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Studying for a test and there is one single question we couldn't answer, and we have multiple results, no-one knows which one the correct one is. Image:

Circular to linear motion diagram

D is 1.3m, C is 0.6m and the angle a is 30 degrees.

Rate of change of a (a') is 150 rad/s. The rate of change of M is asked.

My solution is as follows:

$M=\cos(30^\circ)\cdot0.6+\cos(13.3424)\cdot1.2$

$M'=-(\sin(30^\circ)\cdot0.6\cdot a'+\sin(13.3424)\cdot1.3\cdot B')$

We know that $a'=150$, so we calculate $B'$ ($b=13.3424$)

$\sin(30)\cdot0.6=\sin(b)\cdot1.4$ and thus

$B'=(\cos(30)\cdot0.6\cdot150)/(\cos(13.3424)\cdot1.3)\,\to\,B'=61.62$ rad/s

Plug into $M'$ formula and $M'=63.485688$ rad/s OR m/s?

Did I do something wrong in my approach? I calculated in radians and converted the angles, does this matter?

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    I think it would have been easier to use the law of cosines and treat it as a related rates problem as that's what it looks like to me, unfortunately I can't read that language (Dutch?) And the formatting of the math here is unhelpful – 2017-01-22
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    But how do I corelate rad/s with meter and get a linear speed? – 2017-01-22
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    Is this a math course or a mechanics course, just wondering. – 2017-01-22
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    Math, example exam of the integral course – 2017-01-22
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    I think I'm going to have to try and get an English version of the question just so I can see it in full, it will help me understand your question better – 2017-01-22
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    I embedded the diagram and translated the Dutch to English. – 2017-01-22

1 Answers 1

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Since we wish to find the rate of change of $M$ and we are given rate of change information about $a$ it would be useful to find the relationship between $M$ and $a$ and the fixed parameters $C$ and $D$. This would also allow us to find the answers to the questions regarding specific values of the parameter $a$.

From the Law of Sines and the fact that $b$ will always be an acute angle we see that

\begin{equation} b=\arcsin\left(\frac{C}{D}\sin a\right)\tag{1} \end{equation}

If we let $\mu$ represent the angle opposite $M$ we have

\begin{equation} M=\frac{D}{\sin a}\cdot\sin\mu\tag{2} \end{equation}

But

\begin{eqnarray} \mu&=&180^\circ-(a+b)\\&=&180^\circ-\left(a+\arcsin\left(\frac{C}{D}\sin a\right)\right)\\ \sin\mu&=&\sin\left(a+\arcsin\left(\frac{C}{D}\sin a\right)\right)\\ &=&\sin a\cos\left(\arcsin\left(\frac{C}{D}\sin a\right)\right)+\cos a\cdot\frac{C}{D}\sin a\\ &=&\frac{\sin a}{D}\sqrt{D^2-C^2\sin^2 a}+\frac{\sin a}{D}\cdot C\cos a \end{eqnarray}

Substituting this value of $\sin\mu$ into equation $(2)$ gives

\begin{equation} M=\sqrt{D^2-C^2\sin^2 a}+C\cos a \tag{3} \end{equation}

Given $D=13,\,C=6, a=30^\circ$ equation $(1)$ gives the result $b=13.3424^\circ$ and equation $(3)$ gives the result $M=4\sqrt{10}+3\sqrt{3}=17.845\,$dm.

To find the rate of change of $M$ we take the time derivative of equation $(3)$ and simplify it to obtain

\begin{equation} \frac{dM}{dt}=-C\sin a\left(\frac{\cos a}{\sqrt{D^2-C^2\sin^2 a}}+1\right)\cdot\frac{da}{dt}\tag{4} \end{equation}

Since the radial velocity is already given in terms of "per seconds" it is unnecessary to convert. Substituting all the given values into equation $(4)$ yields an answer of $\frac{dM}{dt}=-480.8$ dm/sec.

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    Why the sinu=sin(a+arcsin(Csina/D)? – 2017-01-22
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    The jump in block 2 isn't clear to me – 2017-01-22
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    nevermind I got it, i cant use gamma to derivate & sinus 180 is 0. I'm to tired. Thank you for your help :) – 2017-01-22
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    $\sin(180^\circ-\theta)=\sin\theta$ – 2017-01-22