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Making Sure Physics Calculations Are Correct For A Rotational Motion / Torque Problem

  • Thread starter jumi
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This isn't really a homework problem, more of a "I'm-building-something-and-need-to-make-sure-the-physics-are-right-before-I-build-something-impractical" problem.

Homework Statement


Homework Equations


The Attempt at a Solution


(I grouped it all together, easier for me to not separate each segment)

I need to spin a motor handle/crank at 25,000 rpm. I need to show that using a single flywheel attached to the spinning motor handle is unpractical.

I know that 25,000rpm is ~2618 rad/sec; this value should be angular frequency, ω, correct?

Then relating that to torque, tau = Iω, where I = mr^2 for a ring of radius r and mass m. And torque also equals tau = rFsinθ, where θ = 90°.

Therefore, Iω = rF

and F = mrω.

Assuming m = 1 kg, r = 0.5 m, F = ~1309 N

Therefore torque = ~655 Nm.

Are the physics relationships correct?

Thanks in advance.

EDIT: Also, tangential velocity is v = rω, correct?
 
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Answers and Replies

  • #2
gneill
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This isn't really a homework problem, more of a "I'm-building-something-and-need-to-make-sure-the-physics-are-right-before-I-build-something-impractical" problem.

Homework Statement


Homework Equations


The Attempt at a Solution


(I grouped it all together, easier for me to not separate each segment)

I need to spin a motor handle/crank at 25,000 rpm. I need to show that using a single flywheel attached to the spinning motor handle is unpractical.
What's a motor handle/crank? What's a motor handle? Are they the same thing? What's the purpose of the flywheel? What is to be considered "unpractical"?

At 25,000 rpm you're going to have to be awfully careful about symmetry and balance. Material stresses and strains could lead to "an unfortunate incident" type of behavior.
I know that 25,000rpm is ~2618 rad/sec; this value should be angular frequency, ω, correct?
Yes.
Then relating that to torque, tau = Iω, where I = mr^2 for a ring of radius r and mass m. And torque also equals tau = rFsinθ, where θ = 90°.
Is the flywheel a ring or a solid disk?
Therefore, Iω = rF

and F = mrω.
No, mrω has the units of linear momentum. And what force are you describing by F? How is this force supposed to be applied?
Assuming m = 1 kg, r = 0.5 m, F = ~1309 N

Therefore torque = ~655 Nm.

Are the physics relationships correct?
You need to describe the physical system in more detail. Because the description is vague it's not possible to comment with any accuracy.
Thanks in advance.

EDIT: Also, tangential velocity is v = rω, correct?
Yup.
 
  • #3
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What's a motor handle/crank? What's a motor handle? Are they the same thing? What's the purpose of the flywheel? What is to be considered "unpractical"?

At 25,000 rpm you're going to have to be awfully careful about symmetry and balance. Material stresses and strains could lead to "an unfortunate incident" type of behavior.
Yes.
Is the flywheel a ring or a solid disk?
No, mrω has the units of linear momentum. And what force are you describing by F? How is this force supposed to be applied?

You need to describe the physical system in more detail. Because the description is vague it's not possible to comment with any accuracy.

Yup.
Thanks for the reply.

I'm powering an 18V DC Motor using an electric dremel connected to the motor shaft / "handle" (idk if you're familiar with small DC motors, there's a little central shaft the sticks out of the body that is rotated to produce voltage) that's spinning at 25,000rpm. I need to show that trying to build a hand crank that one can rotate by hand is unpractical due to the high angular velocity required.

The flywheel is a ring.

The force in question is supposed to be the force applied at the edge of the flywheel (i.e. where a handle might go) perpendicular to the radius, since tau = r x F.

Does that clear some stuff up? Let me know.

Thanks.
 
  • #4
gneill
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20,797
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Force won't determine the rotation speed. For a frictionless system, ANY force that provides torque will eventually lead to infinite angular speed! The trick will be applying the torque!

I think you'll have to look elsewhere for the limiting factors. For example, how might you determine what the maximum rate of motion of a human arm might be for the desired action?
 
  • #5
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Well I reasoned that since v = rω, given a flywheel with a radius of 0.5 m (anything larger would be impractical since 0.5 is pushing the practical size limit), the tangential velocity would be 1309 m/s or 2928 mph (for a value that's easier to visualize).

Therefore, no matter what speed the human arm can generate, it would not get anywhere close to the required speed, thereby making a hand crank impractical.

Is the physics of this reasoning valid?

Thanks for all the help, btw. It's been a good while since I've done rotational motion stuff.
 
  • #6
gneill
Mentor
20,797
2,777


Well I reasoned that since v = rω, given a flywheel with a radius of 0.5 m (anything larger would be impractical since 0.5 is pushing the practical size limit), the tangential velocity would be 1309 m/s or 2928 mph (for a value that's easier to visualize).

Therefore, no matter what speed the human arm can generate, it would not get anywhere close to the required speed, thereby making a hand crank impractical.

Is the physics of this reasoning valid?
While I agree with your conclusion, I won't be able to agree with your "physics by assertion" :smile: For a proper physics based argument you need to show why an arm cannot move fast enough, either by calculation or by experiment (You could see how many times you can move your hand, as fast as you can, in the required rotary motion over a given time interval. Have someone time it. What RPM can you achieve? BY what factor does it fall short of the required value? Is it plausible that practice would allow you to make up the difference?)
Thanks for all the help, btw. It's been a good while since I've done rotational motion stuff.
I'm happy to help!
 

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