Problem 69 - University Physics with Modern Physics

Question

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be 0.640 m. Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an 8.20-kg block of wood from the free end of the rope, as in Fig. E9.45. You release the system from rest and find that the block descends 12.0 m in 4.00 s. What is the moment of inertia of the wheel for this axis?

Step-by-Step Solution

Answer

The moment of inertia is approximately 5.50 kg·m².

Step 1: Understand the Problem

The problem asks for the moment of inertia of a wheel that is rotating about an axis through its center. We need a relation between the given downward motion of a block and the rotational motion of the wheel.

Step 2: Calculate the Acceleration of the Block

Using the equation of motion, the block descends 12 meters in 4 seconds can be described by the equation: \( s = ut + \frac{1}{2} a t^2 \). Here, the initial velocity \( u = 0 \), the distance \( s = 12 \, \text{m} \), and \( t = 4 \, \text{s} \). Solving for \( a \) gives: \( 12 = \frac{1}{2} a (4)^2 \) \( a = \frac{24}{16} = 1.5 \, \text{m/s}^2 \).

Step 3: Relate Linear and Angular Acceleration

The linear acceleration \( a \) and angular acceleration \( \alpha \) are related through the radius \( r \) of the wheel: \( a = r \alpha \). The radius is half the diameter, thus \( r = 0.320 \, \text{m} \). Therefore, \( \alpha = \frac{1.5}{0.320} \approx 4.69 \, \text{rad/s}^2 \).

Step 4: Apply Torque Equation and Solve for Moment of Inertia

The torque \( \tau \) caused by the hanging block is \( \tau = mg r \) where \( m = 8.2 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( r = 0.320 \, \text{m} \). Compute \( \tau = 8.2 \times 9.8 \times 0.320 = 25.792 \, \text{Nm} \). Using \( \tau = I \alpha \), the moment of inertia \( I \) is: \( I = \frac{\tau}{\alpha} = \frac{25.792}{4.69} \approx 5.50 \, \text{kg} \cdot \text{m}^2 \).

Step 5: Final Answer

The moment of inertia of the wheel for the given axis is approximately \( 5.50 \, \text{kg} \cdot \text{m}^2 \).

Key Concepts

Rotational Motion

Rotational motion refers to the movement of an object around an axis. Imagine a spinning wheel: every point on the wheel rotates around the wheel's center. This concept is essential in many natural and engineered systems. When analyzing rotational motion, we consider several factors, like angular velocity, angular acceleration, and rotational inertia. In the case of the spinning wheel in the exercise, we specifically focus on how it rotates around an axis through its center.

Understanding rotational motion requires an appreciation of how linear motion concepts adapt in circular systems. For instance, while objects moving in a straight line have a speed, rotating objects have an angular speed—how quickly they spin around their axis.

In our scenario, the block's downward movement creates a rotational motion in the wheel. This happens because as the block moves down, it pulls a rope attached to the wheel, causing the wheel to turn. This example demonstrates how linear motion can transfer into rotational motion, a fundamental principle in physics that connects the two types of movement.

Torque

Torque is a force that causes an object to rotate. You can think of it as a push or pull that adds a twist. The amount of torque depends on how strong the force is and how far from the pivot point, or axis, it's applied. In mathematical terms, torque (τ) is the product of the force and the radius at which the force acts: \( \tau = F \cdot r \).

In our exercise, the hanging block creates torque on the wheel. The formula \( \tau = mg \cdot r \) is used, where \( m \) is the block's mass, \( g \) is the acceleration due to gravity, and \( r \) is the radius of the wheel. This torque is responsible for turning the wheel and is crucial for calculating the moment of inertia.

Understanding torque helps us comprehend how different forces result in rotation. If you've ever used a wrench, you've applied torque. The longer the wrench, the more torque you can apply with the same force because you increase the radius \( r \). This principle is at play in our rotating wheel scenario.

Angular Acceleration

Angular acceleration is how quickly an object's rotational speed changes. It's like acceleration in linear motion, but for rotations. It tells us how fast an object goes from a slow spin to a fast spin. Angular acceleration (\( \alpha \)) is related to both the torque applied and the moment of inertia (\( I \)), through the equation \( \tau = I \alpha \).

In the problem, we connect the block's linear acceleration to the wheel's angular acceleration. Since the wheel's motion is directly influenced by the block pulling on it, we use the formula: \( a = r \alpha \). This derives the wheel's angular acceleration from the block's linear movement.

By understanding angular acceleration, we see how changes in speed during rotations occur. Whether it's a car turning a corner or a planet orbiting the sun, angular acceleration plays a vital role in how circular motions evolve over time. In our exercise, this understanding allows us to determine the wheel's moment of inertia, showing the wheel's resistance to changes in its rotation.