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Gyroscopic Precession - Mass of wheel & Angular Momentum

  • #1

Homework Statement:

Gyroscopic Precession - Mass of wheel & Angular Momentum - Torque on axle

Relevant Equations:

F = MA
Hello,

I have this i am learning. I have been trying to find information online but have struggled to find anything which helps me. YouTube usually has good videos, but doesn't seem to on this. This is one topic i have never learned before. But keen to.

I was hoping someone could help me complete this, or point me in the right direction please.

I am stuck on determine the mass of the wheel. Possibly some simpe way to do this, but my mind is just blank. Once i have calculated this, i am hoping the next step i will be able to figure out.

Thank you.

Gyroscopic Question.png
 

Answers and Replies

  • #2
BvU
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Hi,

You'll have to find a path from the given data to the mass of the wheel. Obviously, the last step will be from moment of inertia to mass, since that is the only relationship where ##m## features.

What is a good candidate for the link (relationship) between the moment of inertia and the other givens ?
 
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  • #3
Hello,

Would it be m = f / a to find the mass?
 
  • #4
which would then be:

To find f:

f = t/r

f = 4Nm / 0.6m

f = 6.66Nm

T = 6.66Nm x 0.6m

T = 4Nm

So:

m = a/f

m = 27 rads/s / 4Nm = 6.75kg

Mass = 6.75kg

Is this correct?

If not could you help as i am really struggling to learn this.
 
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  • #5
BvU
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Is this correct?
Did you check he dimensions ?

You need a relevant formula for rotation that is equivalent to ##v = v_0+at## in uniformly accelerated linear motion (*). Your
struggling to learn this
makes me wonder if you have anything at all to work with for rotations. Are you familiar with the relationships in the link ?

(*) With a constant acceleration from 4N, what mass would reach 27 m/s in 24 seconds when starting from standstill ?
 
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  • #6
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...
I was hoping someone could help me complete this, or point me in the right direction please.

I am stuck on determine the mass of the wheel...
Ben,
This is a complex combination of torques about three axes and rotations about two of those.
Nevertheless, determining the value of the mass of that bicycle wheel only involves one torque and one rotation.

Let's say a hand applies a constant tangential force onto the initially at rest vertical wheel during those 24 seconds.
The wheel should increase its rotational speed about its axis at a constant rate during that time.

That rate is limited by its rotational inertia (or moment of inertia), which depends on its mass and the square of its radius.
At the end of those 24 seconds of applied torque, the wheel has gained angular impulse.

The result of your post #4 is not correct because you did not use the Newton's second law related equations for rotational movement.
You can find those in this link:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

:cool:
 
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  • #7
Thank you to you both for replying. Very much appreciated.

BvU I am not familiar with the relationships no. I am starting from scratch.
I have watched a lot of videos, and read loads online. But all problems give the mass without calculating it. This is why i am struggling on this part.

Would the relevant formula be Delta x = (v + V0 / 2) * t ?

Lnewqban thank you for the link.

From the information you guys have given me, and the link. I have these calculations. Are these correct? Or on the right track?

Delta x = (v + V0 / 2) * t

Delta x = (27rad/s + 0) / 2) * 24

Delta x = 324 m


t = I * Alpha

I = t / Alpha

I = 324 / 4 = 81

I = mr^2

M = I / m^2

M = 81 / 0.36

M = 225

So mass = 225.

Is this correct?

If not where have i gone wrong?
 
  • #8
BvU
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Would the relevant formula be Delta x = (v + V0 / 2) * t ?
No. That has nothing to do with rotations.
##x## is a length, measured and expressed in meters.
27 radians/second is an angular speed.
To achieve 27 rad/s in 24 seconds with a constant torque, the constant angular acceleration is (27 rad/s ) / (24 s) = 1.125 rad/s2. The formula is ##\omega = \omega_0 + \alpha t## -- in both links it is right next to the linear motion equivalent ##v = v_0 + a\,t##

Are these correct? ##\qquad\qquad## no
Or on the right track? ##\qquad\qquad## no

Delta x = (v + V0 / 2) * t ##\qquad\qquad## no. what is this supposed to mean ?

