Problem 43 - University Physics with Modern Physics

Question

The Spinning Figure Skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center (Fig. Elo.43). When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled, hollow cylinder. His hands and arms have a combined mass of 8.0 \(\mathrm{kg} .\) When outstretched, they span 1.8 \(\mathrm{m}\) ; when wrapped, they form a cylinder of radius 25 \(\mathrm{cm} .\) The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . If his original angular speed is \(0.40 \mathrm{rev} / \mathrm{s},\) what is his final angular speed?

Step-by-Step Solution

Answer

The final angular speed of the skater is approximately 2.84 rad/s.

Step 1: Analyze the Problem

We need to find the final angular speed of the skater after his arms are wrapped around his body. Initially, the skater's arms are outstretched, and then they are wrapped around, changing the distribution of mass and thus the moment of inertia.

Step 2: Understand Moment of Inertia

For the arms stretched out as a slender rod pivoting about its center: \[ I_{arms, stretched} = \frac{1}{12} m L^2 \] where:- \( m = 8.0 \, \mathrm{kg} \) (mass of arms),- \( L = 1.8 \, \mathrm{m} \) (length when outstretched).

Step 3: Calculate Initial Moment of Inertia

Using the formula for a slender rod, compute the moment of inertia of the arms when outstretched:\[ I_{arms, stretched} = \frac{1}{12} \times 8.0 \, \mathrm{kg} \times (1.8 \, \mathrm{m})^2 \] \[ I_{arms, stretched} = 2.16 \, \mathrm{kg} \cdot \mathrm{m}^2 \] The total initial moment of inertia is:\[ I_{initial} = I_{arms, stretched} + I_{body} = 2.16 \, \mathrm{kg} \cdot \mathrm{m}^2 + 0.40 \, \mathrm{kg} \cdot \mathrm{m}^2 \] \[ I_{initial} = 2.56 \, \mathrm{kg} \cdot \mathrm{m}^2 \]

Step 4: Understand Changed Moment of Inertia

After wrapping the arms, they form a hollow cylinder. Its moment of inertia is:\[ I_{arms, wrapped} = m r^2 \] where:- \( m = 8.0 \, \mathrm{kg} \) (mass of arms),- \( r = 0.25 \, \mathrm{m} \) (radius of cylinder).

Step 5: Calculate Final Moment of Inertia

Using the formula for a hollow cylinder, compute the wrapped moment of inertia:\[ I_{arms, wrapped} = 8.0 \, \mathrm{kg} \times (0.25 \, \mathrm{m})^2 \] \[ I_{arms, wrapped} = 0.5 \, \mathrm{kg} \cdot \mathrm{m}^2 \] The total final moment of inertia is:\[ I_{final} = I_{arms, wrapped} + I_{body} = 0.5 \, \mathrm{kg} \cdot \mathrm{m}^2 + 0.40 \, \mathrm{kg} \cdot \mathrm{m}^2 \] \[ I_{final} = 0.9 \, \mathrm{kg} \cdot \mathrm{m}^2 \]

Step 6: Apply Conservation of Angular Momentum

Angular momentum before and after the arms are wrapped remains constant:\[ I_{initial} \cdot \omega_{initial} = I_{final} \cdot \omega_{final} \] where:- \( \omega_{initial} = 0.40 \, \mathrm{rev/s} \times \frac{2\pi}{1} \times \frac{1}{1} \, \mathrm{rad/s} = 0.40 \times 2\pi \, \mathrm{rad/s} \)- Convert revolutions per second to radians per second for calculations.

Step 7: Solve for Final Angular Speed

Substitute known values into the conservation equation:\[ 2.56 \cdot 0.40 \times 2\pi = 0.9 \cdot \omega_{final} \] Solve for \( \omega_{final} \):\[ \omega_{final} = \frac{2.56 \cdot 0.40 \times 2\pi}{0.9} \] Calculate \( \omega_{final} \):\[ \omega_{final} \approx 2.84 \, \mathrm{rad/s} \]

Key Concepts

Moment of Inertia

The concept of Moment of Inertia is crucial in understanding rotational motion. It refers to how much resistance an object has to changes in its rotation. Think of it like mass in linear motion, but for spinning or rotational movement. It depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation.In the case of the figure skater, when his arms are outstretched, they can be modeled as a slender rod. The formula for the moment of inertia of a slender rod rotating about its center is \( I = \frac{1}{12} mL^2 \), where \( m \) is the mass and \( L \) is the length of the rod. When he wraps his arms, they turn into a hollow cylinder, for which the moment of inertia becomes \( I = mr^2 \), where \( r \) is the radius. This change in moment of inertia is what affects the skater's angular speed eventually.

Rotational Dynamics

Rotational Dynamics involves the analysis of rotational motion. It's akin to the linear dynamics we learn in basic physics, extending it to rotating objects. The key principle here is the conservation of angular momentum. Like linear momentum, angular momentum must be conserved unless acted upon by an external force. For the figure skater, his total angular momentum remains constant before and after he tucks his arms. This is represented mathematically as:

• Initial angular momentum: \( L_i = I_{initial} \cdot \omega_{initial} \)
• Final angular momentum: \( L_f = I_{final} \cdot \omega_{final} \)

Because no external torques are acting on him, \( L_i \) equals \( L_f \). This relationship helps calculate the skater’s new angular speed after altering his moment of inertia by changing the position of his arms.

Angular Speed Calculation

Calculating angular speed involves recognizing how changes in moment of inertia affect rotational speed due to conservation laws.For the figure skater, his initial angular speed is given (\(0.40 \text{ rev/s}\)), which needs to be converted into the standard unit, radians per second, for calculations. Using the conversion \(1 \text{ rev} = 2\pi \text{ rad}\), the initial speed in radians per second is \(0.40 \times 2\pi \). To find his final angular speed, \( \omega_{final}\), we use the equation from the conservation of angular momentum: \[ I_{initial} \cdot \omega_{initial} = I_{final} \cdot \omega_{final} \]Substituting the respective values, you solve for \( \omega_{final} \), illustrating how the skater's rotational speed increases as his moment of inertia decreases when he pulls his arms inwards.