## University Physics with Modern Physics (14th Edition)

$v = (5000 ~rpm)(2\pi ~r)(\frac{1~min}{60~s})$ $v = \frac{500~\pi~r}{3}~m/s$ We can use the radial acceleration to find the required radius $r$. $a_{rad} = \frac{v^2}{r} = 3000~g$ $\frac{(\frac{500~\pi~r~}{3}~)~^2}{r} = 3000~g$ $r = \frac{(9)(3000)(9.80~m/s^2)}{(500~\pi ~s^{-1})^2}$ $r = 0.107~m$ Since the required diameter is 0.214 meters, while the bench space is claimed to be 0.127 meters, the claim seems unrealistic.