Delta x = (27rad/s + 0) / 2) * 24 ##\qquad\qquad## no

Delta x = 324 m ##\qquad\qquad## no. The dimension you get multilpying rad/s with s can never be m

t = I * Alpha ##\qquad\qquad## no, That is not t (time, seconds) but ##\tau## (torque, Newton.m )

I = t / Alpha

I = 324 / 4 = 81 ##\qquad\qquad## no. First the 324 was a delta x, now it is a ##\tau## ? And Alpha is not 4.
I suggest you try to find a textbook to your liking that disscusses angular motion.
 
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  • #9
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I actually was unable to determine the mass of the wheel from the limited data given.
It's pretty trivial if one can assume that the entire mass of the wheel is concentrated at the radius 0.6m given, but the description doesn't say that. Identical radius wheels with the same angular moments can have different masses given different distributions of mass.

One could determine it from the rate of precession if the axle is assumed to be weightless, but neither detail is provided.
 
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  • #10
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Thank you to you both for replying. Very much appreciated.
...
Is this correct?

If not where have i gone wrong?
You are very welcome, Ben. :smile:
Please, remember what we discussed in your previous thread about the use of units and how to combine those in your calculations.

Unfortunately, the last attempt is still not correct.
Take a look at these basic equations about rotational movement and let us know if you can see how the expressions ##a=F/m## and ##\alpha=\tau/I## are related:

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

Units for given torque are Newtons-meter.
Units for angular acceleration should be radians/second square.
Units for moment of inertia should be kilograms-meter square.
1 Newton = 1 Kilogram-meter/second square.
 
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  • #11
BvU i am going to get some books on it. So from your equation i can calculate the mass? But as Halc has said he was unable to determine the mass for the wheel. This is why i am struggling. As i can find no way to calculate this.

So how would this question be completed? It is written to be completed. But i can find no way to complete it. Even through searching online, no one shows this.

Thanks for keep helping me.
 
  • #12
etotheipi
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It is a strange question, I can't tell what it's asking. If you take a cylindrical coordinate system centred where the axle meets the cord, then the equation of motion is something like $$\vec{\tau} = \vec{r} \times M\vec{g} = (Mgd)\hat{\theta} = \frac{d\vec{L}}{dt} = \frac{d}{dt} \left(Md^2 \omega_p \hat{z} + I_{cm}\omega_s \hat{r} \right)$$For steady state (!) the first term on the RHS drops out and you can use that ##\dot{\hat{r}} = \dot{\theta} \hat{\theta}##. The total angular momentum about the origin is obtained via. the König theorem. Also ##d## is the distance between the origin and centre of mass of the wheel. That will get you the precession equation.

Anyway I wouldn't spend so much time worrying about this question, it's not clear what we're supposed to do IMO.
 
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  • #13
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So from your equation i can calculate the mass? ##\qquad\qquad## yes, via another relationship
Thanks for keep helping me.##\qquad\qquad## you are welcome
It's pretty trivial if one can assume that the entire mass of the wheel is concentrated at the radius 0.6m given, but the description doesn't say that.
Actually, it does:

1593428037016.png


@Ben_Walker1978 : as in post #2, I try to bring acrosss that the solution for an exercise involves finding a path from the givens to what is asked.

In this exercise the first part is to find the mass. Only ocurrence is in ##I= mR^2## with ##R## given.
So finding ##I## might be a good idea.
At that point you need to ' translate' ##\ F = m\,a\ ## to ##\ \tau = I\,\alpha\ ## -- see the links given.
##\alpha\;## is the next focus, and you need to be able to 'translate' ##\ v = a\,t\ ## to ##\ \omega = \alpha\,t\ ## -- again, see links.

Without some familiarity with these quantities ##\ \phi, \omega, \alpha\ ## and their parallels ##\ x, v, a\ ## you are groping in the dark (which is why I advised a textbook, but the links also provide a pile of material). And of course the relationships. If you understand the parallel, you only need to remember the relationships for one case (the linear motion) and then the 'translation' to rotations is straightforward.

If we can all agree on the first result, ##m##, we can get ready for the second part of the exercise. Such a result is a value plus a checked dimension. Best way is an expression in the givens in the form of symbols -- so only a single calculation and a simultaneous determination/check of the dimension.
 
