## College Physics (4th Edition)

(a) The diver makes $3.02$ turns while falling 10.0 meters in a tuck position. (b) The diver makes $1.56$ turns while falling 10.0 meters in a pike position.
(a) We can find the time it takes to fall 10.0 meters: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(10.0~m)}{9.80~m/s^2}}$ $t = 1.43~s$ We can find the angular speed when in a tuck position: $I~\omega = L$ $\omega = \frac{L}{I}$ $\omega = \frac{106~kg~m^2/s}{8.0~kg~m^2}$ $\omega = 13.25~rad/s$ We can find the number of turns the diver makes while falling 10.0 meters: $(13.25~rad/s)(\frac{1~rev}{2\pi~rad})(1.43~s) = 3.02~rev$ The diver makes $3.02$ turns while falling 10.0 meters in a tuck position. (b) We can find the angular speed when in a pike position: $I~\omega = L$ $\omega = \frac{L}{I}$ $\omega = \frac{106~kg~m^2/s}{15.5~kg~m^2}$ $\omega = 6.84~rad/s$ We can find the number of turns the diver makes while falling 10.0 meters: $(6.84~rad/s)(\frac{1~rev}{2\pi~rad})(1.43~s) = 1.56~rev$ The diver makes $1.56$ turns while falling 10.0 meters in a pike position.