Problem 51 - University Physics with Modern Physics

Question

A wheel with \(c=\frac{4}{9}\), a mass of \(40.0 \mathrm{~kg}\), and a rim radius of \(30.0 \mathrm{~cm}\) is mounted vertically on a horizontal axis. A 2.00 -kg mass is suspended from the wheel by a rope wound around the rim. Find the angular acceleration of the wheel when the mass is released.

Step-by-Step Solution

Answer

#tag_title#Answer#tag_content# When the mass is released, the angular acceleration of the wheel is given by the formula: \(α = \frac{\tau}{I} = \frac{rF}{\frac{1}{2}MR^2}\). First, we calculate the force exerted by the suspended mass: \(F = mg = (2.00 \mathrm{~kg})(9.81 \mathrm{~m/s^2}) = 19.62 \mathrm{~N}\). Next, we calculate the moment of inertia of the wheel: \(I = \frac{1}{2}(40.0 \mathrm{~kg})(0.3 \mathrm{~m})^2 = 1.8 \mathrm{~kg \cdot m^2}\). Now, we substitute the values of torque and moment of inertia into the formula for angular acceleration: \(α = \frac{(0.3 \mathrm{~m})(19.62 \mathrm{~N})}{1.8 \mathrm{~kg \cdot m^2}} = 3.27 \mathrm{~rad/s^2}\). Therefore, the angular acceleration of the wheel when the mass is released is \(3.27 \mathrm{~rad/s^2}\).

Step 1: Identify the forces acting on the system

There are two forces acting on the system: the gravitational force acting on the suspended mass and the tension in the rope. The gravitational force acting on the mass causes it to accelerate downward and exerts a torque on the wheel, causing it to rotate. The tension in the rope opposes the gravitational force and also exerts a torque on the wheel.

Step 2: Determine the moment of inertia of the wheel

To calculate the moment of inertia for the wheel, we can use the formula for a solid cylinder, \(I = \frac{1}{2}MR^2\) where \(M\) is the mass of the wheel, and \(R\) is the rim radius. The moment of inertia of the wheel is: \(I = \frac{1}{2}(40.0 \mathrm{~kg})(0.3 \mathrm{~m})^2\)

Step 3: Calculate the torque acting on the wheel due to the suspended mass

The torque acting on the wheel due to the suspended mass can be estimated using the formula: \(\tau = rF\), where \(r\) is the radius of the rim, and \(F\) is the force exerted by the suspended mass (the gravitational force acting on the mass). The force exerted by the suspended mass is: \(F = mg\), where \(m = 2.00 \mathrm{~kg}\) (the mass of the suspended weight), and \(g = 9.81 \mathrm{~m/s^2}\) (acceleration due to gravity).

Step 4: Use Newton's second law for rotation to find the angular acceleration

Newton's second law for rotation states that the angular acceleration (α) is equal to the torque (τ) divided by the moment of inertia (I): \(α = \frac{\tau}{I}\). Substituting the expressions for torque and moment of inertia, we find the angular acceleration when the mass is released.

Key Concepts

Moment of Inertia

When dealing with rotational motion, the concept of moment of inertia is crucial, as it determines how much torque is necessary for a desired angular acceleration. Think of it as the rotational version of mass in linear motion. Just like mass affects how difficult it is to accelerate or decelerate a car, moment of inertia affects how difficult it is to start or stop a spinning object.

In the context of our problem, the moment of inertia can be calculated using the provided equation for a solid cylinder, a shape that closely matches that of our wheel:

• \( I = \frac{1}{2}MR^2 \)

This formula takes into account the mass distribution relative to the rotation axis. Here, \( M \) is the mass of the wheel, and \( R \) is the radius. Substituting in the given values

• Mass \( M = 40.0 \mathrm{~kg} \)
• Radius \( R = 0.3 \mathrm{~m} \)

we compute the moment of inertia as a specific value, which will later be used to determine the angular acceleration of the wheel.

Newton's Second Law for Rotation

Newton's second law for rotation is a powerful tool when studying rotational dynamics, just like its linear counterpart is used for straight-line motion. The law describes how torque, moment of inertia, and angular acceleration interrelate:

• \( \alpha = \frac{\tau}{I} \)

where \( \alpha \) is the angular acceleration, \( \tau \) is the torque, and \( I \) is the moment of inertia. Unlike linear motion, where forces directly determine acceleration, rotational movement is governed by how much force is applied at a particular point at a distance from the center of rotation.

This formula lets us calculate the rate at which an object's rotational speed changes in response to applied torques. In our current problem:

• Torque \( \tau = \) is the result of the gravitational force pulling the suspended mass downwards.
• Moment of Inertia \( I \) is determined previously.

By plugging in these values, we can directly solve for the angular acceleration \( \alpha \), giving us insight into how quickly the wheel spins as the mass is released.

Torque

Torque is the rotational equivalent of linear force; it dictates how effectively a force can rotate an object around an axis. It is essentially a measure of the force causing an object to rotate, involving both the magnitude of the force and its distance from the pivot point, or axis of rotation.

For our wheel, the torque is generated by the gravitational force on the hanging mass. To compute the torque, you use:

• \( \tau = rF \)

In this formula, \( r \) is the radius of the wheel, which is the distance from the pivot point, and \( F \) is the force equal to the weight of the suspended mass:

• Force, \( F = mg \)

Here, \( m \) is the mass of the suspended object, and \( g \) is the acceleration due to gravity. This approach helps us quantify how much rotational effect the hanging mass will have on the wheel, ultimately allowing us to find the wheel's angular acceleration.