Just looking for confirmation for a torque calculation

[pete94857]

Homework Statement

To rotate a cube shape of 10 mm 180 degrees in 0.001 seconds. The cube weighs 7.5 grams. For simplicity no drag or friction, force applied equally from all sides. Stopping force not included.

Relevant Equations

Moment of inertia.. 1.25 × 10-7 kg . m2

Angular velocity.. (w) 3141.59 rad/s for 180 degrees rotation in 0.001 seconds.

Rotational kinetic energy.. using the values I and w , it equals 6.5 mircojoules.

TL;DR Summary: I've done a torque calculation but as I'm new to this I would appreciate if someone could confirm my answer so I know I'm doing it right.

The energy it would take to rotate a 10 mm cube weighting 7.5 grams 180 degrees in 0.001 seconds. To simplify it I didn't include drag or friction.

My answer is just 0.0000617 Joules

Seems rather low.

 

[kuruman]

How can we tell if it's right or wrong if you don't show how you came up with that number?

Note that energy does not rotate cubes. You need a torque for that and to get a torque you need a force applied appropriately on the object.

 

[Baluncore]

What strategy? Do you apply a constant torque, to accelerate to 90°, then reverse that torque, decelerating to stop at 180° ?

What moment of inertia 'I' value do you use? Do you rotate about an axis through the centre of two opposite faces, or about the longest diagonal, an axis through two diagonally opposite corners?

 

[pete94857]

How can we tell if it's right or wrong if you don't show how you came up with that number?

Note that energy does not rotate cubes. You need a torque for that and to get a torque you need a force applied appropriately on the object.

I know you can apply torque with no energy but to perform a movement I believe does actually require energy.

 

[jbriggs444]

I know you can apply torque with no energy but to perform a movement I believe does actually require energy.

It depends on how exactly you define the problem. I can rotate a cube by 180 degrees with zero net energy expenditure.

A strategy is to spend energy for the first 90 degrees of rotation and then harvest it on the remaining 90 degrees.

 

[pete94857]

It depends on how exactly you define the problem. I can rotate a cube by 180 degrees with zero net energy expenditure.

A strategy is to spend energy for the first 90 degrees of rotation and then harvest it on the remaining 90 degrees.

Sorry I don't understand what you mean by the first paragraph. To your second paragraph then the question would need to change to energy needed to rotate it 90 degrees as the other 90 degrees would be negligible with respect to energy input. I'm try to input the energy to rotate it 180 degrees in 0.001 seconds. , I suppose to your first paragraph I could kind of relate it in the sense of a net zero torque environment maybe. I'm not sure. But if I accelerate it to full speed in 90 degrees it would need to accelerate roughly twice as fast as the other 90 degrees would be deceleration. So in essence if you were to try and harvest the energy back in the second half essentially you'd end up roughly using the same energy as to accelerate it in 180 degrees.

 

[kuruman]

I think the required calculation is to find the rotational kinetic energy if the cube rotates from rest by $\pi$ in 1 ms. That should be straightforward if one assumes constant acceleration.

To @pete94857:
Use the kinematic equation $$\Delta \theta=\dfrac{\omega_0+\omega_f}{2}\Delta t$$ with $\omega_0=0$ to find the final angular speed $\omega_f$. Your calculation assumes that the angular speed is the same from the beginning to the end of the 1 ms interval and is off by a factor of 2.

 

[jbriggs444]

Sorry I don't understand what you mean by the first paragraph. To your second paragraph then the question would need to change to energy needed to rotate it 90 degrees as the other 90 degrees would be negligible with respect to energy input.

Your analysis of the second half is not correct. The energy during the second half is not negligible. It is negative. You can recoup all of the energy that you initially used. For a net of zero.

 

[jbriggs444]

I think the required calculation is to find the rotational kinetic energy if the cube rotates from rest by $\pi$ in 1 ms. That should be straightforward if one assumes constant acceleration.

