Problem 47 - College Physics

Question

A size-5 soccer ball of diameter \(22.6 \mathrm{~cm}\) and mass \(426 \mathrm{~g}\) rolls up a hill without slipping, reaching a maximum height of \(5.00 \mathrm{~m}\) above the base of the hill. We can model this ball as a thin-walled, hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it then have?

Step-by-Step Solution

Answer

Angular velocity: 68.1 rad/s; Rotational kinetic energy: 8.34 J.

Step 1: Find moments of inertia

The moment of inertia for a thin-walled hollow sphere is given by the formula: \[ I = \frac{2}{3} m r^2 \]where: - \(m\) is the mass of the ball,- \(r\) is the radius which is half of the diameter.Given:- Mass, \(m = 426 \text{ g} = 0.426 \text{ kg}\)- Radius, \(r = \frac{22.6 \text{ cm}}{2} = 11.3 \text{ cm} = 0.113 \text{ m}\)Substitute these values into the formula:\[ I = \frac{2}{3} \times 0.426 \times (0.113)^2 \approx 0.0036 \text{ kg} \cdot \text{m}^2 \]

Step 2: Apply conservation of energy

Since the soccer ball rolls up without slipping, conservation of energy applies: the total mechanical energy at the bottom is equal to the total mechanical energy at the top of the hill.The total mechanical energy at the base of the hill is the sum of the translational kinetic energy and rotational kinetic energy. At the top of the hill, all the energy is potential.\[ \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = mgh \]where:- \( \omega \) is the angular velocity,- \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity,- \( h = 5.00 \text{ m} \) is the height of the hill.

Step 3: Express velocity in terms of angular velocity

Because the ball rolls without slipping:\[ v = r \omega \]where \( v \) is the linear velocity and \( \omega \) is the angular velocity at the base of the hill.

Step 4: Simplify energy conservation equation

Substitute \( v = r\omega \) into the energy conservation equation:\[ \frac{1}{2} m (r\omega)^2 + \frac{1}{2} I \omega^2 = mgh \]Substituting for the known values of \( I \) and simplifying:\[ \frac{1}{2} \cdot 0.426 \cdot 0.113^2 \cdot \omega^2 + \frac{1}{2} \cdot 0.0036 \cdot \omega^2 = 0.426 \cdot 9.8 \cdot 5 \]

Step 5: Solve for angular velocity

Now solve the equation from Step 4 for \( \omega \):\[ \frac{1}{2} \cdot 0.426 \cdot 0.113^2 \cdot \omega^2 + \frac{1}{2} \cdot 0.0036 \cdot \omega^2 = 20.865 \approx mgh \]Calculate and solve:\[0.0027 \omega^2 + 0.0018 \omega^2 = 20.865\]\[ 0.0045 \omega^2 = 20.865 \Longrightarrow \omega^2 = \frac{20.865}{0.0045} \]\[ \omega \approx \sqrt{4636.7} = 68.1 \text{ rad/s} \]

Step 6: Calculate rotational kinetic energy

Using the angular velocity found in Step 5 and the rotational kinetic energy formula:\[ \text{Rotational Kinetic Energy (RKE)} = \frac{1}{2} I \omega^2 \]Substituting the known values:\[ \text{RKE} = \frac{1}{2} \times 0.0036 \times (68.1)^2 \approx 8.34 \text{ J} \]

Key Concepts

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation. Just like translational kinetic energy involves an object's mass and linear velocity, rotational kinetic energy depends on the object's moment of inertia and angular velocity. The formula for calculating rotational kinetic energy is \( \text{RKE} = \frac{1}{2} I \omega^2 \). - \( I \) represents the moment of inertia, a measure of an object's resistance to changes in its rotational motion. - \( \omega \) is the angular velocity, which indicates how fast the object is rotating.In the case of the soccer ball from the exercise, we used this formula to find the rotational kinetic energy when it was at the base of the hill. This calculation helps in understanding how energy is distributed within the rolling ball, contributing to both its translation and rotation. By knowing this, we can analyze the mechanics of motion more effectively.

Moment of Inertia

The moment of inertia is a crucial concept when it comes to understanding rotational motion. It's like the rotational analogy to mass in linear motion. It plays a significant role in determining how an object will behave when subjected to rotational forces. In this exercise, the moment of inertia for the thin-walled, hollow sphere (soccer ball) is calculated using the formula \( I = \frac{2}{3} m r^2 \), where:- \( m \) is the mass of the ball, and- \( r \) is the radius.For the soccer ball, we substituted given values to find \( I \approx 0.0036 \text{ kg} \cdot \text{m}^2 \). Understanding how to calculate the moment of inertia allows us to predict how difficult it is to start or stop the object's rotation, influencing the entire energy distribution in the rolling or spinning object.

Angular Velocity

Angular velocity represents the rate of change of the angular displacement, indicating how quickly an object is revolving. It is similar to linear velocity but applies to rotational movement. The relation between linear and angular velocity is crucial in problems involving rolling objects. In the provided exercise, the linear velocity \( v \) of the soccer ball at the base of the hill is linked to the angular velocity \( \omega \) through the equation \( v = r \omega \).When we worked through the exercise, solving for \( \omega \) was a key step. We derived that the ball had an angular velocity of approximately 68.1 rad/s as it started to climb the hill. This value helps in determining the rotational kinetic energy the ball had, giving insights into its dynamic behavior.

Mechanical Energy

Mechanical energy is the sum of potential and kinetic energy in a system. It is a key concept in physics as it is conserved in the absence of non-conservative forces like friction.In the soccer ball exercise, mechanical energy conservation was used to relate the ball's initial energy at the base (kinetic and rotational) to its energy at the top of the hill (potential). The equation used was:\[ \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = mgh \]where:- \( mgh \) is the gravitational potential energy at the top,- \( \frac{1}{2} m v^2 \) is the translational kinetic energy,- \( \frac{1}{2} I \omega^2 \) is the rotational kinetic energy.By understanding mechanical energy, students can better grasp how different forms of energy transition in a system, which is essential for solving various physics problems involving motion.

Physics Problem Solving

Problem-solving in physics often involves breaking down complex situations into more manageable parts. For instance, in this exercise, the process was structured into clear steps:

• Calculating the moment of inertia to understand resistance to rotational motion.
• Applying conservation of energy to relate different forms of energy.
• Linking linear and angular velocity to understand relationships in rotational dynamics.
• Solving the derived equations to find unknown quantities like angular velocity.
• Calculating rotational kinetic energy using found values.

By following these systematic approaches, students can solve physics problems more effectively and build a strong understanding of physical principles. This structured method also helps in identifying which formulas and concepts to apply in various contexts, leading to accurate and insightful solutions.