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, [itex] a_{T} [/itex], is constant, how long after the ball starts moving is radial acceleration, [itex] a_r [/itex], equal to tangential acceleration?

Knowns:

[itex] v_f = v_{\text{final}} = 4.00\, \text{m/s} [/itex]

[itex] v_i = v_{\text{initial}} = 0.00\, \text{m/s} [/itex]

[itex] r = 0.72\, \text{m} [/itex]

relevant equations:

1. [itex] C = 2\pi r [/itex]

2. [itex] {v_f}^2 – {v_i}^2 = 2{a_{T}}d\cos (\theta) [/itex]

3. [itex] a_r = \frac{v^2}{r} [/itex]

4. [itex] v = a_{T}t + v_0 [/itex]

Attempt:

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

[itex] d = (3)C = 6\pi (0.72\, \text{m}) \approx 13.56\, \text{m} [/itex]

So then I rearrange 2. noting that [itex] \theta = 6\pi [/itex];

[itex] 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 [/itex]

then setting [itex] a_r = a_{T} [/itex] and substituting into 3. I get:

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

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

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