I`m making a flywheel and with the math

[wheelslave1]

Simon

You have to understand I am new at this and doing the best I can..

You see this sort of statement leads me to suspect you don't understand your own figures

Yes, I do understand my own figures. I may not use the correct terms but I am trying to learn. Thats why I came here.

Anyway - taking Δt=0.25s to accelerate: the applied torque would be τ=IΔω/Δt just for that one unit that is accelerating.

No idea what that means.

So...your saying that my f=m*a was incorrect. :(

 

[Simon Bridge]

Simon

You have to understand I am new at this and doing the best I can..

No worries - this is why it is important to respond to direct questions even if they seem odd to you. The people helping you have been doing this for a long time and know what they need.

Yes, I do understand my own figures. I may not use the correct terms but I am trying to learn. Thats why I came here.

Willingness to learn is the important part.
One of the things I have been trying to teach you is communication ... you have been used to using technical terms in a non-standard way: very common with self-taught people.

So...your saying that my F=ma was incorrect. :(

No - that is quite correct - just not so useful. It's more useful to use the rotational equivalent since things are spinning here.

That would be $\tau = I\alpha$ ... which is to say that the torque is equal to the moment of inertia multiplied by the angular acceleration.

angular speed ($\omega$) would be "rpm" so angular acceleration ($\alpha$) is rpm per second ... but we prefer to define angular speed in terms of angle-units of "radiens-per-second" (rad/s) because it makes the math easier. So 60rpm is $2\pi$rad/s.

You have an angular acceleration of $(80-20)\text{rpm}/(0.25)\text{s}=240\text{rpm/s} = 8\pi\text{rad/s}^2$

The moment of the inertia of the flywheel is the sum of the moments of inertia of the bits.
You can look up the formulas on wikipedia. The two weights on the ends would be $mr^2$ each while a cylindrical rod (length 2r) connecting them would be $Mr^2/3$ so:

$I=(2m+\frac{1}{3}M)r^2$

I'd put the values in but the list is now over the page :)

Like I said before - once you have the torque, you can work out the linear force to applied at a given radius - that should help you figure how strong to make those gears.

 

[wheelslave1]

Can`t you just give me the answer?

 

[Simon Bridge]

Where's the fun in that? Don't you want to understand your project?

You have (from post #1)
1) A single rod 2 ft across. Axel in the middle.

That would make r=1' = 0.3m

2) A 2 lb weight on both ends of the arm.

2lbs is about 1kg ... you only need ballpark figures right?
so m=1kg

3) Rod weight 1 lb.

M=0.5kg

So $\tau = (2m+\frac{1}{3}M)r^2\alpha = (2+1/6)(0.3)^2(6\pi)=3.68\text{N.m}$

Looks like about 0.25ft-lbs. if I did the conversion right... I'm not used to imperial measures.

You need to multiply this by the distance from the center (in feet) to where you are applying the torque ... that would be the outside of the cog-wheel ... for the 6" one that is 0.125 lbs would be the force on the cog's teeth.

Doesn't look too bad.
You are cautioned to check these figures before going into production ... no responsibility etc etc.

 

[CWatters]

Sounds much lower than I expected but I haven't checked your calculations.

If that's the torque on the output gear shaft I believe you need to multiply up by the gear ratio to get the torque on the input gear shaft (because it's a step up).

 

[CWatters]

Had a quick look and the sums seem ok.

If the output torque is 3.68 N.m then I think the required input torque is going to be four times that or about 10 N.m

If the input is 20rpm (= 0.7∏ rads/s) then I make the power required about

P = 0.7∏ * 10 = 22W

but just for the 0.25 seconds it takes to spin up.

Perhaps best design it assuming the current to your motor might spike to say 100W briefly to give plenty of margin?

 

[wheelslave1]

Why is there such a big difference between linear acceleration and circular acceleration?

So...the answer is 7.4ft-lbs of torque on the 6in main cog. Also need to order a motor that can spike 100W.

 

[jbriggs444]

Why is there such a big difference between linear acceleration and circular acceleration?

There's not much difference. It's just a matter of leverage.

If you keep a wheel the same size and push farther out toward the rim, you get more leverage on the wheel.

If you make the wheel bigger and keep pushing at the same point, the wheel gets more leverage on you.

 

[mfb]

Ok, I think I get the idea now.

And it leads to an interesting theorem: No gear can have a perfect axial symmetry.Consider the following setup: Three wheels on top of each other, with the same axis, but not locked in their rotation. Assume that there is no loss when they spin. The middle wheel B can be moved vertically to slide on either the lower (A) or the upper (C) wheel. Build some gear to force a 4:1-ratio of the angular velocity of the wheels C and A.

Let the middle wheel B slide on the lower wheel A, accelerate them until they rotate with 20 rpm - the top wheel C will rotate with 80 rpm now. Now lift B until it slides on C: B will accelerate, A and C will slow down. Friction leads to some heat. As soon as the system is in equilibrium, lower the (fast-rotating) B until it slides on A: B will slow down, A and C will accelerate. Friction leads to some heat, so the total energy is lower than before. But at the same time, the setup is identical to the initial conditions - A and B move with the same angular velocity and C with 4 times that value. It is isolated from the environment, so momentum conservation gives 20 rpm for A and B and 80 rpm for C.
What is wrong?

Spoiler

I did not account for the gear. It has to have some external anchor which is not on the same axis, and can exchange angular momentum with the environment.

 

[CWatters]

Why is there such a big difference between linear acceleration and circular acceleration?

There is no difference really. If you drop a steel ball onto a thick steel plate the g forces can be very very high.

Consider designing your mechanisim to prevent the "white rod" stopping rapidly. Perhaps using a soft rubber bump stop. The longer the stopping distance (or time) the lower the torque.