Let's solve this problem step-by-step, applying the principles of the conservation of angular momentum.
### Step 1: Initial Information
- The initial rate of revolution is 1 revolution per second.
- Let's denote the initial moment of inertia by [tex]\( I_1 \)[/tex].
### Step 2: Moment of Inertia Decrease
- The problem states that the moment of inertia decreases by 40% when she folds her arms.
- If the initial moment of inertia [tex]\( I_1 \)[/tex] is 1 unit (for simplicity), the final moment of inertia [tex]\( I_2 \)[/tex] would be:
[tex]\[
I_2 = I_1 \times (1 - 0.4) = I_1 \times 0.6
\][/tex]
Therefore, if [tex]\( I_1 = 1 \)[/tex] (as a presumed unit), then:
[tex]\[
I_2 = 1 \times 0.6 = 0.6
\][/tex]
### Step 3: Conservation of Angular Momentum
- For a closed system where no external torques act, the angular momentum before and after folding her arms stays constant.
- The angular momentum is given by:
[tex]\[
L = I \times \omega
\][/tex]
where [tex]\( L \)[/tex] is the angular momentum, [tex]\( I \)[/tex] is the moment of inertia, and [tex]\( \omega \)[/tex] is the angular velocity.
- Initially, the dancer's angular momentum [tex]\( L_1 \)[/tex] with her arms stretched (at [tex]\( \omega_1 = 1 \)[/tex] revolution per second) is:
[tex]\[
L_1 = I_1 \times \omega_1 = 1 \times 1 = 1
\][/tex]
### Step 4: Final Angular Velocity Calculation
- Let the final angular velocity when her arms are folded be [tex]\( \omega_2 \)[/tex]. The conservation of angular momentum states:
[tex]\[
L_1 = L_2
\][/tex]
Therefore,
[tex]\[
I_1 \times \omega_1 = I_2 \times \omega_2
\][/tex]
Substituting the known values:
[tex]\[
1 \times 1 = 0.6 \times \omega_2
\][/tex]
[tex]\[
1 = 0.6 \times \omega_2
\][/tex]
- Solving for [tex]\( \omega_2 \)[/tex]:
[tex]\[
\omega_2 = \frac{1}{0.6} = 1.6666666666666667
\][/tex]
### Conclusion
The new rate of revolution when the ballet dancer folds her arms, causing her moment of inertia to decrease by 40%, is approximately [tex]\( 1.67 \)[/tex] revolutions per second.