It is also straightforward if one optimizes the acceleration profile to minimize energy expenditure under the constraints that 180 degrees rotation is achieved within 1 ms and that torque is always positive.

One then finds that the optimum acceleration profile involves an impulsive torque at the beginning of the interval. This should cut the energy requirement by a factor of 4 over the constant acceleration approach.

 

[pete94857]

It is also straightforward if one optimizes the acceleration profile to minimize energy expenditure under the constraints that 180 degrees rotation is achieved within 1 ms and that torque is always positive.

One then finds that the optimum acceleration profile involves an impulsive torque at the beginning of the interval. This should cut the energy requirement by a factor of 4 over the constant acceleration approach.

Can you show me this ?

 

[jbriggs444]

Can you show me this ?

It follows from an energy argument. The final kinetic energy is equal to the energy supplied to the cube.

We know that the average rotation rate will be 180 degrees in 1 millisecond.

In the constant acceleration case, the final rotation rate will be twice the average rotation rate.

In the initial acceleration case, the rotation rate will be constant after the initial acceleration. So the final rotation rate will match the average rotation rate.

So the final rotation rate in the constant acceleration case is twice the final rotation rate in the initial acceleration case. Twice the rotation rate. Four times the energy.

The constraint on torque being "always positive" is there to prevent cheating with regenerative braking.

Is that what you wanted shown? Or are you concerned with something else?

 

[pete94857]

I think the required calculation is to find the rotational kinetic energy if the cube rotates from rest by $\pi$ in 1 ms. That should be straightforward if one assumes constant acceleration.

To @pete94857:
Use the kinematic equation $$\Delta \theta=\dfrac{\omega_0+\omega_f}{2}\Delta t$$ with $\omega_0=0$ to find the final angular speed $\omega_f$. Your calculation assumes that the angular speed is the same from the beginning to the end of the 1 ms interval and is off by a factor of 2.

Thanks, I understand your words and accept what you mean. Unfortunately I don't understand this $$\Delta \theta=\dfrac{\omega_0+\omega_f}{2}\Delta t$$ with $\omega_0=0$ to find the final angular speed $\omega_f$.

 

[jbriggs444]

Thanks, I understand your words and accept what you mean. Unfortunately I don't understand this $$\Delta \theta=\dfrac{\omega_0+\omega_f}{2}\Delta t$$ with $\omega_0=0$ to find the final angular speed $\omega_f$.

Let us try to justify that equation.

$\Delta \theta$ is the angle through which the cube is rotated. 180 degrees in this case.
$\Delta t$ is the elapsed time for the rotation. 1 millisecond in this case.
$\omega_0$ is the initial rotation rate. We expect that it is zero. The cube starts at rest.
$\omega_f$ is the final rotation rate. We are trying to calculate this.

The average rotation rate will be given by $\frac{\omega_0 + \omega_f}{2}$.
The rotation angle will be given by average rotation rate multiplied by elapsed time.

Which gives us the equation in question. Someone just needs to do the algebra to solve for $\omega_f$.

 

[pete94857]

It follows from an energy argument. The final kinetic energy is equal to the energy supplied to the cube.

We know that the average rotation rate will be 180 degrees in 1 millisecond.

In the constant acceleration case, the final rotation rate will be twice the average rotation rate.

In the initial acceleration case, the rotation rate will be constant after the initial acceleration. So the final rotation rate will match the average rotation rate.

So the final rotation rate in the constant acceleration case is twice the final rotation rate in the initial acceleration case. Twice the rotation rate. Four times the energy.

The constraint on torque being "always positive" is there to prevent cheating with regenerative braking.

Is that what you wanted shown? Or are you concerned with something else?

Why 4 times the energy ? Yes I understand now what you mean about trying to recoup the energy in, I suppose there would need to be a conversion efficiency probably not 100 %.

 

[jbriggs444]

Why 4 times the energy ?

