The figure shows a rotating wheel with radius 40 cm and a connecting rod AP with length 1.2 m. The pin P slides back and forth along the x-axis as the wheel rotates counterclockwise at a rate of 360 revolutions per minute.

enter image description here

  1. Find the angular velocity of the connecting rod, $\frac{d\alpha}{dt}$, in radians per second, when $\theta = \frac{\pi }{3}$.
  2. Express the distance $x = \left|OP \right|$ in terms of $\theta$.
  3. Find an expression for the velocity of the pin P in terms of $\theta$.

I have found a solution for all three questions, but my answer to QUESTION 1 is different from the given solution from the textbook and I am struggling to figure out why.


I use the law of sines to solve this.

$$\frac{\sin\alpha }{OA} = \frac{\sin\theta }{AP}$$

from which I find that:

$$\sin\alpha =\frac{OA}{AP}\sin\theta =\frac{40}{120}\sin\theta =\frac{1}{3}\sin\theta$$

$$\alpha = \arcsin\left(\frac{1}{3}\sin\theta \right)$$

$$\frac{d\alpha }{dt} = \frac{1}{\sqrt{1 - \frac{{\sin\theta }^{2}}{9}}} \cdot \frac{1}{3}\cos\theta \cdot \frac{d\theta }{dt}$$

and this is equal to $$\frac{d\alpha }{dt} = \frac{3\pi }{\sqrt{33}}$$ when $$\theta = \frac{\pi }{3}$$

The problem here is that according to the textbook the right answer is $\frac{4\pi \sqrt{3}}{\sqrt{11}}$ which I quite do not understand. Furthermore I am not supposed, at this stage, to know the derivative of inverse functions including arcsin. Is there any other way to solve this without using the arcsin funcion?

  • $\begingroup$ The book's answer is exactly 4 times yours. Did you have $\d\theta/dt=12\pi$, or something else? Did you divide by 2 somewhere when you should have multiplied? $\endgroup$
    – David K
    Feb 14, 2017 at 7:14

1 Answer 1


We have $$ \sin(\theta) = \frac{s}{r} \\ \sin(\alpha) = \frac{s}{l} $$ such that equating on $s$ we get $$ r \sin(\theta) = l \sin(\alpha) \iff \\ \alpha = \arcsin\left( \frac{r}{l} \sin(\theta) \right) $$ Then the time derivative is $$ \dot{\alpha} = \frac{1}{\sqrt{1-\left( \frac{r}{l} \sin(\theta)\right)^2}} \frac{r}{l}\cos(\theta) \dot{\theta} $$ with $r=0.4\,\text{m}$, $l=1.2\,\text{m}$, $\dot{\theta}=6\cdot 2\pi\,\text {rad}/\text{s}$. Then for $\theta=\pi/3$ we get: $$ \sin(\pi/3) = \sqrt{3}/2 \\ \cos(\pi/3) = 1/2 \\ \dot{\alpha} = \frac{1}{\sqrt{1-(\frac{1}{3} \sqrt{3}/2)^2}} \frac{1}{3} \cdot \frac{1}{2} \cdot 12 \pi = \sqrt{\frac{36}{33}} 2 \pi = \sqrt{\frac{9}{33}} 4\pi = \sqrt{\frac{3}{11}} 4\pi $$

Alternative without $\arcsin$: $$ \sin(\alpha) = \frac{r}{l} \sin(\theta) $$ Differentiating $$ \cos(\alpha) \dot{\alpha} = \frac{r}{l} \cos(\theta) \dot{\theta} \iff \\ \dot{\alpha} = \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\cos(\alpha)} = \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\sqrt{1-\sin(\alpha)^2}} = \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\sqrt{1-(\frac{r}{l} \sin(\theta))^2}} $$

  • $\begingroup$ Thanks, I am glad to see that my method was correct. Any other idea on how to solve this without using the inverse function arcsin? $\endgroup$
    – Jashin_212
    Feb 15, 2017 at 1:18
  • $\begingroup$ I added that way, you still end up with the same formula. $\endgroup$
    – mvw
    Feb 15, 2017 at 8:15

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