Just looking for confirmation for a torque calculation

[jbriggs444]

There is another less important matter that I did not want to bring up while we were still trying to agree upon a solution.

When working a problem, it is good practice to work with symbols and algebra all the way to the end and only substitute numbers in once one has a finished formula in hand.

It is a hard practice to follow. My own inclination is much like yours. I try to figure out one piece at a time, putting numbers to everything and doing some sanity checks as I go to make sure that I've not fallen off the rails halfway through.

Let us try to work the problem symbolically. I will be thinking and writing as I go, so please forgive any wrong turns along the way.

Our strategy will be the one that you have chosen. Figure the rotation rate. Use that to compute the acceleration rate. Use that to compute torque. And use that to compute rotational work.

The average rotation rate ($\omega_\text{avg}$) is given by the angle rotated through ($\Delta \theta$) divided by the elapsed time ($\Delta t)$:$$\omega_\text{avg} = \frac{\Delta \theta}{\Delta t}$$For constant acceleration, the average rotation rate is also given by:$$\omega_\text{avg} = \frac{\omega_f - \omega_0}{2}$$We can relax our rule about plugging in numbers and simply take $\omega_0$ as 0. It is more of a fixed feature of the problem than an adjustable parameter. We can solve and get:$$\omega_f = \frac{2 \Delta \theta}{\Delta t}$$We know that angular acceleration ($\alpha$) is given by the change in angular velocity divided by the change in time. Or:$$\alpha = \frac{\omega_f - \omega_0}{\Delta t} = \frac{\omega_f}{\Delta t} = \frac{2 \Delta \theta}{{\Delta t}^2}$$We need to convert that angular acceleration to a torque ($\tau$) using:$$\tau = I \alpha$$For a cube with side length $l$ about an axis through the centers of two opposing faces, we have the formula:$$I = \frac{1}{6}ml^2$$Substituting, we get:$$\tau = \frac{1}{6}ml^2 \times \frac{2 \Delta \theta}{{\Delta t}^2} = \frac{ml^2 \Delta \theta}{3 {\Delta t}^2}$$To get work from torque, we need to multiply by the rotation angle ($\Delta \theta$). This gives:$$E = \tau \Delta \theta = \frac{ml^2 {\Delta \theta}^2}{3 {\Delta t}^2}$$Now goodness only knows, I have the ability to screw up the algebra. So let us plug in some numbers.

$m$ = 0.0075 kg
$l$ = 0.01 m
$\Delta \theta$ = 180 degrees = $\pi$ radians = 3.14159 radians
$\Delta t$ = 0.001 seconds.

I get 2.46740 kg m²/sec²

Wowsers. And we can see the inverse quadratic scaling with $\Delta t$ right there in the formula.

 

[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.

So increasing the time to 0.01 seconds instead of 0.001 seconds the final energy will be 0.02467 J

 

[pete94857]

There is another less important matter that I did not want to bring up while we were still trying to agree upon a solution.

When working a problem, it is good practice to work with symbols and algebra all the way to the end and only substitute numbers in once one has a finished formula in hand.

It is a hard practice to follow. My own inclination is much like yours. I try to figure out one piece at a time, putting numbers to everything and doing some sanity checks as I go to make sure that I've not fallen off the rails halfway through.

Let us try to work the problem symbolically. I will be thinking and writing as I go, so please forgive any wrong turns along the way.

Our strategy will be the one that you have chosen. Figure the rotation rate. Use that to compute the acceleration rate. Use that to compute torque. And use that to compute rotational work.

The average rotation rate ($\omega_\text{avg}$) is given by the angle rotated through ($\Delta \theta$) divided by the elapsed time ($\Delta t)$:$$\omega_\text{avg} = \frac{\Delta \theta}{\Delta t}$$For constant acceleration, the average rotation rate is also given by:$$\omega_\text{avg} = \frac{\omega_f - \omega_0}{2}$$We can relax our rule about plugging in numbers and simply take $\omega_0$ as 0. It is more of a fixed feature of the problem than an adjustable parameter. We can solve and get:$$\omega_f = \frac{2 \Delta \theta}{\Delta t}$$We know that angular acceleration ($\alpha$) is given by the change in angular velocity divided by the change in time. Or:$$\alpha = \frac{\omega_f - \omega_0}{\Delta t} = \frac{\omega_f}{\Delta t} = \frac{2 \Delta \theta}{{\Delta t}^2}$$We need to convert that angular acceleration to a torque ($\tau$) using:$$\tau = I \alpha$$For a cube with side length $l$ about an axis through the centers of two opposing faces, we have the formula:$$I = \frac{1}{6}ml^2$$Substituting, we get:$$\tau = \frac{1}{6}ml^2 \times \frac{2 \Delta \theta}{{\Delta t}^2} = \frac{ml^2 \Delta \theta}{3 {\Delta t}^2}$$To get work from torque, we need to multiply by the rotation angle ($\Delta \theta$). This gives:$$E = \tau \Delta \theta = \frac{ml^2 {\Delta \theta}^2}{3 {\Delta t}^2}$$Now goodness only knows, I have the ability to screw up the algebra. So let us plug in some numbers.

$m$ = 0.0075 kg
$l$ = 0.01 m
$\Delta \theta$ = 180 degrees = $\pi$ radians = 3.14159 radians
$\Delta t$ = 0.001 seconds.

I get 2.46740 kg m²/sec²

Wowsers. And we can see the inverse quadratic scaling with $\Delta t$ right there in the formula.

Yes I see, it makes sense, I'm still at the point of getting used to the symbols and positioning like when to multiply when no sign is shown or the symbol for pi being quite like a zero , I'm going to try and learn a bit of something new each day and everyday practice what I learnt the day before. It's always been my dream to learn the algebra, calculus, physics , math, I've always been a tinkerer and so often I wish I could have just calculated things out more in depth. Please feel free to continue teaching me anything you believe I need to know, thanks again

 

[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.

A) The cars initial speed is zero as it is starting from rest.

The cars average speed 1 mile divided in 100 m is 1609.344 metres / 100 = 16.09344

16.09344 x 10 sec = 160.9344 sec per mile speed = 22.3693629 mile per hour
= 10 metres per second

Final speed of 20 metres per second.

KE = m x v = 1000 kg x 20 m/s = 200, 000 joules.

Can you please explain further a non uniformed acceleration

 

[Steve4Physics]

A) The cars initial speed is zero as it is starting from rest.

Yes

The cars average speed 1 mile divided in 100 m is 1609.344 metres / 100 = 16.09344

16.09344 x 10 sec = 160.9344 sec per mile speed = 22.3693629 mile per hour
= 10 metres per second

Not sure what happened!

You are told that the car covers a distance of 100 metres in 10 seconds. So average speed $= \frac {total~distance}{total~time} = \frac {100m}{10s} = 10m/s$.

Simple as that! No need to even think about miles and hours!

Final speed of 20 metres per second.

Yes.

KE = m x v = 1000 kg x 20 m/s = 200, 000 joules.

No. The correct formula for the kinetic energy of an object of mass $m$ moving at speed $v$ is KE $= \frac 12 mv^2$.

And 1000 x 20 is not 200,000.

Can you please explain further a non uniformed acceleration

Not sure what you are asking but see if this helps.

For uniform acceleration, speed changes at a steady rate, e.g. increases by 3m/s each second:
At t = 0, speed = 0
At t = 1s, speed = 3m/s
At t = 2s, speed = 6m/s
At t = 3s, speed = 9m/s

For non-uniform acceleration, speed doesn't change at a steady-rate, e.g.
At t = 0, speed = 0
At t = 1s, speed = 2m/s
At t = 2s, speed = 6m/s
At t = 3s. speed = 7m/s

Sometimes, calculations with non-uniform acceleration use calculus.

 

[pete94857]

Yes


Not sure what happened!

You are told that the car covers a distance of 100 metres in 10 seconds. So average speed $= \frac {total~distance}{total~time} = \frac {100m}{10s} = 10m/s$.