$E= \frac{1}{2} I \omega^2$ [repaired to add fraction -- thanks, @kuruman]

Double the rotation rate and you quadruple the energy requirements.
Halve the rotation rate and you divide the energy requirements by four.

Yes I understand now what you mean about trying to recoup the energy in, I suppose there would need to be a conversion efficiency probably not 100 %.

Probably not. If we were designing a real piece of equipment, there would be additional constraints to labor under. None of which have been specified. In order to make the question answerable I've assumed an ideal situation.

 

[pete94857]

I think the required calculation is to find the rotational kinetic energy if the cube rotates from rest by $\pi$ in 1 ms. That should be straightforward if one assumes constant acceleration.

To @pete94857:
Use the kinematic equation $$\Delta \theta=\dfrac{\omega_0+\omega_f}{2}\Delta t$$ with $\omega_0=0$ to find the final angular speed $\omega_f$. Your calculation assumes that the angular speed is the same from the beginning to the end of the 1 ms interval and is off by a factor of 2.

2 millijoules or 2 joules ?

 

[pete94857]

$E=I \omega^2$

Double the rotation rate and you quadruple the energy requirements.
Halve the rotation rate and you divide the energy requirements by four.


Probably not. If we were designing a real piece of equipment, there would be additional constraints to labor under. None of which have been specified. In order to make the question answerable I've assumed an ideal situation.

I see, the formula has just decided to show itself properly before it wasn't showing on my screen it was just ##*_ etc

 

[pete94857]

Thanks, I understand your words and accept what you mean. Unfortunately I don't understand this $$\Delta \theta=\dfrac{\omega_0+\omega_f}{2}\Delta t$$ with $\omega_0=0$ to find the final angular speed $\omega_f$.

It's just decided to load correctly on my screen before it showed my random symbols.

 

[pete94857]

OK thank to all I'll give it another go.

 

[berkeman]

I see, the formula has just decided to show itself properly before it wasn't showing on my screen it was just ##*_ etc

That can happen if you or someone else edits the LaTeX in a thread and does not refresh their browser an extra time to re-render the LaTeX. If you see double-# or double-$ LaTeX delimiters in a post, just refresh your browser to force the LaTeX to be (re-)rendered.

 

[kuruman]

2 millijoules or 2 joules ?

When one of two numbers A and B is off by a factor of 2 this means that one is 2 times larger than the other. In other words, either A = 2B or B = 2A. There are no units involved.

 

[kuruman]

$E=I \omega^2$

Did you mean $E=\frac{1}{2}I\omega^2~$? I know you did but I am pointing it out for OP's sake.

 

[pete94857]

When one of two numbers A and B is off by a factor of 2 this means that one is 2 times larger than the other. In other words, either A = 2B or B = 2A. There are no units involved.

0.0247

 

[kuruman]

0.0247

No, the answer is 42.

 

[pete94857]

No, the answer is 42.

I don't believe it is.

 

[Baluncore]

I don't believe it is.

Then what is the exact question?

 

[pete94857]

Then what is the exact question?

This is the exact question.. To rotate a cube shape of 10 mm 180 degrees in 0.001 seconds. The cube weighs 7.5 grams. For simplicity no drag or friction, force applied equally from all sides. Stopping force not included.

 

[pete94857]

No, the answer is 42.

Unless you are referring to the answer to the world, universe and everything

#hitchhickersguidetothegalaxy

 

[kuruman]

This is the exact question.. To rotate a cube shape of 10 mm 180 degrees in 0.001 seconds. The cube weighs 7.5 grams. For simplicity no drag or friction, force applied equally from all sides. Stopping force not included.

There is no question and no task assigned in the above statement.

Unless you are referring to the answer to the world, universe and everything

#hitchhickersguidetothegalaxy

And since there is no question to be answered, the answer might as well be the answer to everything, namely 42.

 

[pete94857]

There is no question and no task assigned in the above statement.

