Problem 15 - University Physics with Modern Physics

Question

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 \(\mathrm{N}\) is applied to the rim of the wheel. The wheel has radius 0.120 \(\mathrm{m} .\) Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

Step-by-Step Solution

Answer

The moment of inertia of the wheel is approximately 0.255 kg⋅m².

Step 1: Convert Units for Angular Speed

First, convert the angular speed from revolutions per second to radians per second because our calculations will involve standard units. Since 1 revolution is equal to \(2\pi\) radians, \(12.0 \text{ rev/s}\) can be converted to radians per second by calculating: \[\omega = 12.0 \times 2\pi \approx 75.4\, \text{rad/s}\]

Step 2: Calculate Angular Acceleration

Next, determine the angular acceleration \( \alpha \) of the wheel using the formula \( \omega = \alpha t \), where \( t \) is the time. Rearrange the formula to solve for \( \alpha \): \[\alpha = \frac{\omega}{t} = \frac{75.4}{2.00} \approx 37.7 \, \text{rad/s}^2\]

Step 3: Determine Torque from Tangential Force

The torque \( \tau \) applied to the wheel can be calculated by the formula \( \tau = F \cdot r \), where \( F \) is the force applied and \( r \) is the radius of the wheel: \[\tau = 80.0 \, \text{N} \times 0.120 \, \text{m} = 9.6 \, \text{Nm}\]

Step 4: Solve for Moment of Inertia

The relationship between torque, moment of inertia \( I \), and angular acceleration \( \alpha \) is given by \( \tau = I \alpha \). Rearrange this to solve for \( I \): \[I = \frac{\tau}{\alpha} = \frac{9.6}{37.7} \approx 0.255 \, \text{kg} \cdot \text{m}^2\]

Key Concepts

Angular Speed

Angular speed is a measure of how quickly an object rotates or spins. It tells us the rate at which an object moves through an angle. In this context, it is expressed in revolutions per second or radians per second. To convert from revolutions per second to radians per second, we use the fact that one revolution equals \(2\pi\) radians.

• Formula: If \( ext{revolutions per second} = ext{rev/s} \), then \( ext{radians per second} = ext{rev/s} \times 2\pi \).

In this exercise, a wheel achieves an angular speed of 12 revolutions per second, which translates to approximately 75.4 radians per second. This conversion is essential for calculations following standard units used in physics problems. Understanding angular speed helps in configuring and predicting the motion characteristics of rotational bodies, be it wheels, gears, or planets.

Angular Acceleration

Angular acceleration refers to how quickly the angular speed of a rotating object changes over time. It provides insight into how the speed of rotation is increasing or decreasing. Mathematically, angular acceleration \( \alpha \) is defined as the change in angular speed \( \omega \) over a certain period \( t \).

• Formula: \( \alpha = \frac{\omega}{t} \)

In this example, the wheel starts from rest, meaning its initial speed is zero. Over two seconds, its angular speed rises to 12 revolutions per second. Converting the final speed to radians, we calculate an angular acceleration of roughly 37.7 radians per second squared.This concept is crucial in understanding real-world scenarios where objects accelerate or decelerate, and it's the cornerstone of analyzing rotational dynamics.

Torque

Torque is the measure of the rotational force applied to an object. It determines how effectively a force causes an object to rotate. The magnitude of torque depends on two main factors: the force applied and the distance from the point of rotation. This distance is often referred to as the radius.

• Formula: \( \tau = F \cdot r \)

In the exercise, an 80.0 N force is applied at the rim of a wheel with a radius of 0.120 m. This results in a torque of 9.6 Newton meters (Nm).Torque is a key concept in mechanics, and understanding it helps us grasp how levers work or how engines transmit force to rotate wheels.

Tangential Force

Tangential force refers to the force applied in a direction tangential to the point of contact on the rotating object. It is responsible for making the body spin or rotate.

• This is simply the force exerted along the circumference of the object.
• Its effectiveness in generating motion depends on the distance from the axis of rotation—further away means more torque.

In our exercise, the tangential force is given as 80.0 N, exerted perpendicularly to the radius of the wheel, thus maximizing its effect. Understanding tangential force is critical for designing systems involving gears and pulley mechanisms, where precise force application leads to efficient motion.