A diver comes off a board with arms straight up and legs straight down, giving h

Question

A diver comes off a board with arms straight up and legs straight down, giving her a rotational mass of 18 kg*m2. She then tucks into a small ball, decreasing her rotational mass to 3.6 kg*m2. If she initial was moving in a way where she completed just 1 revolution every second, what is her angular speed after she tucks? Hint: Recall that there are 2* radians in 1 revolution.

A diver comes off a board with arms straight up and legs straight down, giving her a rotational mass of 18 kg*m2. She then tucks into a small ball, decreasing her rotational mass to 3.6 kg*m2. If she initial was moving in a way where she completed just 1 revolution every second, what is her angular speed after she tucks? Hint: Recall that there are 2* radians in 1 revolution.

Explanation / Answer

I1 = 18 Kg.m^2

w1 = 1 rev/s

= 2*pi rad/s

I2 = 3.6 Kg.m^2

Use conservation of angular momentum:

I1*w1 = I2*w2

18*2*pi = 3.6*w2

w2 = 5*pi rad/s

Answer: 5*pi rad/s

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