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A disk with a rotational inertia of \(7.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates like a merry-go-round while undergoing a variable torque given by \(\tau=(5.00+2.00 t) \mathrm{N} \cdot \mathrm{m}\). At time \(t=1.00 \mathrm{~s}\), its angular momentum is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its angular momentum at \(t=3.00 \mathrm{~s}\) ?

Short Answer

Expert verified
The angular momentum at \( t=3.00 \) s is \( 23.00 \mathrm{~kg} \cdot \mathrm{m}^{2}/\mathrm{s} \).

Step by step solution

01

Understanding the Relationship between Torque and Angular Momentum

Torque \( \tau \) is related to the rate of change of angular momentum \( L \) through \( \tau = \frac{dL}{dt} \). The problem gives a time-dependent torque function \( \tau(t) = (5.00 + 2.00t) \). We use this to find the change in angular momentum.
02

Formulate the Integral to Find Angular Momentum Change

We need to integrate the torque over time to find the change in angular momentum, \( \Delta L \), from \( t = 1.00 \) s to \( t = 3.00 \) s:\[ L(t) = L_0 + \int_{t_0}^{t} \tau(t') \, dt' \]Here, \( L_0 \) is the initial angular momentum at \( t = 1.00 \) s. We know \( L_0 = 5.00 \). Thus:\[ L(3.00) = 5.00 + \int_{1.00}^{3.00} (5.00 + 2.00t') \, dt' \]
03

Calculate the Integral

Compute the definite integral:\[ \int_{1.00}^{3.00} (5.00 + 2.00t') \, dt' = \left[ 5.00t' + 2.00 \cdot \frac{(t')^2}{2} \right]_{1}^{3} \]Solving it:- At \( t' = 3.00 \): \( 5.00 \times 3.00 + \frac{2.00 \cdot 3.00^2}{2} = 15.00 + 9.00 = 24.00 \)- At \( t' = 1.00 \): \( 5.00 \times 1.00 + \frac{2.00 \cdot 1.00^2}{2} = 5.00 + 1.00 = 6.00 \)Subtract the two outcomes: \[ 24.00 - 6.00 = 18.00 \]
04

Compute Final Angular Momentum

The change in angular momentum \( \Delta L = 18.00 \). Therefore, add this to \( L_0 \):\[ L(3.00) = 5.00 + 18.00 = 23.00 \, \text{kg} \cdot \text{m}^{2}/\text{s} \].

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Key Concepts

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

Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a fundamental concept in rotational dynamics. It measures how much resistance an object has to rotational acceleration about a particular axis. Rotational inertia depends on two major factors:
  • The mass of the object
  • The distribution of this mass around the axis of rotation

In the exercise, the given rotational inertia for the disk is \(7.00 \, \text{kg} \cdot \text{m}^{2}\). This value indicates that it requires a specific amount of torque to change the rotational speed of the disk. Just as mass does for linear motion, rotational inertia plays a critical role in determining how easily an object rotates. The larger the rotational inertia, the harder it is to change the rotational state.
Variable Torque
Torque is like the rotational equivalent of force. It represents how much a force acting on an object causes it to rotate. When the torque is constant, calculations might seem straightforward. However, when torque changes over time, as seen in the given exercise, it introduces complexity to the problem.
The problem describes a torque that changes with time: \(\tau(t) = 5.00 + 2.00t \, \text{N} \cdot \text{m}\). This means that the amount of torque applied to the disk not only depends on time but also increases as time advances.
This variable torque affects the angular momentum since the rate of change in angular momentum is directly connected to the applied torque. Understanding how to handle this variation through integration, as indicated in the solution, is essential in navigating real-world applications.
Integration in Physics
In physics, integration is a powerful mathematical tool used to find quantities when variables are changing continuously. For problems involving variable torque, integration helps us solve for changes in angular momentum by considering the entire interval of time.
To find the change in angular momentum from time \( t = 1.00 \) to \( t = 3.00 \), we use the integral:\[ L(3.00) = L_0 + \int_{1.00}^{3.00}(5.00 + 2.00t') \, dt' \]Where \( L_0 \) is the initial angular momentum. Integration here sums up the continuous effect of torque over time, accounting for all variations within the given period.
By solving the integral, we capture the complete picture of how the torque changes the disk's angular momentum, something that isn't possible with simple algebraic methods.
Rate of Change of Angular Momentum
Angular momentum, like linear momentum, is a crucial concept in understanding motion. The rate at which angular momentum changes is governed by torque. Mathematically, this relationship is often expressed through the equation:\[ \tau = \frac{dL}{dt} \]
In simpler terms, this means torque is the rate of change of angular momentum over time. For the given problem, the variable torque causes this rate to fluctuate, necessitating the use of calculus to determine how much the angular momentum changes.
When the torque applied is time-dependent, calculating the overall change in angular momentum involves integrating the torque over the relevant time interval. This defines the difference between the initial and final angular momenta. Thus, a comprehensive understanding of both the initial state, variable effects over time, and final states are integrated into one solution.

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Most popular questions from this chapter

A uniform wheel of mass \(10.0 \mathrm{~kg}\) and radius \(0.400 \mathrm{~m}\) is mounted rigidly on a massless axle through its center (Fig. 11-62). The radius of the axle is \(0.200 \mathrm{~m}\), and the rotational inertia of the wheel- axle combination about its central axis is \(0.600 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) The wheel is initially at rest at the top of a surface that is inclined at angle \(\theta=30.0^{\circ}\) with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by \(2.00 \mathrm{~m}\), what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

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A \(1200 \mathrm{~kg}\) airplane is flying in a straight line at \(80 \mathrm{~m} / \mathrm{s}\), \(1.3 \mathrm{~km}\) above the ground. What is the magnitude of its angular momentum with respect to a point on the ground directly under the path of the plane?

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to "take up" the angular momentum (Fig. 11-18). In \(0.700 \mathrm{~s}\), one arm sweeps through \(0.500\) rev and the other arm sweeps through \(1.000\) rev. Treat each arm as a thin rod of mass \(4.0 \mathrm{~kg}\) and length \(0.60 \mathrm{~m}\), rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

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