Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was trying this problem and i couldnt come up with a solution, it doesnt seem to hard but I cant think of what to do.

There is a point on a wheel that is rotating at a constant speed. The bottom of the wheel is 1 meters above the ground. At a point Q the point on the wheel is 16 meters above the ground. After 4 seconds its reaches the top of the wheel. After another 8 seconds it reaches a point P. If the wheel has a diameter of 16 meters how do we find out how high the point is at point P.

Thanks!

share|improve this question
1  
If you can use coordinates, they might help greatly. Center the rotating circle on the vertical axis; trigonometry might be helpful. –  J. M. Apr 15 '12 at 16:03
add comment

1 Answer

To follow the argument below, you will need to draw a diagram.

Let $O$ be the centre of the wheel, and $T$ the top of the wheel.

Draw a perpendicular from $Q$ to the vertical line $OT$, meeting that line at $X$. The height of the point $Q$ (and therefore of $X$) is $16$, and the height of $T$ is $17$. It follows that $XT=1$ and therefore $OX=7$.

Let $\theta=\angle QOX$. We have $\cos\theta=\frac{7}{8}$. In $4$ seconds the wheel travels through an angle $\theta$. So in $8$ seconds it travels through an angle $2\theta$.

We can find $\cos 2\theta$ approximately, using a calculator, or exactly, using $\cos 2\theta=2\cos^2\theta-1$. Let's go for exactly. We get $\cos 2\theta=\frac{34}{64}$.

Now we can find the height of point $P$. The picture is much like the one we worked with for $Q$. Draw a perpendicular from $P$ to the vertical line $OT$, meeting that line at $Y$. Then $OY=8\cos 2\theta$. So the height of $P$ above the centre of the wheel is $8\cos 2\theta$, and therefore the height above the ground is $9+8\cos 2\theta$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.