Angular acceleration, inertia, torque, mass, force and velocity

[akkamaan]

Homework Statement

I have an angular acceleration question. I have two different scenarios I want to compare. The difference is the distance from rotational center point.

The input torque is 1Nm in both scenarios. I want to accelerate a mass of 2kg over an 90 degree angle (0.5PI rad).

My question is, what is the peripheral velocity and the kinetic energy of the 2 kg mass, at the end of the 0.5PI rad rotation?

Scenario 1. The 2kg mass, have it's center of gravity 1.0m from the rotational center point.

Scenario 2. The 2kg mass, have it's center of gravity 1.2m from the rotational center point.

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Doc Al]

What's the rotational inertia of the 2 kg mass? (Is it a point mass?)

 

[akkamaan]

What's the rotational inertia of the 2 kg mass? (Is it a point mass?)

Yes...lets say it is the head of a sledge hammer

Thanks for the quick response!

 

[Doc Al]

OK, then how do you find the rotational inertia of a point mass?

 

[akkamaan]

OK, then how do you find the rotational inertia of a point mass?

It must be from the rotational work and Newton's 2nd law ??

 

[Doc Al]

It must be from the rotational work and Newton's 2nd law ??

No, I'm talking about the moment of inertia of a point mass. How is it defined?

But to compare the two scenarios, do consider the rotational work done by the torque.

 

[akkamaan]

No, I'm talking about the moment of inertia of a point mass. How is it defined?

that must be the mass*r²

But to compare the two scenarios, do consider the rotational work done by the torque.

Yes

 

[Doc Al]

that must be the mass*r²

Right.

 

[akkamaan]

Right.

so
scenario 1. I=2x1.0²=2Nm²?
scenario 2. I=2x1.2²=2.88Nm²?

But how about the velocity after a 0.5PI rad swing??

 

[Doc Al]

so
scenario 1. I=2x1.0²=2Nm²?
scenario 2. I=2x1.2²=2.88Nm²?

Right. Except for the units. (The units should be Kg-m².)

But how about the velocity after a 0.5PI rad swing??

There are several ways to figure that out:
(1) Use dynamics to figure out the acceleration. Then kinematics to figure out the speed.
(2) Use energy methods.

Try it several ways and compare.

 

[akkamaan]

Right. Except for the units. (The units should be Kg-m².)


There are several ways to figure that out:
(1) Use dynamics to figure out the acceleration. Then kinematics to figure out the speed.
(2) Use energy methods.

Try it several ways and compare.

Ok I gave it a try...
I did set the torque to 25Nm
that made a force on the mass at the end of the
1) 1.0m 2kg F=25N
2) 1.2m 2kg F=20.83N

F=m x a

a1=F/m=12.50 m/s2
a2=F/m=10.42 m/s2

With the radius of 1.0m alt 1.2m the distance for a 0.5PI rad angle will be
d1=1.57m
d2=1.88m

Now i figured out with testing in a spread sheet that i takes
0.67 sec for the 1.0m arm to travel the 0.5pi rad, the final velocity will be 8.39 m/s
and
0.82 sec for the 1.2m arm to travel the 0.5pi rad, the final velocity 8.57 m/s

Does thes number make any sense?
Or did I miss something (which I have a feeling of )??

 

[Doc Al]

Ok I gave it a try...
I did set the torque to 25Nm
that made a force on the mass at the end of the
1) 1.0m 2kg F=25N
2) 1.2m 2kg F=20.83N

F=m x a

a1=F/m=12.50 m/s2
a2=F/m=10.42 m/s2

With the radius of 1.0m alt 1.2m the distance for a 0.5PI rad angle will be
d1=1.57m
d2=1.88m

OK. But even easier would have been to use rotational dynamics: Torque = I*alpha instead of F = m*a.

Now i figured out with testing in a spread sheet that i takes
0.67 sec for the 1.0m arm to travel the 0.5pi rad, the final velocity will be 8.39 m/s
and
0.82 sec for the 1.2m arm to travel the 0.5pi rad, the final velocity 8.57 m/s

Not sure how you figured out the time. Just use kinematics.

To check your answer, solve it using work and energy methods.

