Problem 20 - University Physics with Modern Physics

Question

A string is wrapped several times around the rim of a small hoop with radius 8.00 \(\mathrm{cm}\) and mass 0.180 \(\mathrm{kg} .\) The free end of the string is held in place and the hoop is released from rest (Fig. E10.20). After the hoop has descended \(75.0 \mathrm{cm},\) calculate (a) the angular speed of the rotating hoop and (b) the speed of its center.

Step-by-Step Solution

Answer

(a) Angular speed \( \omega \approx 19.2 \ \mathrm{rad/s} \). (b) Center speed \( v \approx 1.54 \ \mathrm{m/s} \).

Step 1: Understanding the Moment of Inertia of the Hoop

To calculate the angular speed of the rotating hoop, we need to understand the moment of inertia. For a hoop, the moment of inertia \( I \) about its central axis is given by \( I = MR^2 \), where \( M = 0.180 \, \mathrm{kg} \) and \( R = 0.08 \, \mathrm{m} \).

Step 2: Applying Energy Conservation

The hoop is released from rest, descending 75.0 cm in height. Using energy conservation, the potential energy lost is equal to the kinetic energy gained. Initial potential energy: \( U_i = Mgh = 0.180 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 0.75 \, \mathrm{m} \). Final kinetic energy: \( K_f = \frac{1}{2}I \omega^2 + \frac{1}{2}Mv^2 \).

Step 3: Relating Linear Speed to Angular Speed

Since the string unwinds without slipping, the linear speed \( v \) of the center of the hoop and the angular speed \( \omega \) are related by \( v = R\omega \).

Step 4: Solving for Angular Speed \( \omega \)

Substitute \( v = R\omega \) into the kinetic energy equation: \( \frac{1}{2} I \omega^2 + \frac{1}{2} M\left(R\omega\right)^2 = Mgh \). Substitute \( I = MR^2 \) into the equation and solve for \( \omega \).

Step 5: Calculating the Angular Speed

Solve the equation: \( \frac{1}{2} (MR^2)\omega^2 + \frac{1}{2} M (R\omega)^2 = Mgh \). Simplified, it becomes \( M(R^2 \omega^2) = 2Mgh \). Then, \( \omega = \sqrt{2gh/R^2} \). Substituting the values gives \( \omega = \sqrt{2 \times 9.8 \times 0.75/0.08^2} \approx 19.2 \, \mathrm{rad/s} \).

Step 6: Calculating the Speed of the Center

Using the relationship \( v = R \omega \), where \( R = 0.08 \, \mathrm{m} \) and \( \omega \approx 19.2 \, \mathrm{rad/s} \), calculate \( v = 0.08 \times 19.2 = 1.54 \, \mathrm{m/s} \).

Key Concepts

Moment of Inertia

The moment of inertia is a fundamental concept in rotational dynamics that quantifies an object's resistance to changes in its rotational motion. For a hoop or ring, it is calculated using the formula: \( I = MR^2 \). This means the moment of inertia depends on both the mass \( M \) and the square of the radius \( R \) of the hoop.
This relationship helps us understand how mass is distributed with respect to the rotational axis. The further the mass is from the axis, the larger the moment of inertia, which implies more resistance to rotational change.
The moment of inertia is crucial when analyzing rotational motion because it appears in equations involving torque, angular acceleration, and kinetic energy. When the hoop in our exercise descends, knowing its moment of inertia lets us relate its potential energy to rotational kinetic energy, helping us find the hoop's angular speed.

Energy Conservation

Energy conservation is a key principle in physics stating that the total energy in a closed system remains constant. It's particularly useful in analyzing mechanical systems where energy may change forms but the total amount stays consistent.
As the hoop falls 75 cm, it loses potential energy. This energy doesn't vanish; it converts into kinetic energy, both translational (linear) and rotational. The initial potential energy is calculated with \( U_i = Mgh \), where \( g \) is the acceleration due to gravity, and \( h \) is the height.
Energy conservation allows us to set this initial potential energy equal to the total kinetic energy, which includes both the hoop's rotational energy \( \frac{1}{2}I\omega^2 \) and its linear speed component \( \frac{1}{2}Mv^2 \). Solving these equations helps us understand the speeds related to the hoop's motion.

Angular Speed

Angular speed, denoted \( \omega \), describes how fast an object rotates—from the Latin "angulus" meaning "angle." It's measured in radians per second (rad/s) and is similar to linear speed but for rotation.
In our exercise, angular speed is determined by how quickly the hoop spins as it descends. The linear speed of the hoop's center and its angular speed are related by \( v = R\omega \). Since the string unwinds with no slip, knowing the radius \( R \) and calculating \( \omega \) shows how fast the hoop rotates.
By using energy conservation and the moment of inertia, we solve for \( \omega \). This angular speed reflects the rate of change of the angle—how much the hoop spins as it unrolls the string. Understanding angular speed is pivotal in calculations for rotating objects.

Linear Speed

Linear speed, denoted as \( v \), refers to the distance covered per unit of time by an object moving along a path. Unlike angular speed which covers rotations, linear speed explains movement over a linear distance.
For the hoop descending in the exercise, we find its linear speed using the relationship \( v = R\omega \). Here, \( R \) is the hoop's radius and \( \omega \) its angular speed. As the hoop falls, understanding this link helps calculate how fast its center moves.
Calculating linear speed in our scenario provides insights into both the full motion path of the hoop and its kinetic state at any point. Linear speed is crucial in practical applications where it's important to compare with rotational characteristics, ensuring a thorough understanding of motion dynamics.