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  • #14
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Actually, it does:

Moment of inertia = mR^2
Ah, I didn't read it that way, taking R to be the size of the wheel, but of course it's a bit bigger than that. R defines its moment. Yes, the mass is easily determined from this.
 
  • #15
From reading the replies, and reading on the internet i have got this. I have also ordered 2 books to read and learn.

Moment of Inertia = $$mR^2$$

So $$ I = mR^2$$

To find I:

Translate $$F = Ma$$ to $$ Torque = I \cdot \alpha$$

I will complete this when i receive my books.

So now i will solve $$\alpha$$

$$\alpha = \frac{w2 - w1}{t}$$

$$\alpha = \frac{27 rad/s}{24 seconds}$$
$$\alpha = 1.125rad/s^2$$

I am quite confident up to now this is correct.

So now solve:

$$torque = I \cdot \alpha$$

Rearrange to:

$$I = \frac{torque}{\alpha}$$

$$I = \frac{4Nm}{1.125rad/s^2}$$

$$I = 3.55kgm^2$$

Now Rearrange Moment of Inertia to find the mass:

$$M = \frac{I}{R^2}$$

$$M = \frac{3.55kgm^2}{0.6m^2}$$

$$M = 9.86kg$$

So the Mass is 9.86kg.

Is this now correct?

If not where have i gone wrong. I feel i am now so close to understanding.

Any replies much appreciated.
 
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  • #16
BvU
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I am quite confident up to now this is correct.
Good. And I agree.
So the Mass is 9.86 kg.
Correct, although rounding off in the complete expression yields 9.88 kg.

(minor detail, since R is only given in one decimal. Since the other givens are in two decimals, I don't think more than two digits is quite correct; nevertheless I would write down 9.88 kg too :rolleyes:
The picture shows a pretty heavy wheel -- definitely not a racing bike :biggrin: )

Well done so far. Next step is easy: angular momentum.

And looking ahead: http://steamexperiments.com/experiment/gyroscopic-precession/

and a Walter Lewin show (at approx 24')
 
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  • #17
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... Is this now correct?

If not where have i gone wrong. I feel i am now so close to understanding.

Any replies much appreciated.
Excellent progress, Ben!
Your result is close to mine (9.87 Kg).

Please, note that the actual mass of the wheel differs some from the calculated value, which corresponds to an equivalent hoop of 0.6 m radius, for which the used equation of moment of inertia fully applies.
We are approximating our real world wheel, which mass is less concentrated, to that ideal hoop.

Please, keep going. :smile:
 
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  • #18
Thank you both for your help!! Very much appreciated! I am now going to attempt to finsh the question off.

My books have now arrived aslo, so i can read them.

Thank you again.
 
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  • #19
Now i have finished the rest of the question.

Angular Momentum:

$$L = I \cdot w$$

$$I = 3.55kgm^2$$
$$w = 27Rad/s$$

$$ L = 3.55kgm^2 \cdot 27Rad/s$$
$$ L = 9.85kgm^2/s$$

Torque in Axle:

$$Torque = mgl$$
$$Torque = 9.86kg \cdot 9.81m/s \cdot0.12m$$
$$Torque = 11.6N$$

Explain energy transfer that is produced during gyroscopic procession. Referring to the conservation of momentum in this instance:

When the torque is applied, then the torque is topped. The angular momentum of the system can not change. This is because their is no torque on the system. Spin angular momentum will always move in the direction of the torque.
When torque is applied to a wheel. We get procession. Conservation of angular momentum will always be maintained as long as we have no net torque acting on the object.

Are these correct?

I though i would do the writing to ensure i understand.

If not where have i gone wrong?

Thank you.
 
  • #20
BvU
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## L = 3.55\ {\rm kg\, m^2} \cdot 27\ {\rm Rad/s}= 9.85 \ {\rm kg\,m^2/s}##
Nope. Estimate a result before using a calculator... Even then, the answer is wrong. Previously you had
$$ I = {\tau \over \alpha} = {\tau \,\Delta t\over \Delta \omega} $$ so now you have $$ L = I\,\omega = \tau\, \Delta t = 96 \ {\rm kg\, m^2/s}$$ exactly.

