A diver begins a high dive with an angular velocity of 6.3 rad/s. By tucking himself into a ball, he decreases his moment of inertia by a factor of 9.0. What is his new angular speed?

Question

A diver begins a high dive with an angular velocity of 6.3 rad/s. By tucking himself into a ball, he decreases his moment of inertia by a factor of 9.0. What is his new angular speed?

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2 answers
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Answer 1

bobpursley answered
14 years ago

angular momentum is conserved:

angmom= I w If I goes down to 1/5, w increases by 5.

Answer 2

Explain Bot answered
11 months ago

To find the new angular speed of the diver, we need to use the conservation of angular momentum. Angular momentum is given by the equation:

L = I * ω

Where:
L is the angular momentum,
I is the moment of inertia,
ω (omega) is the angular velocity.

Since the diver decreases his moment of inertia by a factor of 9.0, the new moment of inertia (I') will be 1/9.0 times the initial moment of inertia (I).

So, I' = (1/9.0) * I

Since angular momentum is conserved, we can equate the initial angular momentum (L) to the final angular momentum (L'):

I * ω = I' * ω'

Substituting the values we have:

I * ω = (1/9.0) * I * ω'

Simplifying the equation, we can cancel out the moment of inertia (I):

ω = (1/9.0) * ω'

Now we can solve for ω', the new angular speed of the diver:

ω' = 9.0 * ω

Substituting the given value ω = 6.3 rad/s, we can calculate the new angular speed:

ω' = 9.0 * 6.3 rad/s = 56.7 rad/s

Therefore, the diver's new angular speed is 56.7 rad/s.