ANgular Velocity

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  • #1
Alexyboy

Homework Statement


The 15kg plate shown above is able to roate on its axle which is supported on the the journal bearings at A and B. A bullet is fired into the plate providing it with an initial angular momentum of 9kgm/s. Find the angular velocity of the plate when it has rotated thropugh 180 degrees from the postion shown. Neglect the mass of the bullet.



Homework Equations


Moment of Inertia of a Plate = [tex]\frac{1}{3} \times Mass \times width^2[/tex]
[tex]L = I \omega[/tex]

Then I'm not sure, some equations I know for rotation about a fixed axis are:
[tex]\Sigma F_{n}=m\omega^2 r_{g}[/tex]
[tex]\Sigma F_{t}=m\alpha r_{g}[/tex]
[tex]\Sigma M_{g}=I_{G}\alpha[/tex]

The Attempt at a Solution


I calculated I as 0.1125 [tex]kgm^{2}...[/tex] ([tex]\frac{1}{3} \times 15 \times 0.15^{2}[/tex])


so using [tex] \omega= \frac{L}{I}[/tex] this gives me a [tex]\omega[/tex] of 80... however I'm not sure what the units as both 80 rad s^-1 and 80ms^-1 seem unreasonable. I am also unsure how to calulate [tex]\alpha[/tex].
 

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  • #2
Hootenanny
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Welcome to PF Alex,

It would be useful if you could post your diagram, then I can verify you moment of inertia.
 
  • #3
Alexyboy
Image 'Pending Approval'. Sorry I accidently press post (rather than preview) before I had finished and sorry about some of the dodgy latexing as I've never used it before.
 
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  • #4
Alexyboy
diagram of plate now viewable
 
  • #5
Hootenanny
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Nope, 80 rad/s sounds good to me. All your working is correct thus far, now all you need to do is determine the torque about the y-axis.

HINT: It may be useful to introduce an angular variable measuring the angle between the plate and the z-axis.
 
  • #6
Alexyboy
Torque is the rate of change of angular momenutum correct.

Therefore [tex]Torque = \frac{dL }{dt}[/tex]

Using your hint

Therefore [tex]Torque = I \frac{d \omega }{dt}[/tex]

[tex]Torque = I \frac{d^2 \theta}{dt^2}[/tex]

[tex]Torque = I \alpha[/tex]

as [tex] \omega [/tex] is the rate of change of angle

my knowns are at Time = 0, [tex] \theta [/tex] = 0 and [tex] \omega [/tex] = 80 rad/s

do I cross multiply and intergrate both side? Can is the torque constant can I Intergrate it with respect to t?
 
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  • #7
Hootenanny
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Torque is the rate of change of angular momenutum correct.

Therefore [tex]Torque = \frac{dL }{dt}[/tex]

Using your hint

Therefore [tex]Torque = I \frac{d \omega }{dt}[/tex]

[tex]Torque = I \frac{d^2 \theta}{dt^2}[/tex]

[tex]Torque = I \alpha[/tex]

as [tex] \omega [/tex] is the rate of change of angle

my knowns are at Time = 0, [tex] \theta [/tex] = 0 and [tex] \omega [/tex] = 80 rad/s
All good :approve:. Can you now write the torque in terms of the angular variable?

HINT: The only force creating a torque about the y-axis is the weight of the lamina.

Edit: One minor point, it is quite easy to get confused with the direction of rotation, so I would suggest that you indicate the direction using vector notation.
 
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  • #8
Alexyboy
This is where my maths let me down, second order diffentials are kinda fuzzy .

[tex] \tau = I \frac{d^2 \theta}{dt^2}[/tex]

[tex] \frac {\tau}{I} = \frac{d \theta^2}{dt^2}[/tex]


[tex] \frac {\tau}{I} \int\int dt = \int\int d \theta[/tex]

[tex] \frac{\tau}{I} \frac{t^2}{2} = \frac{\theta^2}{2} + C [/tex]

when t = 0 [tex]\theta[/tex] = 0 therefore C = 0

[tex]\tau = \frac I{t^2}{\theta^2}[/tex]


OK obviously I'm missunderstanding something here as i've still got t involved which is a unknown. I haven't used vector notation as the way I've been taught it doesn't seemed to be supported by latex
 
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  • #9
Hootenanny
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This is where my maths let me down, second order diffentials are kinda fuzzy .

[tex] \tau = I \frac{d^2 \theta}{dt^2}[/tex]

[tex] \frac {\tau}{I} = \frac{d \theta^2}{dt^2}[/tex]


[tex] \frac {\tau}{I} \int\int dt = \int\int d \theta[/tex]

[tex] \frac{\tau}{I} \frac{t^2}{2} = \frac{\theta^2}{2} + C [/tex]

when t = 0 [tex]\theta[/tex] = 0 therefore C = 0

[tex]\tau = \frac I{t^2}{\theta^2}[/tex]


OK obviously I'm missunderstanding something here as i've still got t involved which is a unknown. I haven't used vector notation as the way I've been taught it doesn't seemed to be supported by latex
You can't solve the ODE yet since the torque is a function of the angular variable, [itex]\tau = \tau(\theta)[/itex]. Therefore, as I said in my previous post, you need to determine the net torque about the y-axis in terms of the plate's weight and [itex]\theta[/itex] before you can proceed.
 
