# A piston is connected by a rod of 14 cm to a crankshaft at a point 5 cm away from the axis of rotation. Determine how fast the crankshaft is rotating when the piston is 11 cm away from the axis of rotation and is moving toward it at 1200 cm/s?

Jun 3, 2017

$\frac{d \theta}{\mathrm{dt}} = 325.114 \frac{\textcolor{w h i t e}{\text{." "rad}}}{s}$

#### Explanation:

This problem is asking us to find $\frac{d \theta}{\mathrm{dt}}$, or the angular velocity of the piston.

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Let $Q = \overline{O Q}$

So we know that:

(dQ)/dt]_(Q= 11) = -1200"cm"/s

Before we start, let's write out a formula for theta given what we know, using the Law of Cosines:

${14}^{2} = {5}^{2} + {Q}^{2} - 2 \left(5\right) \left(Q\right) \cos \left(\theta\right)$

$10 Q \cos \theta = {Q}^{2} - 171$

$10 \cos \theta = Q - \frac{171}{Q}$

Now, take the derivative of this with respect to time:

$- 10 \sin \theta \cdot \frac{d \theta}{\mathrm{dt}} = \frac{\mathrm{dQ}}{\mathrm{dt}} + \frac{171}{Q} ^ 2 \cdot \frac{\mathrm{dQ}}{\mathrm{dt}}$

Since we already have values for $Q$ and $\frac{\mathrm{dQ}}{\mathrm{dt}}$, all we need to do is find $\sin \theta$ when $Q = 11$.

To do that, let's use the Law of Cosines again, but manipulate it further until we get $\sin \theta$. (This time, we know $Q = 11$)

${14}^{2} = {5}^{2} + {11}^{2} - 2 \left(5\right) \left(11\right) \cos \left(\theta\right)$

$50 = - 110 \cos \left(\theta\right)$

$- \frac{5}{11} = \cos \left(\theta\right) = \sqrt{1 - {\sin}^{2} \theta}$

$\frac{25}{121} = 1 - {\sin}^{2} \theta$

$\frac{96}{121} = {\sin}^{2} \theta$

$\frac{4 \sqrt{6}}{11} = \sin \theta$

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Now we know that when $Q = 11$, $\frac{\mathrm{dQ}}{\mathrm{dt}} = - 1200$ and $\sin \theta = \frac{4 \sqrt{6}}{11}$.

This is all we need to solve for $\frac{d \theta}{\mathrm{dt}}$:

$- 10 \sin \theta \cdot \frac{d \theta}{\mathrm{dt}} = \frac{\mathrm{dQ}}{\mathrm{dt}} + \frac{171}{Q} ^ 2 \cdot \frac{\mathrm{dQ}}{\mathrm{dt}}$

$- 10 \left(\frac{4 \sqrt{6}}{11}\right) \cdot \frac{d \theta}{\mathrm{dt}} = - 1200 + \frac{171}{121} \left(- 1200\right)$

$- \frac{40 \sqrt{6}}{11} \cdot \frac{d \theta}{\mathrm{dt}} = - 1200 \left(\frac{292}{121}\right)$

$\frac{d \theta}{\mathrm{dt}} = \frac{1200 \cdot 292 \cdot 11}{40 \sqrt{6} \cdot 121} \frac{\text{ ""rad}}{s}$

$\frac{d \theta}{\mathrm{dt}} = \frac{8760}{11 \sqrt{6}} = \frac{8760 \sqrt{6}}{66} = \frac{1460 \sqrt{6}}{11} \frac{\text{ ""rad}}{s}$

Or if you prefer decimal form:

$\frac{d \theta}{\mathrm{dt}} = 325.114 \frac{\text{ " "rad}}{s}$