Question

An object made of two disk-shaped sections, \(\mathrm{A}\) and \(\mathrm{B}\), as shown in the figure, is rotating about an axis through the center of disk A. The masses and the radii of disks \(A\) and \(B\). respectively are, \(2.00 \mathrm{~kg}\) and \(0.200 \mathrm{~kg}\) and \(25.0 \mathrm{~cm}\) and \(2.50 \mathrm{~cm}\). a) Calculate the moment of inertia of the object. b) If the axial torque due to friction is \(0.200 \mathrm{~N} \mathrm{~m}\), how long will it take for the object to come to a stop if it is rotating with an initial angular velocity of \(-2 \pi \mathrm{rad} / \mathrm{s} ?\)

Short answer

Question: Calculate the time it takes for the object (a combination of two disks) to come to a stop, given the following data. Disk A has a mass of 2.00 kg and a radius of 25.0 cm. Disk B has a mass of 0.200 kg and a radius of 2.50 cm. The initial angular velocity is -2π rad/s, and the torque due to friction is 0.200 N.m. Answer: To find the time it takes for the object to stop, follow these steps: 1. Calculate the moments of inertia for Disk A and Disk B using the formula \(I = \frac{1}{2}MR^2\). 2. Add the moments of inertia of both disks to find the total moment of inertia of the object. 3. Calculate the angular acceleration needed to stop the object using the formula \(\alpha = \frac{\tau}{I}\). 4. Use the equation \(t = \frac{\omega_f - \omega_0}{\alpha}\) to find the time it takes for the object to stop. Using the given data, follow each step to determine the time it takes for the object to come to a stop.

Step-by-step solution

Moment of Inertia for a disk

The moment of inertia for a disk rotating around its central axis can be calculated using the formula \(I = \frac{1}{2}MR^2\), where \(I\) is the moment of inertia, \(M\) is the mass of the disk, and \(R\) is the radius of the disk.

Moment of Inertia for Disk A

First, we need to calculate the moment of inertia for disk A using the given mass and radius. The mass of Disk A is \(2.00\,\mathrm{kg}\), and the radius is \(25.0\,\mathrm{cm}\), which is equal to \(0.25\,\mathrm{m}\) in SI units. So, the moment of inertia for Disk A will be: \(I_A = \frac{1}{2} (2.00\,\mathrm{kg}) (0.25\,\mathrm{m})^2\).

Moment of Inertia for Disk B

Similarly, we need to calculate the moment of inertia for disk B using the given mass and radius, which are \(0.200\,\mathrm{kg}\) and \(2.50\,\mathrm{cm}\), respectively. To convert the radius into SI units, we need to divide it by 100, i.e., \(2.50\,\mathrm{cm}\) = \(0.025\,\mathrm{m}\). So, the moment of inertia for Disk B will be: \(I_B = \frac{1}{2} (0.200\,\mathrm{kg}) (0.025\,\mathrm{m})^2\).

Combined Moment of Inertia

The moment of inertia of the object is the sum of the individual moments of inertia of the two disks. Thus, the total moment of inertia of the object will be: \(I = I_A + I_B\). #b)# Time to Come to a Stop

Calculating Angular Acceleration

We know that the torque due to friction (\(\tau\)) is equal to the moment of inertia (\(I\)) multiplied by the angular acceleration (\(\alpha\)), or \(\tau = I \alpha\). We are given the torque due to friction as \(0.200\,\mathrm{N}\,\mathrm{m}\). We can find the angular acceleration required to stop the object by rearranging the equation: \(\alpha = \frac{\tau}{I}\).

Calculating the Time

Since the final angular velocity (\(\omega_f\)) of the object will be zero when it comes to a stop, we can use the equation \(\omega_f = \omega_0 + \alpha t\) to find the time (\(t\)) it will take for the object to stop. We know the initial angular velocity (\(\omega_0\)) is \(-2\pi\,\mathrm{rad/s}\). By rearranging the equation to solve for \(t\), we get \(t = \frac{\omega_f - \omega_0}{\alpha}\).

Key concepts

Understanding Angular Velocity

Angular velocity, denoted by the symbol \( \omega \), is a vector quantity that represents how fast an object rotates or revolves around a fixed axis. It is measured in radians per second (\( \text{rad/s} \)). In essence, it tells us the angle through which an object turns in a unit of time.

For instance, when a disk-shaped object like the one in the textbook exercise rotates, every point on the object moves through the same angle in the same time interval. This uniform motion describes the angular velocity of the object. When calculating the initial angular velocity \( \omega_0 = -2 \pi \ \text{rad/s} \) (negative indicating the direction of rotation), it gives us a complete understanding of how fast the object was spinning before any forces acted upon it.

When studying physics problems, knowing the angular velocity can help us understand the dynamic behavior of rotating systems and predict the outcomes of torque and angular acceleration applications.

Exploring Angular Acceleration

Angular acceleration is the rate at which an object's angular velocity changes with time. Denoted by \( \alpha \), it is measured in radians per second squared (\( \text{rad/s}^2 \)). If an object's rotation speeds up or slows down, angular acceleration is present.

In our textbook problem, when an external torque due to friction is applied to the rotating object, it causes the object to experience angular acceleration opposite to its direction of rotation. The angular acceleration can be calculated by dividing the torque by the moment of inertia (\( \alpha = \frac{\tau}{I} \)), a key relationship that quantifies how the rotational motion is affected by external influences.

If an object moves with a constant angular velocity, there is no angular acceleration. However, when a force (i.e., a frictional torque) acts on the object, as in the case with our example, the decelerative angular acceleration affects how quickly the object comes to a stop. Understanding angular acceleration allows us to solve for time, given initial conditions and forces at play.

The Role of Torque

Torque, represented by the Greek letter \( \tau \), is a measure of the force that causes an object to rotate around an axis. It is the rotational equivalent of linear force and is calculated as the product of the force and the distance from the axis of rotation to the point where the force is applied, formatted as \( \tau = rF \sin(\theta) \), where \( r \) is the lever arm, \( F \) is the force, and \( \theta \) is the angle between the force and the lever arm direction.

In the given exercise, friction creates a torque that opposes the rotation of the object. The magnitude of this torque is \( 0.200 \ \text{N} \cdot \text{m} \) and acts to decelerate the object. Torque directly influences the angular acceleration of an object, since \( \tau = I \alpha \), where \( I \) is the moment of inertia. The larger the torque, the greater the object's change in rotational motion.

By analyzing torque, students can gain insights into the forces involved in rotational dynamics and the effects these forces have on an object's rotation, such as how long it takes for an object to come to a complete stop under a given torque.