Answer
We can rank the flywheels in order of radial acceleration of a point at the rim, from largest to smallest:
$c \gt d \gt a \gt b \gt e$
Work Step by Step
In general: $\omega = \frac{2\pi}{T}$
$v = \omega~r$
$a_c = \frac{v^2}{r}$
We can find the radial acceleration of a point at the rim of each flywheel:
(a) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0040~s} = 1571~rad/s$
$v = (1571~rad/s)(0.080~m) = 125.7~m/s$
$a_c = \frac{(125.7~m/s)^2}{0.080~m} = 197,000~m^2/s$
(b) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0040~s} = 1571~rad/s$
$v = (1571~rad/s)(0.020~m) = 31.4~m/s$
$a_c = \frac{(31.4~m/s)^2}{0.020~m} = 49,000~m^2/s$
(c) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0010~s} = 6283~rad/s$
$v = (6283~rad/s)(0.080~m) = 502.6~m/s$
$a_c = \frac{(502.6~m/s)^2}{0.080~m} = 3,158,000~m^2/s$
(d) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0010~s} = 6283~rad/s$
$v = (6283~rad/s)(0.020~m) = 125.7~m/s$
$a_c = \frac{(125.7~m/s)^2}{0.020~m} = 790,000~m^2/s$
(e) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0040~s} = 1571~rad/s$
$v = (1571~rad/s)(0.010~m) = 15.7~m/s$
$a_c = \frac{(15.7~m/s)^2}{0.010~m} = 25,000~m^2/s$
We can rank the flywheels in order of radial acceleration of a point at the rim, from largest to smallest:
$c \gt d \gt a \gt b \gt e$
