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Kinetic wood log splitter

Kinetic wood log splitter

Kinetic wood log splitter

(OP)
Hello.
I am facing a problem that I have tried to solve so hard but I keep getting unreasonable numbers.
I am trying to design a kinetic wood log splitter. For this design, I made a flywheel that has a radius of 0.30m and a mass moment of inertia of 1.25 kg.m2 and I assumed after breaking the angular velocity would be 70% of the initial angular velocity. In other terms
πœ”2=0.7*πœ”1
During impulse, it was mentioned that the maximum force to break the log is 9000N
Using the principle of angular impulse and momentum that states πΌπœ”1+Ξ£βˆ«π‘€.𝑑𝑑=πΌπœ”2
I am trying here to compute the angular velocity but I don't have the moment change with respect to time (Ξ£βˆ«π‘€.𝑑𝑑). I have tried to assume that Ξ£βˆ«π‘€.𝑑𝑑 = Σ∫r.F.𝑑𝑑 where Σ∫F.𝑑𝑑 is equal to the linear impulse that was provided to be 1200 N.s and r is the outer radius of the flywheel. Therefore, πœ”1=π‘Ÿπ‘βˆ«πΉ.𝑑𝑑/0.3 𝐼 = 0.3*1200/(0.3*1.25) = 960 rad/s!!!! which is extremely large. Can someone please tell me what is not right here or what is the unrealistic approach that I am using in my designing calculation? I am truly losing my brain Please.

RE: Kinetic wood log splitter

You have mixed the problem, particularly units. The force required is 9000N, but that's not the amount of energy that is required. The flywheel, on the other hand, stores energy and not force.

For most wood the log splits completely after the first 10% or so separates. I haven't measured, but I have swung a splitting ax enough to estimate this is the case when the grain is straight. If not, and I've recently split twisted grain wood from near the roots, impulse won't do it. In either case, you need to know how far the force is required to be applied.

For the numbers - you need to figure out the energy required to perform the split Force * distance and set the difference in energy content between πœ”1 and πœ”2 equal to that. Since the mass moment of inertia is the same and you have the limit that 70% of πœ”1 should be retained then you have an equation with one unknown, which is 30% of πœ”1.

RE: Kinetic wood log splitter

(OP)
Thanks for the answer
Actually in my calculations i didn't use the 9000 value. I only mentioned it because it was reported in the problem statement. I have the equation with one unknown which is πœ”1 but still I'm getting large number
πœ”1=π‘Ÿπ‘βˆ«πΉ.𝑑𝑑/0.3 𝐼 = 0.3*1200/(0.3*1.25) = 960 rad/s
I'm using 1200 which is the impulse here.
Maybe you can elaborate bit more for me please? Just regarding how to use the equation you mentioned correctly.

RE: Kinetic wood log splitter

Impulse has the units of work. The change in rotation is the change in energy - that is the work done by the flywheel.

That is that 1/2 I (πœ”1^2-πœ”2^2) = work extracted.

You need to work through the units on your equation - for one thing, the radius doesn't figure into solving your problem. I have no idea what "p" is. Just appears and then disappears.


Well, that's not right.

RE: Kinetic wood log splitter

Back to units; there was an impulse, not work. Sigh. Old habits. Anyway:

1200 N*sec * [(1kg*m/sec^2)/N] = 1200 kg*m/sec (because sec/sec^2 leaves "/sec"] so what is needed is to eliminate everything but the /sec

Then (1200 kg*m/sec / 1.2 kg*m^2)*.3 m = (1200*.3) kg*m^2/sec)/(1.2 kg*m^2) = 300 radians/sec. [kg cancels and so does m^2]


What is "p" in your equation?

RE: Kinetic wood log splitter

(OP)
Thanks a lot, man. I appreciate it.
I was confused about what you meant by 'p' as well. It is a subscript rp which stands for the flywheel radius.
So do you think the number is logical? I believe I should have a larger flywheel in this case however the mass will be very large or have two flywheels which each of them will have a large mass moment of inertia.

RE: Kinetic wood log splitter

I think my answer is logical - I don't think your equation was, but it's up to you to check the units.

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