# Thread: Extra credit word problem

1. ## Extra credit word problem

I need some help with this bonus problem. Any help would be appreciated.

A train rolls down a track at 60mph. It has wheels with radius = 15 in. The wheels contain a flange designed to hold the train on the track that extends 1 in from the radius of the riding surface

Since Point A (on the flange) is farther from the center of the wheel than Point B (the wheel), it is moving faster than B.

Due to this phenomenon, there is portion of the rotation that Point A is moving backward with respect to the ground.

Find the duration of this time period in seconds (Accurate to 10 decimals).

Hint: The answer is very near one hundredth of a second.

2. Originally Posted by anna_sims
...
A train rolls down a track at 60mph. It has wheels with radius = 15 in. The wheels contain a flange designed to hold the train on the track that extends 1 in from the radius of the riding surface
Since Point A (on the flange) is farther from the center of the wheel than Point B (the wheel), it is moving faster than B....
Hi,

I've attached a diagram of the described situation.

R is the radius pointing to A resp. A'. R = 16"
r is the radius pointing to B. r = 15"

When the wheel performs the angle $\displaystyle \alpha$, the the point A moves back.

You've got right triangles so you can calculate $\displaystyle \alpha$:

$\displaystyle \cos(\frac{1}{2} \alpha)=\frac{15}{16}$. Thus $\displaystyle \alpha$ = 40.728&#176;

The perimeter of the (active) wheel = $\displaystyle p=2 \pi \cdot r=30\pi ft$

Now calculate the number of revolutions per second. I've got 8.8837885 rps. (Please check my result carefully. I'm the king of mistakes here!).

That means the wheel performs 8.8837885 * 360&#176; = 3198.16&#176; per second.

The time which is needed to perform the angle $\displaystyle \alpha$:

$\displaystyle t=\frac{40.728 ^\circ}{3198.16^\circ} \approx 0.012734807 s$

(By the way: Why do you want a time exact to billionst second?)

EB