2014 James Ruse MX2 Trial Question 16(a) (1 Viewer)

porcupinetree · May 28, 2015

This is the question: http://imgur.com/9t4UWJO

My solution(s): http://imgur.com/p28GCzW

Me and my teacher both had similar approaches to parts i and ii, however, we had different approaches to part iii. I am probably more inclined to trust his method (Method 2 on my solution sheet) as there was a step in my method which we were both a bit unsure about. I think it may have something to do with the fact that alpha is actually a function of theta, rather than simply a constant, so when we differentiate sin(alpha + theta) wrt theta, it is more complicated than simply saying cos(alpha + theta) as I have given.

Can anyone verify either answer or provide some commentary? Help is much appreciated.

Thanks.

InteGrand · May 28, 2015

Re: 2014 James Ruse Question 16a

porcupinetree said:
This is the question: http://imgur.com/9t4UWJO

My solution(s): http://imgur.com/p28GCzW

Me and my teacher both had similar approaches to parts i and ii, however, we had different approaches to part iii. I am probably more inclined to trust his method (Method 2 on my solution sheet) as there was a step in my method which we were both a bit unsure about. I think it may have something to do with the fact that alpha is actually a function of theta, rather than simply a constant, so when we differentiate sin(alpha + theta) wrt theta, it is more complicated than simply saying cos(alpha + theta) as I have given.

Can anyone verify either answer or provide some commentary? Help is much appreciated.

Thanks.

Yes, I also got $-\frac{\pi}{5}$ rad s^-1 (using partial derivatives).

InteGrand · May 28, 2015

Re: 2014 James Ruse Question 16a

Of course, we didn't need partial derivatives, we could just differentiate the original equation with respect to t (and you're right, $\alpha$ is a function of $\theta$ ; it is easy to see this on the diagram by varying $\theta$ , and it's also seen from the equation they asked to prove, which shows that $\theta$ and $\alpha$ depend on each other).

$We know $2\sin \alpha = \sin(\alpha+ \theta)$, where $\alpha \equiv \alpha (t)$ and $\theta \equiv \theta (t)$ (i.e. they are functions of time). Note that by the sum rule for differentiation, $\frac{\mathrm{d}}{\mathrm{d}t}(\alpha + \theta)=\frac{\mathrm{d}\alpha}{\mathrm{d}t} + \frac{\mathrm{d}\theta}{\mathrm{d}t}$.$

$Use (implicit) differentiation on both sides with respect to $t$:$

$2\cos \alpha \cdot \frac{\mathrm{d}\alpha}{\mathrm{d}t}=\cos (\alpha+ \theta) \cdot \left( \frac{\mathrm{d}\alpha}{\mathrm{d}t} + \frac{\mathrm{d}\theta}{\mathrm{d}t}\right)$

$$\Rightarrow (2\cos \alpha - \cos (\alpha+ \theta))\cdot \frac{\mathrm{d}\alpha}{\mathrm{d}t} = \pi \cos (\alpha+ \theta)$ (since we are given that $\dot{\theta}=\pi$)$

$$\Rightarrow \frac{\mathrm{d}\alpha}{\mathrm{d}t} = \frac{\pi \cos (\alpha+ \theta)}{2\cos \alpha - \cos (\alpha+ \theta)}$.$

$Now note that when $\theta =\frac{\pi}{2}$, $\triangle OPC$ is right-angled, so using right-angle trigonometry and Pythagoras in this triangle yields $\cos (\alpha _1)=\frac{2}{\sqrt{5}},\sin \alpha _1=\frac{1}{\sqrt{5}}$, where $\alpha _1$ is the value of $\alpha$ at the instant when $\theta = \frac{\pi}{2}$. Also note that when $\theta = \frac{\pi}{2}$, $\cos (\alpha+ \theta) = \cos \left(\alpha _1 + \frac{\pi}{2}\right) = -\sin \alpha _1 = -\frac{1}{\sqrt{5}}$.$

$Putting all this information into our formula $\frac{\mathrm{d}\alpha}{\mathrm{d}t} = \frac{\pi \cos (\alpha + \theta)}{2\cos \alpha - \cos (\alpha+ \theta)}$, we find that the required angular velocity is $\omega _1 = \frac{\pi \cdot \left(\frac{-1}{\sqrt{5}} \right)}{2\cdot \frac{2}{\sqrt{5}} - \left(\frac{-1}{\sqrt{5}} \right)}$, which simplifies to $-\frac{\pi}{5}$.$

InteGrand · May 28, 2015

Re: 2014 James Ruse Question 16a

porcupinetree said:
This is the question: http://imgur.com/9t4UWJO

My solution(s): http://imgur.com/p28GCzW

Me and my teacher both had similar approaches to parts i and ii, however, we had different approaches to part iii. I am probably more inclined to trust his method (Method 2 on my solution sheet) as there was a step in my method which we were both a bit unsure about. I think it may have something to do with the fact that alpha is actually a function of theta, rather than simply a constant, so when we differentiate sin(alpha + theta) wrt theta, it is more complicated than simply saying cos(alpha + theta) as I have given.

Can anyone verify either answer or provide some commentary? Help is much appreciated.

Thanks.

Basically the mistake in your working was where you thought it was on your sheet: you essentially said $\frac{\mathrm{d}}{\mathrm{d}\theta}(\sin (\alpha + \theta))= \cos (\alpha + \theta))\cdot \frac{\mathrm{d}\theta}{\mathrm{d}\theta}$ , but as $\alpha$ depends on $\theta$ , the step should instead have been: $\frac{\mathrm{d}}{\mathrm{d}\theta}(\sin (\alpha + \theta))= \cos (\alpha + \theta))\cdot \frac{\mathrm{d}}{\mathrm{d}\theta}(\alpha + \theta) = \cos (\alpha + \theta))\cdot \left(\frac{\mathrm{d}\alpha}{\mathrm{d}\theta} +1 \right).$

If you do this and follow through, you'll get to the right answer. (So you basically erroneously assumed $\frac{\mathrm{d}\alpha}{\mathrm{d}\theta} = 0$ on the RHS, although you did the right thing on the LHS.)

porcupinetree · May 28, 2015

Re: 2014 James Ruse Question 16a

InteGrand said:
Of course, we didn't need partial derivatives, we could just differentiate the original equation with respect to t (and you're right, $\alpha$ is a function of $\theta$ ; it is easy to see this on the diagram by varying $\theta$ , and it's also seen from the equation they asked to prove, which shows that $\theta$ and $\alpha$ depend on each other).

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Thank you so much for your help; I understand it a lot better now

InteGrand · May 28, 2015

Re: 2014 James Ruse Question 16a

porcupinetree said:
Thank you so much for your help; I understand it a lot better now

You're welcome

Drongoski · May 29, 2015

Re: 2014 James Ruse Question 16a

InteGrand said:
Yes, I also got $-\frac{\pi}{5}$ rad s^-1 (using partial derivatives).

Just had a crack at it and also got the same answer; thought I got it wrong, until I saw yours. Maybe my approach a bit different.

Quite a challenging question, imo.

2014 James Ruse MX2 Trial Question 16(a) (1 Viewer)

porcupinetree

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InteGrand

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InteGrand

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InteGrand

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porcupinetree

not actually a porcupine

InteGrand

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Drongoski

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