Problem 84 - College Physics

Question

A stationary ice skater with a mass of \(80.0 \mathrm{~kg}\) and a moment of inertia (about her central vertical axis) of \(3.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) catches a baseball with her outstretched arm. The catch is made at a distance of \(1.00 \mathrm{~m}\) from the central axis. The ball has a mass of \(145 \mathrm{~g}\) and is traveling at \(20.0 \mathrm{~m} / \mathrm{s}\) before the catch. (a) What linear speed does the system (skater \(+\) ball) have after the catch? (b) What is the angular speed of the system (skater \(+\) ball) after the catch? (c) What percentage of the ball's initial kinetic energy is lost during the catch? Neglect friction with the ice.

Step-by-Step Solution

Answer

a) 0.922 m/s; b) 0.922 rad/s; c) 95.4% energy lost.

Step 1: Calculate the Initial Angular Momentum of the Ball

The angular momentum (L) of an object about an axis is calculated as L = r \cdot m \cdot v, where r is the distance from the axis, m is the mass, and v is the linear velocity. For the ball:\[L = 1.00 \, \mathrm{m} \times 0.145 \, \mathrm{kg} \times 20.0 \, \mathrm{m/s} = 2.90 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\]This is the initial angular momentum before the catch.

Step 2: Apply Conservation of Angular Momentum

According to the conservation of angular momentum, the total angular momentum before the catch (just of the ball) is equal to the total angular momentum after the catch (ball plus skater):\[L_{\text{initial}} = I_{\text{skater}} \cdot \omega + r^2 \cdot m_{\text{ball}} \cdot \omega\]Setting this equation equal to the initial angular momentum:\[2.90 = 3.00 \cdot \omega + 1^2 \cdot 0.145 \cdot \omega\]Solve for \omega (angular speed after the catch).

Step 3: Solve for System Angular Speed \(\omega\)

By simplifying the equation from step 2:\[2.90 = (3.00 + 0.145) \cdot \omega\]\[\omega = \frac{2.90}{3.145} \approx 0.922 \, \mathrm{rad/s}\]Thus, the angular speed of the system after the catch is 0.922 rad/s.

Step 4: Determine the Linear Speed of the Skater

The linear speed v of the skater can be related to the angular speed using the relationship v = r \cdot \omega for a point on a rotating object:\[v = 1.00 \, \mathrm{m} \times 0.922 \, \mathrm{rad/s} = 0.922 \, \mathrm{m/s}\]Thus, the linear speed of the skater (plus the ball) after the catch is 0.922 m/s.

Step 5: Calculate the Initial and Final Kinetic Energies

Calculate the initial kinetic energy KE_\text{initial} of the ball:\[KE_\text{initial} = \frac{1}{2} \cdot m \cdot v^2 = \frac{1}{2} \cdot 0.145 \, \mathrm{kg} \cdot (20.0 \, \mathrm{m/s})^2 = 29.0 \, \mathrm{J}\]Calculate the final kinetic energy KE_\text{final} of the system using I and \omega:\[KE_\text{final} = \frac{1}{2} \cdot (I_\text{skater} + r^2 \cdot m_\text{ball}) \cdot \omega^2 = \frac{1}{2} \cdot 3.145 \cdot (0.922)^2 \approx 1.33 \, \mathrm{J}\]Thus, the kinetic energy is lost in the process.

Step 6: Calculate the Percentage of Kinetic Energy Lost

The percentage of kinetic energy lost is determined by:\[\text{Percentage lost} = \left(\frac{KE_\text{initial} - KE_\text{final}}{KE_\text{initial}}\right) \times 100\%\]\[\text{Percentage lost} = \left(\frac{29.0 - 1.33}{29.0}\right) \times 100\% \approx 95.4\%\]Thus, 95.4% of the ball's initial kinetic energy is lost.

Key Concepts

Kinetic Energy

Kinetic energy is a form of energy that an object possesses due to its motion. The classic formula to calculate kinetic energy \( KE \) is given by: \[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of the object, and \( v \) is its velocity. In the exercise, the initial kinetic energy of the ball is calculated by plugging its mass \( 0.145 \, \mathrm{kg} \) and initial velocity \( 20.0 \, \mathrm{m/s} \) into this formula.- Initial kinetic energy of the ball: \( KE_{\text{initial}} = \frac{1}{2} \times 0.145 \, \mathrm{kg} \times (20.0 \, \mathrm{m/s})^2 = 29.0 \, \mathrm{J} \)After the ice skater catches the ball, the system's kinetic energy is lower than the initial kinetic energy of the ball alone. This indicates that some kinetic energy is lost as the collision between the ball and skater is inelastic. In inelastic collisions, kinetic energy is not conserved because some energy is transformed into other forms such as heat or sound. The final kinetic energy of the system is calculated using the moment of inertia and angular speed, with the equation:\[ KE_{\text{final}} = \frac{1}{2} I \omega^2 \]This loss manifests as approximately 95.4% of the initial kinetic energy being lost, pointing out the inefficiency in converting the ball's motion energy into rotational energy.

Angular Speed

Angular speed is the rate at which an object rotates or spins around a central point. It contrasts with linear speed, which measures straight-line movement. Angular speed is usually measured in radians per second \( \mathrm{rad/s} \).In the problem, after catching the ball, the skater and ball rotate together, requiring us to find their combined angular speed. We use the conservation of angular momentum, expressed as:\[ L_{\text{initial}} = L_{\text{final}} \]where the initial angular momentum of the ball before the catch equals the total angular momentum of the skater-ball system after.The equation for angular momentum \( L \) is:\[ L = I \omega \]or in the ball's case before contact:\[ L = r \cdot m \cdot v \]Given that \( r = 1.00 \, \mathrm{m} \), the conservation equation is set to solve for \( \omega \):- Initial ball angular momentum: \( L = 2.90 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} \)- Post-catch system angular speed: \[ \omega = \frac{2.90}{3.145} \approx 0.922 \, \mathrm{rad/s} \]Solving with respect to \( \omega \), we find the angular speed of the system is approximately \( 0.922 \, \mathrm{rad/s} \), which neatly demonstrates the principle of conservation of angular momentum in rotational systems.

Linear Speed

Linear speed refers to how quickly an object moves along a path, measured in units like meters per second \( \mathrm{m/s} \). For objects in rotation, linear speed can be derived from the angular speed using the formula:\[ v = r \cdot \omega \]where \( v \) is the linear speed, \( r \) is the radius or the distance from the rotation axis, and \( \omega \) is the angular speed.In the case of the skating example, once the skater catches the ball, the skater and ball system's overall linear speed is determined by calculating:- Given \( r = 1.00 \, \mathrm{m} \) and \( \omega = 0.922 \, \mathrm{rad/s} \), the linear speed \( v \) becomes: \[ v = 1.00 \, \mathrm{m} \times 0.922 \, \mathrm{rad/s} = 0.922 \, \mathrm{m/s} \]This result shows that the linear speed of the skater-ball system post-catch is \( 0.922 \, \mathrm{m/s} \). The skater's speed is tied directly to the radius from the axis and how fast the system spins, encapsulating the relationship between linear and angular motion. This interconnectedness is fundamental in understanding dynamics wherever rotational movement is present.