  • #10
Hootenanny
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Edit: In actual fact you're never going to have to solve and ODE since [itex]\alpha = \dot{\omega}=\boldmath{\ddot{\theta}}[/itex], which is what you wish to determine.
 
  • #11
Alexyboy
aah ok, I missunderstood one your earlier hints.

[tex] \tau = -mgsin\theta [/tex]?
 
  • #12
Hootenanny
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aah ok, I missunderstood one your earlier hints.

[tex] \tau = -mgsin\theta [/tex]?
You're almost correct, but your missing one vital piece of information. What is the definition of torque?
 
  • #13
Alexyboy
yes I'm missing the distance at which the force acts, I'm assuming this is the plates centre of gravity... Thanks for all your help!
 
  • #14
Hootenanny
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yes I'm missing the distance at which the force acts, I'm assuming this is the plates centre of gravity...
Correct, so your torque is,

[tex]\tau = -0.075mg\sin\theta\left(\underline{j}\right)[/tex]

And adding the vector notation to your previous answer,

[tex]\tau = I \frac{d^2 \theta}{dt^2}\left(-\underline{j}\right)[/tex]

So can you finish the question off yourself?

Thanks for all your help!
No problem :smile:
 
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  • #15
Alexyboy
I might be able to...

I need to find the total torque from 0 to 180 degrees?

This is done by intergrating:

[tex]\tau = -0.075mg\sin\theta[/tex](j)

with repect to tau between 0 and 180

which gives

[tex]\tau = 0.075mg\cos180-0.075mg\cos 0[/tex]

which simplefies to

[tex]\tau = 0.150mg[/tex](-j)

giving a total torque of 0 t 180 degrees of 22.1Nm

then I'm not sure, can I use that to find [tex]\alpha[/tex] and if so how do I use it find the final velocity? I've managed to lose my train of thought from yesterday
 
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  • #16
Hootenanny
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You can't integrate directly, because (as you say) that will give you the total torque throughout the rotation from 0 to [itex]\pi[/itex], which isn't very useful. Remember what our goal is, the question wants the angular velocity after a given angle of rotation. Therefore, we should be aiming to write the angular velocity as a function of [itex]\theta[/itex].

When I said that you won't have to solve a second order ODE, it may have been misleading. You will have to solve an ODE, but there is a nice trick that we can do to make it a little easier. So I'll make a start for you, if we equate the two equation we have for torque we obtain,

[tex]I\ddot{\theta} = 0.075mg\sin\theta[/tex]

Where the dots represent time derivatives. Now collecting the constants,

[tex]\ddot{\theta} = A\sin\theta\hspace{2cm}\left(*\right)[/tex]

So we have the second time derivative of the angular variable is equal to some function of theta. Keep in mind that we want [itex]\omega\left(\theta\right) = \dot{\theta}\left(\theta\right)[/itex], this is where a little trick comes in handy.

Using the chain rule consider (noting that [itex]\theta = \theta\left(t\right)[/itex]),

[tex]\frac{d}{d\theta}\left(\dot{\theta}\right)^2 = \frac{d}{dt}\left(\dot{\theta}^2\right)\frac{dt}{d\theta}[/tex]

Once again applying the chain rule to the first differential,

[tex]\frac{d}{d\theta}\left(\dot{\theta}\right)^2 = 2\dot{\theta}\frac{d}{dt}\left(\dot{\theta}\right)\frac{dt}{d\theta} = 2\dot{\theta}\ddot{\theta}\frac{1}{\dot{\theta}}[/tex]

We cancel the [itex]\dot{\theta}[/itex] and divide by two to obtain,

[tex]\ddot\theta = \frac{1}{2}\frac{d}{d\theta}\left(\dot\theta^2\right)[/tex]

Hence (*) can be re-written,

[tex]\frac{1}{2}\frac{d}{d\theta}\left(\dot\theta^2\right) = A\sin\theta[/tex]

Do you follow? Can you now take the next step?
 
  • #17
Alexyboy
[
tex]I\ddot{\theta} = 0.075mg\sin\theta[/tex]
I tried doing it the other way but got in a terrible pickle trying to double intergrate

[tex] \frac{1}{\sin\theta}[t/tex]


and then I realised I had the problem of not knowing time again so I thought 'this is the way to do it.' Thats a really, really nice trick with the chain rule, I'll have to remember it for the future.

As omega is the rate of chage of angle

[tex] \omega = \dot{\theta}[/tex]

[tex]\frac{1}{2}\frac{d}{d\theta} (\omega^2) = A\sin\theta [/tex]

[tex] (\omega^2) = 2A \int \sin\theta d \theta [/tex]

[tex] \omega^2 = -2A cos\theta + C [/tex]


therefore

[tex]C = \omega^2 - 2A\cos\theta[/tex]

at [tex]\theta[/tex] = 0 [tex]\omega[/tex] = 80

[tex]A = \frac{0.075\times15\times\9.807}{0.1125}[/tex]

A = 98.07

therefore

[tex]C = 80^2 + 196.14 \times 1[/tex]

[tex]C = 6596[/tex]

[tex] \omega = \sqrt{-2A cos\theta + 6596}[/tex]

At at theta = 180

[tex] \omega = \sqrt{-196.14\times -1 + 6596}\left(-\underline{i}\right)[/tex]

therefore at 180 degrees the plate will have a angular velocity of -82.4 rad/s

which seems pretty reasonable to me

Thats a great help! Angular Dynamics is a big weakness of mine (I'm alright if its going in a straight line!) If anything like this comes it in the end of year Dynamics Paper I'm sorted!
 
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  • #18
Hootenanny
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I haven't checked your arithmetic, but your method looks good. The only correction I can see is that your angular velocity should be in the (-j) direction (using the right hand rule). The chain rule 'trick' is definitely a good one to know and the method of introducing an angular variable is also good to remember, especially in rotational motion.

Anyway good job on your solution :approve:
 
  • #19
Could anyone possibly elaborate on the bit where you used chain rule to make the differential equation terms of d theta rather than d t because I didn't understand that at all. Thanks
 
  • #20
Hootenanny
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Could anyone possibly elaborate on the bit where you used chain rule to make the differential equation terms of d theta rather than d t because I didn't understand that at all. Thanks
Is it the entire derivation, or just a particular step which you don't understand?
 
  • #21
The entire derivation. I understand that we want a differential equation with [tex]d\theta[/tex] on the bottom, instead of dt. But I didn't really follow any of the getting there (so the whole part marked with the asterisk.
Sorry to be a pain!
 
  • #22
Hootenanny
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The entire derivation. I understand that we want a differential equation with [tex]d\theta[/tex] on the bottom, instead of dt. But I didn't really follow any of the getting there (so the whole part marked with the asterisk.
Sorry to be a pain!
No problem, I'll use Leibzig's notation to try and make it clearer. Let's break it down, firstly let,

[tex]u = \left(\frac{d\theta}{dt}\right)\hspace{2cm}(**)[/tex]

And we want to evaluate,

[tex]\frac{d}{d\theta}u^2[/tex]

So by the http://mathworld.wolfram.com/ChainRule.html" [Broken] we may write,

[tex]\frac{d}{d\theta}u^2 = \frac{d}{du}\left(u^2\right)\frac{du}{d\theta} = 2u\frac{du}{d\theta}\hspace{2cm}(***)[/tex]

Okay, let's just consider the second differential in isolation,

[tex]\frac{du}{d\theta}[/tex]

However, noting that u is a composite function of t and [itex]\theta[/itex], again using the chain rule we can write,

[tex]\frac{du}{d\theta} = \frac{du}{dt}\frac{dt}{d\theta}\hspace{1cm}(****)[/tex]

Using (**) we can now evaluate the first differential,

[tex]\frac{du}{dt} = \frac{d}{dt}\left(\frac{d\theta}{dt}\right) = \frac{d^2\theta}{dt^2}[/tex]

Hence, we can rewrite (****) thus,

[tex]\frac{du}{d\theta} = \left(\frac{d^2\theta}{dt^2}\right)\frac{dt}{d\theta}[/tex]

Noting that,

[tex]\frac{dt}{d\theta} = \frac{1}{d\theta/dt}[/tex]

The previous expression becomes,

[tex]\frac{du}{d\theta} = \left(\frac{d^2\theta}{dt^2}\right)\frac{1}{d\theta/dt}[/tex]

Substituting this result into (***) we obtain,

[tex]\frac{d}{d\theta}u^2 = 2u\left(\frac{d^2\theta}{dt^2}\right)\frac{1}{d\theta/dt}[/tex]

And substituting (**) into the above equation,

[tex]\frac{d}{d\theta}\left(\frac{d\theta}{dt}\right)^2 = 2{\color{blue}\left(\frac{d\theta}{dt}\right)}\left(\frac{d^2\theta}{dt^2}\right){\color{blue}\frac{1}{d\theta/dt}}[/tex]

Noting that we can cancel the two terms in blue, we finally obtain,

[tex]\frac{d}{d\theta}\left(\frac{d\theta}{dt}\right)^2 = 2\left(\frac{d^2\theta}{dt^2}\right)[/tex]

The rest is simply algebraic manipulation. Do you follow?
 
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  • #23
Wayhey, I understand! Thanks alot, really appreciate the help.
 
  • #24
Hootenanny
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Wayhey, I understand! Thanks alot, really appreciate the help.
It was a pleasure :smile:
 
  • #25
quick question, could this problem be solved using conservation of energy, i.e initial kinetic energy at the top plus gravitational potential energy equals kinetic energy at the bottom

KE(1) + GPE = KE(2)

so

1/2 I[ω(1)]^2 + mgΔh = 1/2 I[ω(2)]^2


or am i missing something?
 

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