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A sanding disk with rotational inertia \(8.6 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the initially stationary disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Short Answer

Expert verified
Angular momentum is 0.528 kg m²/s and angular velocity is 61.4 rad/s.

Step by step solution

01

Understand the Given Parameters

Firstly, we need to gather the given information from the problem. We have: - Rotational inertia (moment of inertia), \( I = 8.6 \times 10^{-3} \) kg m², - Torque, \( \tau = 16 \) N m, - Time, \( t = 33 \) ms = 0.033 s. We are required to find the angular momentum and the angular velocity of the disk after 33 ms.
02

Calculate Angular Momentum

Angular momentum (\( L \)) can be calculated using the formula \( L = \tau \times t \), where \( \tau \) is the torque and \( t \) is the time. Substitute the given values:\[L = 16 \text{ N m} \times 0.033 \text{ s} = 0.528 \text{ kg m}^2/\text{s}.\]So, the angular momentum of the disk after 33 ms is 0.528 kg m²/s.
03

Calculate Angular Velocity Using Angular Momentum

To find the angular velocity \( \omega \), we use the relationship between angular momentum \( L \) and angular velocity, given by \( L = I \cdot \omega \). We can rearrange this formula to solve for \( \omega \):\[\omega = \frac{L}{I} = \frac{0.528 \text{ kg m}^2/\text{s}}{8.6 \times 10^{-3} \text{ kg m}^2}\]Calculate \( \omega \):\[\omega \approx 61.4 \text{ rad/s}.\]Thus, the angular velocity of the disk is approximately 61.4 rad/s after 33 ms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Inertia
Rotational inertia, often referred to as the moment of inertia, is a measure of an object's resistance to changes in its rotation. It's a key concept in rotational dynamics, analogous to mass in linear motion.
Unlike mass, rotational inertia depends not only on the amount of matter in an object but also on the distribution of that matter around the axis of rotation. For instance, two objects with the same mass can have different rotational inertias if their mass is distributed differently.
This means that it takes more effort to change the rotation of an object with a high rotational inertia than one with a low rotational inertia. In our problem, the sanding disk has a rotational inertia of \( 8.6 \times 10^{-3} \text{ kg m}^2 \). This value tells us how much torque is needed to achieve a certain angular acceleration. When considering rotational dynamics, it's important to think about both the mass and how far this mass is from the axis of rotation.
Torque
Torque is the rotational equivalent of force. It's what causes an object to start rotating, stop rotating, or change its rotational speed.
The magnitude of torque depends on three factors:
  • The force applied
  • The distance from the axis of rotation to where the force is applied (lever arm)
  • The angle between the force and the lever arm
If you imagine pushing a door, the more torque you apply, the faster the door will swing open.
In our exercise, the motor of the electric drill applies a torque of 16 N m to the sanding disk, which sets it into rotation. The application of torque over a certain amount of time will change the angular momentum of the disk. This is crucial for understanding how the system behaves over time.
Angular Momentum
Angular momentum is like the rotational version of linear momentum. It's a measure of the quantity of rotation an object has, considering its rotational inertia and angular velocity.
The formula to calculate angular momentum \( L \) is given by the product of torque \( \tau \) and time \( t \):
\[ L = \tau \times t \]
This means that the longer a torque is applied or the greater the torque, the higher the angular momentum will be.
In the exercise, we found the angular momentum of the disk after the torque has been applied for 33 milliseconds, resulting in \( 0.528 \text{ kg m}^2/\text{s} \). This shows us how much rotational motion has been imparted to the disk in that brief time.
Angular Velocity
Angular velocity indicates how quickly an object rotates or revolves. It's measured in radians per second \( \text{rad/s} \). Angular velocity tells us how fast the angle between an object and a reference line is changing over time.
In the context of rotational motion, there's a direct relationship between angular velocity and angular momentum. The angular momentum \( L \) is given by the formula:
\[ L = I \cdot \omega \]
which allows us to solve for angular velocity \( \omega \) as:
\[ \omega = \frac{L}{I} \]
In our solution, the angular velocity of the sanding disk, after the torque is applied for 33 ms, is approximately 61.4 rad/s.
This value indicates the rate at which the disk spins, and having a higher angular velocity means the disk spins faster.

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Most popular questions from this chapter

An automobile traveling at \(80.0 \mathrm{~km} / \mathrm{h}\) has tires of \(70.0 \mathrm{~cm}\) diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in \(30.0\) complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?

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