And since there is no question to be answered, the answer might as well be the answer to everything, namely 42.

OK I see your point. I'm missing the first part of the question. How much energy J does it take to....

 

[Steve4Physics]

@pete94857, may I suggest that the question is badly written. Possibly the intended question is this:

​

A solid, uniform cube measures 10mm x 10mm x 10mm, has a mass of 7.5grams and is able to rotate about an axis passing through the centres of a pair of opposing faces. The cube is initially at rest.​

​

How much energy is required to rotate the cube through 180 degrees in 0.001s?​

​

You may assume that the angular speed increases uniformly (i.e. angular acceleration is constant) and that there are no drag or frictional forces.​

If were a student having to hand-in this work, I would write out my (above) interpretation of the question and include it as part of my answer.

 

[jbriggs444]

In addition to what @Steve4Physics has just stated, the important part of the problem is not the numeric answer that is obtained. The important part is the methodology that is used to attack and solve the problem.

We do not particularly care what number you get for the answer.

We care about what principles you invoke.

We want to see the equations you arrive at and understand how they follow from the principles.

We want to see you solve the equations.

We want to see you extract the appropriate given values from the problem statement, apply any necessary unit conversions and substitute them into the solution that you have obtained.

We want to see you carry the units through the calculation.

In a perfect world we would like to see you discuss a strategy before even starting all of the above.

There is a reason for all of this. It helps us figure out where you went wrong. And where you went right. It helps us help you.

 

[pete94857]

@pete94857, may I suggest that the question is badly written. Possibly the intended question is this:

​

A solid, uniform cube measures 10mm x 10mm x 10mm, has a mass of 7.5grams and is able to rotate about an axis passing through the centres of a pair of opposing faces. The cube is initially at rest.​

​

How much energy is required to rotate the cube through 180 degrees in 0.001s?​

​

You may assume that the angular speed increases uniformly (i.e. angular acceleration is constant) and that there are no drag or frictional forces.​

If were a student having to hand-in this work, I would write out my (above) interpretation of the question and include it as part of my answer.

Yes that's much better thanks

 

[pete94857]

There is no question and no task assigned in the above statement.

And since there is no question to be answered, the answer might as well be the answer to everything, namely 42.

By the way it's my number 1 favourite book.

 

[pete94857]

OK here we go.

I have a different value again... but hopefully a correct process.

Firstly break down the info I need

I have a solid 10 mm cube being rotated around it's own axis the axis is in the centre of a side and goes straight through to the opposite side.

It weights 7.5 grams

No drag or friction.

I want to know how much energy J will be used to accelerate it and complete a 180 degree rotation in 0.001 seconds starting from zero velocity.

I don't have the correct symbols etc on my phone fonts so some parts I'll just have to word.

Part one. Finding inertia I need the length of my cube in metres and the weight in kg. Then multiply them then divide by six.

I = 1/6 × (m 0.0075 x length 0.01 squared)

= 0.000000125 m2

Next bit...

Angular velocity.... w = radians/time

W= pi/ 0.001 =3141.59 radians per second

Next is ...

Angular acceleration (a) = changing w / changing t

As initial w is zero it easily is 3141.59
/
0.001 seconds = 3,141,590 rads per second

Then ...

Torque T = l x a

0.000000125 x 3,141,590 = 0.39269875 Nm

Last bit ...

Work W = T x pi radians

0.39269875 x pi radians = 1.2336995081J

The end.

 

[jbriggs444]

I = 1/6 × (m 0.0075 x length 0.01 squared)

= 0.000000125 m2

0.000000125 kg m²

W= pi/ 0.001 =3141.59 radians per second

Agreed. $\omega_\text{avg}$ = 3141.59 radians per second.

Angular acceleration (a) = changing w / changing t

As initial w is zero it easily is 3141.59
/
0.001 seconds = 3,141,590 rads per second

You are going a bit off the rails here.

