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ANgular Velocity

  • Thread starter Alexyboy
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  • #1
Alexyboy
1. 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.



2. 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]

3. 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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