4. a skater is spinning through the final portion of her floor program with her arms initially outstretched…

4. a skater is spinning through the final portion of her floor program with her arms initially outstretched (i = 50 kgm^2). she starts her spin at 6.26 rad/s and then pulls her arms in close to her body so her final i = 43 kgm^2. what is her resulting angular velocity?

4. a skater is spinning through the final portion of her floor program with her arms initially outstretched (i = 50 kgm^2). she starts her spin at 6.26 rad/s and then pulls her arms in close to her body so her final i = 43 kgm^2. what is her resulting angular velocity?

Answer

Explanation:

Step1: Recall the law of conservation of angular momentum

$L_i = L_f$, where $L = I\omega$. So $I_i\omega_i=I_f\omega_f$.

Step2: Rearrange the formula to solve for $\omega_f$

$\omega_f=\frac{I_i\omega_i}{I_f}$

Step3: Substitute the given values

$I_i = 50\ kg\cdot m^2$, $\omega_i=6.26\ rad/s$, $I_f = 43\ kg\cdot m^2$. Then $\omega_f=\frac{50\times6.26}{43}$.

Step4: Calculate the result

$\omega_f=\frac{313}{43}\approx7.28\ rad/s$

Answer:

$7.28\ rad/s$