You have calculated the average rotation rate as 3141.59 radians per second. I agree that you know the initial rotation rate. What you have not calculated yet is the final rotation rate.

You need the final rotation rate so that you can figure out the difference between the final rotation rate and the initial rotation rate.

Alternately, you could figure out the difference between the average rotation rate and the initial rotation rate and then divide by half of the elapsed time since the average rotation rate will be achieved halfway through the interval.

The units are also incorrect. Angular acceleration should be in rads per second per second.

Stopping at the first discovered error.

 

[Steve4Physics]

@pete94857 I’d like to add to what @jbriggs444 has said,. You might want to try this…

A car has mass 1000kg (not ‘weight 1000kg’ by the way, because in physics 'weight' is a gravitational force measured in newtons!).

The car accelerates uniformly from rest and covers a distance of 100m in 10s.

a) What is the car’s initial speed?
b) What is the car’s average speed?
c) What is the car’s final speed?
d) What is the car’s final kinetic energy?

Your original problem isn’t that different from the above if you think about it. You are just using angles rather than distances and moment of inertia rather than mass.

 

[pete94857]

0.000000125 kg m²


Agreed. $\omega_\text{avg}$ = 3141.59 radians per second.


You are going a bit off the rails here.

You have calculated the average rotation rate as 3141.59 radians per second. I agree that you know the initial rotation rate. What you have not calculated yet is the final rotation rate.

You need the final rotation rate so that you can figure out the difference between the final rotation rate and the initial rotation rate.

Alternately, you could figure out the difference between the average rotation rate and the initial rotation rate and then divide by half of the elapsed time since the average rotation rate will be achieved halfway through the interval.

The units are also incorrect. Angular acceleration should be in rads per second per second.

Stopping at the first discovered error.

Isn't the angular velocity the final rotation rate ?

Angular velocity.... w = radians/time

W= pi/ 0.001 =3141.59 radians sec2

If so any function against zero would surely produce the same value. Please explain

Edit ... because it's in 180 degrees, pi radians accounts for it

So now I'm confused

 

[pete94857]

@pete94857 I’d like to add to what @jbriggs444 has said,. You might want to try this…

A car has mass 1000kg (not ‘weight 1000kg’ by the way, because in physics 'weight' is a gravitational force measured in newtons!).

The car accelerates uniformly from rest and covers a distance of 100m in 10s.

a) What is the car’s initial speed?
b) What is the car’s average speed?
c) What is the car’s final speed?
d) What is the car’s final kinetic energy?

Your original problem isn’t that different from the above if you think about it. You are just using angles rather than distances and moment of inertia rather than mass.

I'm trying to do this problem first thank you.

 

[jbriggs444]

Isn't the angular velocity the final rotation rate ?

Which angular velocity are you talking about?

There is the initial angular velocity. That is zero. That is not equal to the final rotation rate.
There is the average angular velocity. That is 3141.59 radians per second. That second is not squared.
There is the final angular velocity. That is something that you have not calculated.

Angular velocity.... w = radians/time

The units for angular velocity are usually radians per time unit, yes. But that does not tell us anything much.

If the angular velocity were constant then one could calculate the one and only angular velocity by dividing the angle rotated through ($\Delta \theta$) by the time elapsed ($\Delta t$).

But the angular velocity here is not constant.

W= pi/ 0.001 =3141.59 radians sec2

That is not correct.

A correct statement would be that $\omega_\text{avg} = \frac{\Delta \theta}{\Delta t} = \frac{\pi \text{ radians}}{0.001 \text{ seconds}}$ = 3141.59 radians/sec

A helpful equation is: $\omega_\text{avg} = \frac{\omega_0 + \omega_f}{2}$. When starting from rest under uniform acceleration, the average rotation rate is half of the final rotation rate.

If so any function against zero would surely produce the same value. Please explain

Explain what?

