Problem 56 - Fundamentals of Physics

Question

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to "take up" the angular momentum (Fig. \(11-18\) ). In \(0.700 \mathrm{~s},\) one arm sweeps through 0.500 rev and the other arm sweeps through 1.000 rev. Treat each arm as a thin rod of mass \(4.0 \mathrm{~kg}\) and length \(0.60 \mathrm{~m}\), rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

Step-by-Step Solution

Answer

The total angular momentum is approximately 6.45 kg m²/s.

Step 1: Understand the problem requirements

We are required to find the total angular momentum of the athlete's arms as they both rotate to counter her body's angular momentum. We will treat each arm as a thin rod rotating about one end, and we need to account for both arms since they have different angular displacements in the specified time.

Step 2: Find the moment of inertia for one arm

For a thin rod rotating about one end, the moment of inertia is given by \( I = \frac{1}{3} m L^2 \).Given the mass \( m = 4.0 \, \text{kg} \) and the length \( L = 0.60 \, \text{m} \), we find the moment of inertia of one arm: \[ I = \frac{1}{3} \cdot 4.0 \, \text{kg} \cdot (0.60 \, \text{m})^2 = 0.48 \, \text{kg} \, \text{m}^2 \].

Step 3: Calculate the angular velocity for each arm

Angular velocity \( \omega \) is given by \( \omega = \frac{\Delta \theta}{\Delta t} \).For the first arm moving through \(0.500\) revolutions in \(0.700\, \text{s}\), convert revolutions to radians: \(0.500 \times 2\pi\, \text{rad/rev} = \pi\, \text{rad}\).Thus, \( \omega_1 = \frac{\pi}{0.700} \, \text{rad/s} \).For the second arm moving through \(1.000\) revolutions: \(1.000 \times 2\pi\, \text{rad/rev} = 2\pi\, \text{rad}\).Thus, \( \omega_2 = \frac{2\pi}{0.700} \, \text{rad/s} \).

Step 4: Calculate the angular momentum for each arm

The angular momentum \( L \) is given by \( L = I \omega \).For the first arm:\[ L_1 = 0.48 \, \text{kg} \, \text{m}^2 \cdot \frac{\pi}{0.700} \, \text{rad/s} = \frac{0.48\pi}{0.700} \, \text{kg m}^2/\text{s} \].For the second arm:\[ L_2 = 0.48 \, \text{kg} \, \text{m}^2 \cdot \frac{2\pi}{0.700} \, \text{rad/s} = \frac{0.96\pi}{0.700} \, \text{kg m}^2/\text{s} \].

Step 5: Calculate the total angular momentum

The total angular momentum \( L_{\text{total}} \) is the sum of the angular momentum of both arms:\[ L_{\text{total}} = L_1 + L_2 = \frac{0.48\pi}{0.700} + \frac{0.96\pi}{0.700} = \frac{1.44\pi}{0.700} \, \text{kg m}^2/\text{s} \].Calculate the numerical value:\[ L_{\text{total}} \approx 6.45 \, \text{kg m}^2/\text{s} \].

Key Concepts

Moment of Inertia

The concept of moment of inertia is crucial in understanding how objects behave under rotational motion. It essentially measures how difficult it is for an object to change its rotational speed, akin to mass in linear motion. For a thin rod, like an athlete's arm, the moment of inertia when rotating around one end is calculated with the formula: \[ I = \frac{1}{3} m L^2 \] where \( m \) is the mass of the rod, and \( L \) is its length.
Think of it like this: heavier or longer objects are harder to spin, just like it's harder to start running with a heavy backpack compared to when you're empty-handed. So, for our athlete, each arm with a mass of 4.0 kg and a length of 0.60 m would have a moment of inertia of 0.48 kg⋅m². This means that this is the resistance each arm possesses to changes in its rotational motion. Without such resistance, objects would rotate at the slightest nudge.
In the case of the athlete, this moment of inertia allows her to control the distribution of her angular momentum and maintain balance during her jump.

Rotational Motion

Rotational motion is what happens when an object spins around a central point or axis. For our athlete, her arms moving are an example of rotational motion around the shoulder joint.
When thinking of rotational motion, it's helpful to imagine the rest of her body as a counterweight that she has to balance out with her arms moving in the opposite direction.
There are several components to understand:

• **Rotation axis**: This is the line around which the rotation occurs. In this case, it's through the athlete's shoulders.
• **Angular momentum**: This is a measure of the rotation's extent and direction. It depends on the object's moment of inertia and angular velocity.
• **Centripetal force**: This force pulls a rotating body toward the center of its circular path.

The athlete's arms being outstretched changes her angular momentum because it alters the distribution of her weight. By changing her arms' position and movement, the athlete adjusts her body's rotation to ensure a safe landing.

Angular Velocity

Angular velocity tells us how quickly an object is spinning. In simpler terms, it's the rate at which an angle changes with respect to time. For heavy lifters, it's crucial because it distinguishes dizzy spins from graceful ones.
The formula for angular velocity \( \omega \) is: \[ \omega = \frac{\Delta \theta}{\Delta t} \] where \( \Delta \theta \) is the change in angle (in radians) and \( \Delta t \) is the timeframe.
For the athlete, as one arm sweeps through 0.500 revolutions in 0.700 seconds, we first convert to radians: \( 0.500 \times 2\pi \) rad/rev = \( \pi \) radians.
Apply the formula: \[ \omega_1 = \frac{\pi}{0.700} \text{ rad/s} \] And for the other arm moving through 1.000 revolutions: \[ \omega_2 = \frac{2\pi}{0.700} \text{ rad/s} \] These calculations are essential as they help determine how her arms' speed impacts her total angular momentum. By understanding and utilizing these velocities, the athlete can control her spin and adjust her body mid-flight.