Problem 42 - University Physics with Modern Physics

Question

A small block on a frictionless, horizontal surface has a mass of 0.0250 \(\mathrm{kg} .\) It is attached to a massless cord passing through a hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 1.75 \(\mathrm{rad} / \mathrm{s}\) . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 \(\mathrm{m} .\) Model the block as a particle. (a) Is the angular momentum of the block conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block. (d) How much work was done in pulling the cord?

Step-by-Step Solution

Answer

(a) Yes, because no external torques act on the block. (b) New angular speed is 7.00 rad/s. (c) Change in kinetic energy is 0.009685 J. (d) Work done is 0.009685 J.

Step 1: Understanding Conservation of Angular Momentum

The angular momentum of a system is given by \( L = I \omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular speed. For a point mass \( m \) at a distance \( r \) from the axis of rotation, \( I = mr^2 \). Initially, \( L_i = m r_i^2 \omega_i \). Since no external torques act on the system as it is frictionless, angular momentum is conserved: \( L_i = L_f \) where \( L_f = m r_f^2 \omega_f \).

Step 2: Calculating New Angular Speed

Given that \( L_i = L_f \), we can write \( m r_i^2 \omega_i = m r_f^2 \omega_f \). Solving for the new angular speed \( \omega_f \), we have:\[ \omega_f = \left(\frac{r_i^2}{r_f^2}\right) \omega_i \].Substitute \( r_i = 0.300 \ \text{m} \), \( r_f = 0.150 \ \text{m} \), and \( \omega_i = 1.75 \ \text{rad/s} \) into the equation.\[ \omega_f = \left( \frac{0.300^2}{0.150^2} \right) \times 1.75 = 7.00 \ \text{rad/s} \].

Step 3: Calculating Change in Kinetic Energy

The kinetic energy of the rotating block is given by \( K = \frac{1}{2} I \omega^2 = \frac{1}{2} m r^2 \omega^2 \). Calculate the initial kinetic energy \( K_i \) and final kinetic energy \( K_f \):\[ K_i = \frac{1}{2} (0.025)(0.300^2)(1.75^2) = 0.00344 \ \text{J} \]\[ K_f = \frac{1}{2} (0.025)(0.150^2)(7.00^2) = 0.013125 \ \text{J} \]The change in kinetic energy is:\[ \Delta K = K_f - K_i = 0.013125 - 0.00344 = 0.009685 \ \text{J} \]

Step 4: Calculating Work Done

The work done on the system is equal to the change in kinetic energy since no energy is lost to friction or external forces:\[ W = \Delta K = 0.009685 \ \text{J} \].

Key Concepts

Conservation of Angular Momentum

The principle of conservation of angular momentum is pivotal in understanding rotational dynamics. Angular momentum is defined as the product of the moment of inertia and angular velocity, expressed mathematically as \( L = I \omega \). For a point mass \( m \) rotating at a distance \( r \) from an axis, the moment of inertia can be represented as \( I = mr^2 \). Thus, the angular momentum for the block revolving in a circle is \( L = mr^2 \omega \).
In our exercise, since the surface is frictionless and no external torques are applied, the system's angular momentum is conserved. This means that the initial angular momentum \( L_i = mr_i^2 \omega_i \) before the cord is pulled remains equal to the final angular momentum \( L_f = mr_f^2 \omega_f \).
By employing conservation laws, we determine that when the radius decreases, the angular speed must increase to maintain the same angular momentum, leading to the equation \( \omega_f = \left(\frac{r_i^2}{r_f^2}\right) \omega_i \). Understanding this conservation principle is crucial to solving the problem of changing circle radii.

Kinetic Energy

Kinetic energy in rotational motion is similar to linear motion, but here it depends on both the mass distribution and the square of the angular velocity. Mathematically, it's expressed as \( K = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia. For a mass orbiting a circle with radius \( r \), the kinetic energy becomes \( K = \frac{1}{2} m r^2 \omega^2 \).
In our situation, initially, the block has an angular speed of \( 1.75 \, \text{rad/s} \) at a radius of \( 0.300 \, \text{m} \). Substituting these values into the equation gives the initial kinetic energy \( K_i = 0.00344 \, \text{J} \).
After pulling the cord, the radius reduces to \( 0.150 \, \text{m} \) and the angular speed increases to \( 7.00 \, \text{rad/s} \), resulting in a new kinetic energy \( K_f = 0.013125 \, \text{J} \). The increased energy stems from the system's rotational adjustment to preserve angular momentum, illustrating energy's dynamic relationship with rotational motion.

Work-Energy Principle

The work-energy principle connects the net work done on an object to its change in kinetic energy. In cases where friction is absent, like in this exercise, all the work done translates directly into changing the system's kinetic energy.
Here, as the cord is pulled and the block's motion changes, the work done on the block is quantifiable using the calculated change in kinetic energy. Initially, the kinetic energy is \( 0.00344 \, \text{J} \) and increases to \( 0.013125 \, \text{J} \). The work done, calculated as \( \Delta K = K_f - K_i \), results in \( 0.009685 \, \text{J} \).
This means all the energy resulting from the work done manifests as kinetic energy, showcasing the work-energy principle's role in rotational dynamics. This principle helps us understand how work done through force application can change an object's energy state without considering energy loss due to friction or other forces.