Edit ... because it's in 180 degrees, pi radians accounts for it

I have no objection to 180 degrees being $\pi$ radians. The difficulty is your confusion between average rotation rate and final rotation rate when the rotation rate is not constant.

So now I'm confused

Yes. We get that.

 

[pete94857]

Which angular velocity are you talking about?

There is the initial angular velocity. That is zero. That is not equal to the final rotation rate.
There is the average angular velocity. That is 3141.59 radians per second. That second is not squared.
There is the final angular velocity. That is something that you have not calculated.


The units for angular velocity are usually radians per time unit, yes. But that does not tell us anything much.

If the angular velocity were constant then one could calculate the one and only angular velocity by dividing the angle rotated through ($\Delta \theta$) by the time elapsed ($\Delta t$).

But the angular velocity here is not constant.

Which angular velocity are you talking about?

There is the initial angular velocity. That is zero. That is not equal to the final rotation rate.
There is the average angular velocity. That is 3141.59 radians per second. That second is not squared.
There is the final angular velocity. That is something that you have not calculated.


The units for angular velocity are usually radians per time unit, yes. But that does not tell us anything much.

If the angular velocity were constant then one could calculate the one and only angular velocity by dividing the angle rotated through ($\Delta \theta$) by the time elapsed ($\Delta t$).

But the angular velocity here is not constant.


That is not correct.

A correct statement would be that $\omega_\text{avg} = \frac{\Delta \theta}{\Delta t} = \frac{\pi \text{ radians}}{0.001 \text{ seconds}}$ = 3141.59 radians/sec

A helpful equation is: $\omega_\text{avg} = \frac{\omega_0 + \omega_f}{2}$. When starting from rest under uniform acceleration, the average rotation rate is half of the final rotation rate.


Explain what?


I have no objection to 180 degrees being $\pi$ radians. The difficulty is your confusion between average rotation rate and final rotation rate when the rotation rate is not constant.


Yes. We get that.

That is not correct.

A correct statement would be that $\omega_\text{avg} = \frac{\Delta \theta}{\Delta t} = \frac{\pi \text{ radians}}{0.001 \text{ seconds}}$ = 3141.59 radians/sec

A helpful equation is: $\omega_\text{avg} = \frac{\omega_0 + \omega_f}{2}$. When starting from rest under uniform acceleration, the average rotation rate is half of the final rotation rate.


Explain what?


I have no objection to 180 degrees being $\pi$ radians. The difficulty is your confusion between average rotation rate and final rotation rate when the rotation rate is not constant.


Yes. We get that.

I apologise for missing bits out ref should be pi radians/ time. I'm finding it difficult to see why the angular velocity is not equal to the final velocity.

 

[Steve4Physics]

I'm trying to do this problem first thank you.

You should find your Post #1 problem easier once you can do the (simpler) Post #37 problem!

 

[pete94857]

I apologise for missing bits out ref should be pi radians/ time. I'm finding it difficult to see why the angular velocity is not equal to the final velocity.

OK I get it....

 

[pete94857]

I apologise for missing bits out ref should be pi radians/ time. I'm finding it difficult to see why the angular velocity is not equal to the final velocity.

OK here we go... again...

I have a solid 10 mm cube being rotated around it's own axis the axis is in the centre of a side and goes straight through to the opposite side.
It weights 7.5 grams
No drag or friction.
I want to know how much energy J will be used to accelerate it and complete a 180 degree rotation in 0.001 seconds starting from zero velocity.

I don't have the correct symbols etc on my phone fonts so some parts I'll just have to word.

Part one. Finding inertia I need the length of my cube in metres and the weight in kg. Then multiply them then divide by six.

I = 1/6 × (m 0.0075 x length 0.01 squared)
= 0.000000125 m2

Next bit...

Average Angular velocity.... w = radians/time

W= pi/ 0.001 =3141.59 radians per second squared.

As the average angular velocity is half the final velocity (uniformed acceleration) if it wasn't a uniformed acceleration I'd need to elaborate further. Maybe.

