## Physics: Principles with Applications (7th Edition)

$a = 3.9 ~m/s^2$
We can find the velocity of the speck as it completes 45 rpm on the wheel through the formula: $v = \frac{(45)(2\pi r)}{T}$ $v = \frac{(45)(2\pi)(0.175 ~m)}{60 ~s}$ $v = 0.825~m/s$ We can then use the velocity of the speck to find the acceleration: $a = \frac{v^2}{r}$ $a = \frac{(0.825 ~m/s)^2}{(0.175 ~m)}$ $a = 3.9 ~m/s^2$