Problem 75 - Fundamentals of Physics

Question

In a playground, there is a small merry-go-round of radius \(1.20 \mathrm{~m}\) and mass \(180 \mathrm{~kg}\). Its radius of gyration (see Problem 79 of Chapter 10 ) is \(91.0 \mathrm{~cm}\). A child of mass \(44.0 \mathrm{~kg}\) runs at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate (a) the rotational inertia of the merry-go-round about its axis of rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c) the angular speed of the merry-go-round and child after the child has jumped onto the merry-go-round.

Step-by-Step Solution

Answer

(a) 150.119 kg·m². (b) 158.4 kg·m²/s. (c) 0.742 rad/s.

Step 1: Calculate the Rotational Inertia of the Merry-go-round

The rotational inertia, or moment of inertia, of an object is given by the formula \( I = mk^2 \), where \( m \) is the mass of the object and \( k \) is the radius of gyration. For the merry-go-round, the mass \( m = 180 \ \mathrm{kg} \) and the radius of gyration \( k = 91.0 \ \mathrm{cm} = 0.91 \ \mathrm{m} \). Therefore, the rotational inertia is:\[I = 180 \times (0.91)^2 = 150.119 \ \mathrm{kg} \cdot \mathrm{m}^2.\]

Step 2: Calculate the Angular Momentum of the Running Child

The angular momentum \( L \) of the child about the axis of rotation of the merry-go-round is given by \( L = mvr \), where \( m = 44.0 \ \mathrm{kg} \) is the mass of the child, \( v = 3.00 \ \mathrm{m/s} \) is the speed of the child, and \( r = 1.20 \ \mathrm{m} \) is the radius of the merry-go-round. Therefore, the angular momentum is:\[L = 44.0 \times 3.00 \times 1.20 = 158.4 \ \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}.\]

Step 3: Calculate the Angular Speed After the Child Jumps On

When the child jumps onto the merry-go-round, we use conservation of angular momentum. The initial angular momentum of the system is the angular momentum of the child, \( L = 158.4 \ \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} \). The final angular momentum \( I_{\text{total}} \omega \), where \( I_{\text{total}} = I + mr^2 \) is the combined rotational inertia (merry-go-round and child), and \( \omega \) is the angular speed.Firstly, calculate \( I_{\text{total}} \):\[I_{\text{child}} = 44.0 \times (1.20)^2 = 63.36 \ \mathrm{kg} \cdot \mathrm{m}^2,\]\[I_{\text{total}} = 150.119 + 63.36 = 213.479 \ \mathrm{kg} \cdot \mathrm{m}^2.\]Now, set the initial angular momentum equal to the final:\[158.4 = 213.479 \omega \Rightarrow \omega = \frac{158.4}{213.479} \approx 0.742 \ \mathrm{rad/s}.\]

Key Concepts

Moment of Inertia

The concept of moment of inertia is central to understanding rotational dynamics, much like mass is essential in linear motion. Moment of inertia, often denoted by \( I \), measures how difficult it is to change the rotational motion of an object around an axis. Essentially, it's a measure of an object's resistance to angular acceleration.In this exercise, the moment of inertia is especially important for the merry-go-round. The formula to find moment of inertia is given by:

• \( I = mk^2 \)

Where \( m \) is the mass of the object, and \( k \) is the radius of gyration. Here, the mass \( m \) of the merry-go-round is \( 180 \, \mathrm{kg} \), and the radius of gyration \( k \) is \( 0.91 \, \mathrm{m} \). When these values are plugged into the formula, we find the rotational inertia to be approximately \( 150.119 \, \mathrm{kg} \cdot \mathrm{m}^2 \).The moment of inertia is a key component in predicting how rotational forces will act upon an object. The higher the inertia, the more force needed to change its rotation state.

Radius of Gyration

The radius of gyration is an abstract yet important concept in understanding rotational inertia. It represents how far from the axis of rotation, the entire mass of an object can be thought to be concentrated, without changing its moment of inertia.We calculate the radius of gyration \( k \) for the merry-go-round using the formula:

• \( k = \sqrt{\frac{I}{m}} \)

Here, the moment of inertia \( I \) is \( 150.119 \, \mathrm{kg} \cdot \mathrm{m}^2 \) and mass \( m \) is \( 180 \, \mathrm{kg} \). From these values, the problem in the exercise provides the radius of gyration as \( 91.0 \, \mathrm{cm} \) or \( 0.91 \, \mathrm{m} \).Understanding radius of gyration simplifies complex calculations by allowing us to work with a single point, representing the distribution of the object’s mass. Often used in structural engineering and biomechanics, this concept aids in simplifying rotational dynamics problems, providing a deeper grasp of how mass affects rotational inertia.

Conservation of Angular Momentum

The principle of conservation of angular momentum is fundamental in rotational dynamics, similar to the conservation of linear momentum. It states that if no external torque acts on a system, the total angular momentum remains constant. In this exercise, the child's movement and jump onto the merry-go-round is an illustration of this principle. Before the child jumps on, the merry-go-round is stationary, and only the child has angular momentum which can be calculated using:

• \( L = mvr \)

Where \( m \) is mass of child, \( v \) is velocity of child, and \( r \) is the radius of path relative to the axis.Upon jumping onto the merry-go-round, the combined system of the child and the merry-go-round must conserve this angular momentum. Therefore:

• Total Angular Momentum Before = Total Angular Momentum After

Initially calculated as \( 158.4 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} \), this needs to equal the final state, which includes rotational inertia of both the merry-go-round and child. This conservation allows us to solve for the new angular speed \( \omega \) using the known values:

• \( \omega = \frac{L}{I_{\text{total}}} \)

This aspect hints at how changes within a system redistribute angular velocities without any interchange with outer forces, a crucial part of rotational dynamics.