3141.59 x 2 = 6,283.18 radians per second squared.

Next is ...

Angular acceleration (a) = changing w / changing t

As initial w is zero it easily is 6,283.18 / 0.001 seconds = 6,283,180 radians per second squared.

Then ...

Torque T = l x a

0.000000125 x 6,283,180 = 0.7853975 Nm

Last bit ...

Work W = T x pi radians

0.7853975 x pi radians = 2.4673990161 J

 

[pete94857]

OK here we go... again...

I have a solid 10 mm cube being rotated around it's own axis the axis is in the centre of a side and goes straight through to the opposite side.
It weights 7.5 grams
No drag or friction.
I want to know how much energy J will be used to accelerate it and complete a 180 degree rotation in 0.001 seconds starting from zero velocity.

I don't have the correct symbols etc on my phone fonts so some parts I'll just have to word.

Part one. Finding inertia I need the length of my cube in metres and the weight in kg. Then multiply them then divide by six.

I = 1/6 × (m 0.0075 x length 0.01 squared)
= 0.000000125 m2

Next bit...

Average Angular velocity.... w = radians/time

W= pi/ 0.001 =3141.59 radians per second squared.

As the average angular velocity is half the final velocity (uniformed acceleration) if it wasn't a uniformed acceleration I'd need to elaborate further. Maybe.

3141.59 x 2 = 6,283.18 radians per second squared.

Next is ...

Angular acceleration (a) = changing w / changing t

As initial w is zero it easily is 6,283.18 / 0.001 seconds = 6,283,180 radians per second squared.

Then ...

Torque T = l x a

0.000000125 x 6,283,180 = 0.7853975 Nm

Last bit ...

Work W = T x pi radians

0.7853975 x pi radians = 2.4673990161 J

If I were to change the time by a factor of 10 example 0.01 could I simply change the final answer to 0.2467 J, I'm assuming yes as looking through it that seems logical.

 

[jbriggs444]

Part one. Finding inertia I need the length of my cube in metres and the weight in kg. Then multiply them then divide by six.

I = 1/6 × (m 0.0075 x length 0.01 squared)
= 0.000000125 m2

The units are kg m². You are still dropping the kg.

Average Angular velocity.... w = radians/time

W= pi/ 0.001 =3141.59 radians per second squared.

As the average angular velocity is half the final velocity (uniformed acceleration) if it wasn't a uniformed acceleration I'd need to elaborate further. Maybe.

3141.59 x 2 = 6,283.18 radians per second squared.

The units for angular velocity are radians per second. Not radians per second squared.

Yes, if this was not uniform acceleration then things would become more difficult.

Angular acceleration (a) = changing w / changing t

As initial w is zero it easily is 6,283.18 / 0.001 seconds = 6,283,180 radians per second squared.

This time the units are right. Angular acceleration is measured in radians per second squared.

Torque T = l x a

0.000000125 x 6,283,180 = 0.7853975 Nm

If you were keeping track of units, that would be kg m² times radians per second squared giving kg m²/second². Which does indeed turn out to be the same thing as a Newton-meter.

Work W = T x pi radians

0.7853975 x pi radians = 2.4673990161 J

Yes. Radians are dimensionless, and a Joule is another name for a Newton-meter.

It would be good to have a sanity check. Let us take that final rotation rate (6283.18 rad/sec) and use it to compute a final kinetic energy.

$KE = \frac{1}{2}I \omega^2 = 0.5 \times 0.000000125 \times 6283.18^2$ = 2.467 J.

Yes, that checks out. Work in is equal to energy out.

 

[jbriggs444]

If I were to change the time by a factor of 10 example 0.01 could I simply change the final answer to 0.2467 J, I'm assuming yes as looking through it that seems logical.

One always needs to be careful when scaling a problem. Does the answer change linearly with the scale factor? Quadratically? Inverse? Inverse quadratically? Something else? A classic example of this kind of issue is the cube-square problem.

