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.