Problem 18 - University Physics with Modern Physics

Question

BIO Gymnastics. We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75 \(\mathrm{kg}\) and diameter 1.0 \(\mathrm{m}\) . If this tumbler rolls forward at 0.50 rev/s, (a) how much total kinetic energy does he have, and (b) what percent of his total kinetic energy is rotational?

Step-by-Step Solution

Answer

The tumbler has a total kinetic energy of approximately 138.61 J, with 33.3% being rotational.

Step 1: Convert Rev/s to m/s

To find the linear velocity, we need to convert the revolutions per second to meters per second. Since 1 revolution is the same as the circumference of the cylinder, the circumference is given by \( \pi \times \text{diameter} = \pi \times 1.0 \, \text{m} = \pi \, \text{m} \). Thus 0.50 rev/s is equivalent to a linear velocity \( v \) of \( 0.50 \times \pi \approx 1.57 \, \text{m/s} \).

Step 2: Calculate the Moment of Inertia

The moment of inertia \( I \) for a solid cylinder is given by \( I = \frac{1}{2} m r^2 \), where \( m = 75 \, \text{kg} \) and the radius \( r = 0.5 \, \text{m} \). So, \( I = \frac{1}{2} \times 75 \times (0.5)^2 = 9.375 \, \text{kg} \cdot \text{m}^2 \).

Step 3: Calculate Linear Kinetic Energy

The linear kinetic energy \( KE_{\text{linear}} \) is given by \( \frac{1}{2} m v^2 \). Using the mass \( m = 75 \, \text{kg} \) and velocity \( v = 1.57 \, \text{m/s} \), the linear kinetic energy is \( \frac{1}{2} \times 75 \times (1.57)^2 \approx 92.43525 \, \text{J} \).

Step 4: Calculate Rotational Kinetic Energy

The rotational kinetic energy \( KE_{\text{rotational}} \) is given by \( \frac{1}{2} I \omega^2 \). Use \( I = 9.375 \, \text{kg} \cdot \text{m}^2 \) and \( \omega = 0.50 \times 2 \pi = \pi \, \text{rad/s} \). Thus, \( KE_{\text{rotational}} = \frac{1}{2} \times 9.375 \times (\pi)^2 \approx 46.1701768 \, \text{J} \).

Step 5: Calculate Total Kinetic Energy

The total kinetic energy \( KE_{\text{total}} \) is the sum of linear and rotational kinetic energy: \( KE_{\text{total}} = 92.43525 \, \text{J} + 46.1701768 \, \text{J} \approx 138.6054268 \, \text{J} \).

Step 6: Determine Percentage of Rotational Energy

To find the percentage of the rotational energy, divide the rotational energy by the total energy and multiply by 100: \( \frac{46.1701768}{138.6054268} \times 100 \approx 33.3\% \).

Key Concepts

Moment of Inertia

When we talk about moment of inertia, we're essentially discussing an object's reluctance to change its rotational state. Just like how mass measures how much an object resists changes in linear motion, moment of inertia does the same for rotation. For a gymnast modeled as a solid cylinder, the moment of inertia is calculated using a formula specific to cylinders. This formula is:

• \( I = \frac{1}{2} m r^2 \)

where \(m\) is the mass of the cylinder (or gymnast in our case) and \(r\) is the radius. In the provided example, the gymnast's mass is 75 kg, and the radius, being half of the diameter, is 0.5 meters. Inserting these values gives us:

• \( I = \frac{1}{2} \times 75 \times (0.5)^2 = 9.375 \text{ kg} \cdot \text{m}^2 \)

This value represents how much rotational motion the gymnast has "resisted." The larger this value, the harder it is to change the gymnast's spin. This concept is central in understanding the energy and dynamics involved when gymnasts perform their routines.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation. It's a crucial aspect of kinetic energy, especially when analyzing objects that are in rotational motion, like a gymnast tumbling. The formula to determine the rotational kinetic energy is:

• \( KE_{\text{rotational}} = \frac{1}{2} I \omega^2 \)

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. The angular velocity \( \omega \) can be found from the rotation speed, given in revolutions per second. In the given scenario, we have an angular velocity of \( \pi \) radians per second. Given the moment of inertia from earlier steps:

• \( KE_{\text{rotational}} = \frac{1}{2} \times 9.375 \times (\pi)^2 \approx 46.17 \text{ J} \)

This energy shows how much energy is used to keep the gymnast spinning. Understanding the rotational kinetic energy can help in assessing performance and safety in activities involving spins or rotations.

Linear Velocity

Linear velocity refers to the rate of change of an object’s position along a path. For a rolling gymnast modeled as a cylinder, linear velocity can be derived from rotational motion. This means converting how many times the gymnast spins per second into how fast they move along the ground. Each revolution corresponds to a distance equal to the cylinder's circumference.

• The circumference is \( \pi \times \text{diameter} = \pi \times 1.0 \text{ m} = \pi \text{ m} \).
• Given 0.50 revolutions per second, converting this to meters per second results in a velocity of \( 0.50 \times \pi \approx 1.57 \text{ m/s} \).

This velocity is crucial because it provides insights into the gymnast's speed as they roll forward. Linear velocity, together with rotational speed, helps fully characterize the gymnast’s motion, combining both translational and rotational motion into one comprehensive picture. It’s important for planning and understanding movements dynamically, especially in sports and physics-related studies.