If you increase the time by a factor of 10, you reduce the final rotation rate by a factor of 10. If you divide this reduced rotation rate by the increased time, you will see that the required acceleration is reduced by a factor of 100. So the final result is reduced by a factor of 100.

Alternately, if you increase the time by a factor of 10, you reduce the final rotation rate by a factor of 10. If you calculate rotational kinetic energy, the result is reduced by a factor of 100 because the $\omega$ is squared in the formula. So again, by this line of reasoning, we get a reduction by a factor of 100.

So in this case the answer scales inverse quadratically with the elapsed time.

 

[pete94857]

One always needs to be careful when scaling a problem. Does the answer change linearly with the scale factor? Quadratically? Inverse? Inverse quadratically? Something else? A classic example of this kind of issue is the cube-square problem.

If you increase the time by a factor of 10, you reduce the final rotation rate by a factor of 10. If you divide this reduced rotation rate by the increased time, you will see that the required acceleration is reduced by a factor of 100. So the final result is reduced by a factor of 100.

Alternately, if you increase the time by a factor of 10, you reduce the final rotation rate by a factor of 10. If you calculate rotational kinetic energy, the result is reduced by a factor of 100 because the $\omega$ is squared in the formula. So again, by this line of reasoning, we get a reduction by a factor of 100.

So in this case the answer scales quadratically with the elapsed time.

Ah yes I see, thanks very much.

The units are kg m². You are still dropping the kg.


The units for angular velocity are radians per second. Not radians per second squared.

Yes, if this was not uniform acceleration then things would become more difficult.


This time the units are right. Angular acceleration is measured in radians per second squared.


If you were keeping track of units, that would be kg m² times radians per second squared giving kg m²/second². Which does indeed turn out to be the same thing as a Newton-meter.


Yes. Radians are dimensionless, and a Joule is another name for a Newton-meter.

It would be good to have a sanity check. Let us take that final rotation rate (6283.18 rad/sec) and use it to compute a final kinetic energy.

$KE = \frac{1}{2}I \omega^2 = 0.5 \times 0.000000125 \times 6283.18^2$ = 2.467 J.

Yes, that checks out. Work in is equal to energy out.

Thanks

 

[pete94857]

The units are kg m². You are still dropping the kg.


The units for angular velocity are radians per second. Not radians per second squared.

Yes, if this was not uniform acceleration then things would become more difficult.


This time the units are right. Angular acceleration is measured in radians per second squared.


If you were keeping track of units, that would be kg m² times radians per second squared giving kg m²/second². Which does indeed turn out to be the same thing as a Newton-meter.


Yes. Radians are dimensionless, and a Joule is another name for a Newton-meter.

It would be good to have a sanity check. Let us take that final rotation rate (6283.18 rad/sec) and use it to compute a final kinetic energy.

$KE = \frac{1}{2}I \omega^2 = 0.5 \times 0.000000125 \times 6283.18^2$ = 2.467 J.

Yes, that checks out. Work in is equal to energy out.

Got it. you're a star

 

[pete94857]

One always needs to be careful when scaling a problem. Does the answer change linearly with the scale factor? Quadratically? Inverse? Inverse quadratically? Something else? A classic example of this kind of issue is the cube-square problem.

If you increase the time by a factor of 10, you reduce the final rotation rate by a factor of 10. If you divide this reduced rotation rate by the increased time, you will see that the required acceleration is reduced by a factor of 100. So the final result is reduced by a factor of 100.

Alternately, if you increase the time by a factor of 10, you reduce the final rotation rate by a factor of 10. If you calculate rotational kinetic energy, the result is reduced by a factor of 100 because the $\omega$ is squared in the formula. So again, by this line of reasoning, we get a reduction by a factor of 100.

So in this case the answer scales inverse quadratically with the elapsed time.

All notes made ,thanks again for your patience and time it means a lot to me.