A ballet dancer spins with 2.4 rev/s with her arms outstretched, when the moment of inertia about axis of rotation is 1 .With her arms folded, the moment of inertia about the same axis becomes 0.61 . calculate the new rate of spin.

Correct question is;

A ballet dancer spins with 2.4 rev/s with her arms outstretched,when the moment of inertia about axis of rotation is I. With her arms folded,the moment of inertia about the same axis becomes 0.6I about the same axis. Calculate the new rate of spin.

4 rev/s

Explanation:

We are given;

Initial Angular velocity; ω_i = 2.4 rev/s

Initial moment of inertia; I_i = I

Final moment of inertia; I_f = 0.6I

From conservation of angular momentum, we have;

I_i × ω_i = I_f × ω_f

Where ω_f is the new rate of spin.

Thus, let's make it the subject to get;

ω_f = (I_i × ω_i/I_f)

Plugging in relevant values, we have;

ω_f = (I × 2.4/0.6I)

I will cancel out to give;

ω_f = 2.4/0.6

ω_f = 4 rev/s

true

explanation:

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle