# non uniform circular motion

Here is the problem I am working on:

A steel ball is accelerated around a circular track of radius 72 cm from rest. after 3 circuits the ball reaches a speed of 4.00 m/s, assuming tangential acceleration, $a_{T}$, is constant, how long after the ball starts moving is radial acceleration, $a_r$, equal to tangential acceleration?

Knowns:

$v_f = v_{\text{final}} = 4.00\, \text{m/s}$

$v_i = v_{\text{initial}} = 0.00\, \text{m/s}$

$r = 0.72\, \text{m}$

relevant equations:

1. $C = 2\pi r$

2. ${v_f}^2 – {v_i}^2 = 2{a_{T}}d\cos (\theta)$

3. $a_r = \frac{v^2}{r}$

4. $v = a_{T}t + v_0$

Attempt:

So what I first did was solve for the total distance travelled by the ball in preparation of using 2. to solve for $a_{T}$;

$d = (3)C = 6\pi (0.72\, \text{m}) \approx 13.56\, \text{m}$

So then I rearrange 2. noting that $\theta = 6\pi$;

$a_{T} =\frac{{v_f}^2}{2d} = \frac{(4.00 \, \text{m/s})^2}{2(13.56\, \text{m})} \approx 0.589\, \text{m}/\text{s}^2$

then setting $a_r = a_{T}$ and substituting into 3. I get:

$v = \sqrt{ra_{T}} = \sqrt{(0.72\, \text{m})(0.589\, \text{m}/\text{s}^2)} \approx 0.65 \, \text{m/s}$ and thus using 4. to solve for time,

$t = \frac{v}{a_{T}} = \frac{0.65\, \text{m/s}}{0.589\, \text{m}/\text{s}^2} \approx 1.1 \, \text{s}$.

However, the answer in the back of the book says i should be getting 0.18 s. any help would be greatly appreciated!

EDIT: I feel this would be better in the hoework section, however i do not know how to move it.

http://ift.tt/1oyq58l