 

[akkamaan]

OK. But even easier would have been to use rotational dynamics: Torque = I*alpha instead of F = m*a.


Not sure how you figured out the time. Just use kinematics.

To check your answer, solve it using work and energy methods.

http://goo.gl/mMYBi"

 

[Doc Al]

Your algorithm for calculating the distance (in column H) is off. For a given time, you've correctly calculated the final velocity at that time. To find the distance traveled, use average velocity*time.

Even easier would be to use a kinematic relationship between velocity, distance, and acceleration.

 

[akkamaan]

Your algorithm for calculating the distance (in column H) is off. For a given time, you've correctly calculated the final velocity at that time. To find the distance traveled, use average velocity*time.

OK, I got it. http://goo.gl/mMYBi" , I left it "open" so you can check the formulas in the cells...

Here is my take now on this comparison...
My theory is that, if one swing as hard as one can, with both "hammers", one use the same torque on both.
And I can now see, that, of course, the result at impact will be the same between the 2 cases, only difference is that it takes longer time with a longer shaft. That make the power (Nm/s) different.

Using the same torque, means using the same energy, and the energy at the other en got to be the same.

Am I right in this?? Makes sense to me just now...LOL

Even easier would be to use a kinematic relationship between velocity, distance, and acceleration.

Sorry, I do not understand what you mean there...hmmmm...

I also want to learn how to do it the other ways you mentioned...

There are several ways to figure that out:
(1) Use dynamics to figure out the acceleration. Then kinematics to figure out the speed.
(2) Use energy methods.

Try it several ways and compare.

Have I already "touched" some of that in my discussion above?

 

[Doc Al]

OK, I got it. Here is my spread sheet again, I left it "open" so you can check the formulas in the cells...

Looks good.

Here is my take now on this comparison...
My theory is that, if one swing as hard as one can, with both "hammers", one use the same torque on both.
And I can now see, that, of course, the result at impact will be the same between the 2 cases, only difference is that it takes longer time with a longer shaft. That make the power (Nm/s) different.

Using the same torque, means using the same energy, and the energy at the other en got to be the same.

Am I right in this?? Makes sense to me just now...LOL

Good! The way I look at it--which is what I meant by using "energy methods"--is that the work done is the same in both cases: Work = Torque*theta. So both cases end up with the same energy and thus velocity. You can find the velocity by setting the work equal to the rotational KE.

Sorry, I do not understand what you mean there...hmmmm...

I also want to learn how to do it the other ways you mentioned...

My point was that I see no need for a spreadsheet. You have the distance and acceleration, thus you can find the speed directly using V² = 2ad.

Even simpler would have been to stick to rotational quantities instead of converting them to linear quantities. You'd find the angular acceleration alpha using Torque = I*alpha.

Have I already "touched" some of that in my discussion above?

Definitely.

 

[akkamaan]

Thank you for all the help!

Truly appreciate the way you help!

So why do we not hold the axes head direct in our hand, that will make us split faster??
My answer...there is a physical body limitation how fast we can accelerate our arm, not only from available torque...and in addition to that, we will have a hard time to absorb the impact of the energy in our hand and arm?
A karate trained guy, can handle this different that a normal person??

I realize you might not be a body physician

I do the math you recommended, and make some derivation of formulas...

I have another, related topic about how different designed axes head penetrate wood...
I need to get some input so I can scientifically "anchor" my theory a little...

Shall I start a new thread or keep on, in this thread??

[PLAIN]http://akkamaan.com/firewood/FSS_vs_X25_penetration.jpg

Per A

 

[Doc Al]

So why do we not hold the axes head direct in our hand, that will make us split faster??
My answer...there is a physical body limitation how fast we can accelerate our arm, not only from available torque...and in addition to that, we will have a hard time to absorb the impact of the energy in our hand and arm?

In addition, swinging an axe by the handle allows us to use gravity to our advantage.

A karate trained guy, can handle this different that a normal person??

I'm not sure what you mean.

I have another, related topic about how different designed axes head penetrate wood...
I need to get some input so I can scientifically "anchor" my theory a little...

Shall I start a new thread or keep on, in this thread??

I suggest that you start a new thread.