Perhaps I've just gone round in a circle, but I find the car analogy easier to understand and the calculation shouldn't be difficult.
curiouschris
#9
Nov25-08, 04:46 PM
P: 116
Hi meta, excellent! that analogy works for me.

So then my question would be, does the radius of the tyre (or flywheel) come in to play when calculating the force applied to the axle.

My logic says that if I have two tyres of different radius's but each having the same stored kinetic energy each size tyre will rotate a different amount based on its radius, thus the energy transfer to the axle would be different (per rotation) dependent on the radius of the tyre.

Given two tyres both storing the same kinetic energy the smaller tyre would have to rotate more to push the vehicle the same distance as the larger tyre.

I have answered my own question. Yes the radius does come into play.

Given that I could probably work out the formula I was looking for in my original post.


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