Here you see the advantage of working with expressions in symbols: things that divide out don't lead to extra rounding errors.

More later...
 
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  • #21
BvU
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Torque=11.6 N
Remark: symbols in expressions are in italic, but units must be in regular (\text{ ... } or {\rm ... } or {\sf ...}

Error: g is in m/s2, not in m/s
Error 2: kg m/s2 times m is not Newton:

1 N = 1 kg m/s2
Torque is in N m (force x arm)​


When the torque is applied, then the torque is topped. ##\qquad\qquad you mean 'stopped' (1) ?
The angular momentum of the system can not change. ##\qquad\qquad ???
This is because their is no torque on the system. ##\qquad\qquad you just calculated a torque of 11.6 Nm ???
...

We get procession##\qquad\qquad It's called precession
(1) the torque of 4 Nm is applied until the 27 rad/s is reached -- if is that the one you mean with 'is topped' then I agree.
'When the torque is applied ' -- you mean when the support on the free end of the axle is removed (as in the videos). The description in the exercise text is a bit vague, but my perception is that you are supposed to read it that way (otherwise it really gets much too complicated)

I don't know what to do with
1593989337196.png

Since ##\tau \perp \omega##, the 11.6 Nm only changes the direction of ##\omega,\ ## I would think no energy transfer occurs, but that sure doesn't feel good: where does the energy for the rotation about a vertical axis come from ? A decrease in ##\omega##, a slight drop of the free end of the axle ?

Pehaps @etotheipi has a suggestion ?
 
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  • #22
etotheipi
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I don't know what to do with
View attachment 265928
Since ##\tau \perp \omega##, the 11.6 Nm only changes the direction of ##\omega,\ ## I would think no energy transfer occurs, but that sure doesn't feel good: where does the energy for the rotation about a vertical axis come from ? A decrease in ##\omega##, a slight drop of the free end of the axle ?

Pehaps @etotheipi has a suggestion ?
I think you are right; for such an idealised gyroscope, the kinetic energy is constant. If ##\omega_p## is the precession angular velocity and ##\omega_s## is the spin angular velocity of the wheel, then by the König theorem the total kinetic energy is $$T = \frac{1}{2}Md^2 {\omega_p}^2 + \frac{1}{2}I_{wheel}{\omega_s}^2$$I'll finish off the analysis I started earlier,$$\vec{\tau} = \vec{r} \times M\vec{g} = (Mgd)\hat{\theta} = \frac{d\vec{L}}{dt} = \frac{d}{dt} \left(Md^2 \omega_p \hat{z} + I_{cm}\omega_s \hat{r} \right) = I_{cm} \omega_s \omega_p \hat{\theta}$$ $$\omega_p = \frac{Mgd}{I_{cm}\omega_s}$$so with constant spin angular velocity ##\omega_s##, we also have constant ##\omega_p##.
 
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  • #23
Hello,

Thank you both for replying.

Both very knowledgeable!

For angular momentum it was a typo. I had:

$$L = I \cdot w$$
$$L = 3.55kgm^2 \cdot 27Rad/s$$
$$L = 95.85kgm^2/s$$

You have just rounded it up to 96.

For Torque i have re calculated to:

$$\text{Torque} = mgl$$
$$\text{Torque} = 9.86kg \cdot9.81m/s^2 \cdot 0.12m$$
$$\text{Torque} = 11.6Nm$$.

With regards to the text. I will get something wrote up. Just want to get the calculations correct so i can learn, and write up neat so i remeber.

Thanks again for all the help.

Also @BvU Thanks for the tip about how to write text. Always wondered how to do that

Ben
 
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  • #24
BvU
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You have just rounded it up to 96
I have not rounded off !$$L = I\,\omega = \tau\, \Delta t = = 4 \times 24 = 96 \ {\rm kg\, m^2/s}$$exactly As I tried to explain. Unclear or not understood ?
And $$ I = {\tau \over \alpha} = {\tau \,\Delta t\over \Delta \omega} = 3.55555555555555555555...$$ when not rounded off, so $$M = I/r^2 = 9.87654321... \approx 9.88 $$not 9.86 or 9.87.

Not that it matters much here, but in other exercises it can cause errors.
 
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