Simple as that! No need to even think about miles and hours!


Yes.


No. The correct formula for the kinetic energy of an object of mass $m$ moving at speed $v$ is KE $= \frac 12 mv^2$.

And 1000 x 20 is not 200,000.


Not sure what you are asking but see if this helps.

For uniform acceleration, speed changes at a steady rate, e.g. increases by 3m/s each second:
At t = 0, speed = 0
At t = 1s, speed = 3m/s
At t = 2s, speed = 6m/s
At t = 3s, speed = 9m/s

For non-uniform acceleration, speed doesn't change at a steady-rate, e.g.
At t = 0, speed = 0
At t = 1s, speed = 2m/s
At t = 2s, speed = 6m/s
At t = 3s. speed = 7m/s

Sometimes, calculations with non-uniform acceleration use calculus.

OK thanks... I converted to mph just for my own visualisation

 

[pete94857]

Yes


Not sure what happened!

You are told that the car covers a distance of 100 metres in 10 seconds. So average speed $= \frac {total~distance}{total~time} = \frac {100m}{10s} = 10m/s$.

Simple as that! No need to even think about miles and hours!


Yes.


No. The correct formula for the kinetic energy of an object of mass $m$ moving at speed $v$ is KE $= \frac 12 mv^2$.

And 1000 x 20 is not 200,000.


Not sure what you are asking but see if this helps.

For uniform acceleration, speed changes at a steady rate, e.g. increases by 3m/s each second:
At t = 0, speed = 0
At t = 1s, speed = 3m/s
At t = 2s, speed = 6m/s
At t = 3s, speed = 9m/s

For non-uniform acceleration, speed doesn't change at a steady-rate, e.g.
At t = 0, speed = 0
At t = 1s, speed = 2m/s
At t = 2s, speed = 6m/s
At t = 3s. speed = 7m/s

Sometimes, calculations with non-uniform acceleration use calculus.

Yes I do yet it I just had a very bad migraine yesterday but didn't want to be ignorant to your efforts I do really appreciate it thank you

 

[Steve4Physics]

Yes I do yet it I just had a very bad migraine yesterday but didn't want to be ignorant to your efforts I do really appreciate it thank you

No problem. Thank you. I hope the migraine is better. Keep well hydrated and avoid alcohol and caffeine!

Your final answer for the kinetic energy (200,00J) is correct. But it's important to realise (which you probably already do) that what you actually wrote:

KE = m x v = 1000 kg x 20 m/s = 200, 000 joules.

is wrong.

More importantly, you should see the correspondence between the orginal rotating cube problem and the accelerating car problem, They are both solved in the same way.

 

[pete94857]

I do thank you, I should have put KE = 1/2 ×I x Wf

 

[jbriggs444]

I do thank you, I should have put KE = 1/2 ×I x Wf

The correct formula is $KE = \frac{1}{2} \times I \times {\omega_f}^2$

Note that the $\omega_f$ is squared.

If you do not apply the factor of $\frac{1}{2}$ and do not square $\omega_f$ then the result will be angular momentum. We use the symbol "L" for angular momentum: $L = I \times \omega_f$.


In the car problem, the same applies. The correct formula is $KE = \frac{1}{2} \times m \times v^2$

If you do not apply the factor of $\frac{1}{2}$ and do not square $v$ then the result will be momentum. We use the symbol "p" for momentum: $p = m \times v$

 

[pete94857]

The correct formula is $KE = \frac{1}{2} \times I \times {\omega_f}^2$

Note that the $\omega_f$ is squared.

If you do not apply the factor of $\frac{1}{2}$ and do not square $\omega_f$ then the result will be angular momentum. We use the symbol "L" for angular momentum: $L = I \times \omega_f$.


In the car problem, the same applies. The correct formula is $KE = \frac{1}{2} \times m \times v^2$

If you do not apply the factor of $\frac{1}{2}$ and do not square $v$ then the result will be momentum. We use the symbol "p" for momentum: $p = m \times v$

Notes taken thank you