Chapter 8, Problem 57P

(a)

To determine

The angular momentum of skater.

Answer to Problem 57P

Angular momentum of the skater is $13.6{\text{kg-m}}^{\text{2}}\text{/s}$

Explanation of Solution

Given:

Angular velocity, $\omega =3.5\text{rev/s = 3}\text{.5}\times \text{2}\pi \text{rad/s = 7}\pi \text{ rad/s}$

Height, $h=1.5m$

Radius, $r=15cm=0.15m$

Mass, $m=55kg$

Formula used:

Moment of inertia, $I=\frac{M{R}^2}{2}$

Angular Momentum, $L=I\omega$

Calculation:

Substituting values in formula of moment of inertia,

  $I=\frac{(55){(0.15)}^2}{2}=0.61875{\text{kg-m}}^{\text{2}}$

Substituting values in angular momentum formula,

  $\begin{array}{l}L=(0.61875)(7\pi )\\ L=13.6{\text{kg-m}}^{\text{2}}\text{/s}\end{array}$

Conclusion:

Angular momentum of the skater is $13.6{\text{kg-m}}^{\text{2}}\text{/s}$

(b)

To determine

The torque required to stop the spinning of skater.

Answer to Problem 57P

The torque required to stop the skater is $2.72N-m$ in direction opposite to her spin.

Explanation of Solution

Given:

Initial angular momentum, $L_i=13.6{\text{kg-m}}^{\text{2}}\text{/s}$

Time, $t=5s$

Final angular momentum, $L_f=0$

Formula used:

Torque, $\tau =\frac{Changeinangularmomentum}{time}=\frac{△L}{t}=\frac{L_f-L_i}{t}$

Calculation:

Substituting values in formula of torque,

  $\begin{array}{l}\tau =\frac{0-13.6}{5}\\ \tau =\frac{-13.6}{5}\\ \tau =-2.72N-m\end{array}$

Conclusion:

Torque required to stop the spinning of skater is $2.72N-m$ in opposite direction to her spin.