Answer
(a) $v = 32.9~m/s$
(b) The acceleration of the head is $27.4~m/s^2$ greater than the acceleration of the feet.
(c) 35.5 rpm
Work Step by Step
(a) $a_R = \frac{v^2}{R} = 12.5~g$
$v^2 = 12.5~g~R$
$v = \sqrt{12.5~g~R} = \sqrt{(12.5)(9.80~m/s^2)(8.84~m)}$
$v = 32.9~m/s$
(b) Let $v_f$ be the velocity of the feet.
$v_f = (32.9~m/s)(\frac{6.84~m}{8.84~m}) = 25.5~m/s$
$a_R = \frac{v^2}{R} = \frac{(25.5~m/s)^2}{6.84~m}$
$a_R = 95.1~m/s^2$
We can find the difference between the two accelerations.
$12.5~g - 95.1~m/s^2
= (12.5)(9.80~m/s^2) - (95.1~m/s^2)
= 27.4~m/s^2$
The acceleration of the head is $27.4~m/s^2$ greater than the acceleration of the feet.
(c) $(32.9~m/s)(\frac{1~rev}{(2\pi)(8.84~m)})(\frac{60~s}{1~min}) = 35.5~rpm$
