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A particle is to move in an \(x y\) plane, clockwise around the origin as seen from the positive side of the \(z\) axis. In unit-vector notation, what torque acts on the particle if the magnitude of its angular momentum about the origin is (a) \(4.0 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\), (b) \(4.0 t^{2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\), (c) \(4.0 \sqrt{t} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\), and (d) \(4.0 / t^{2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) ?

Short Answer

Expert verified
(a) 0, (b) \(8.0 t\), (c) \(\frac{2.0}{\sqrt{t}}\), (d) \(-\frac{8.0}{t^3}\).

Step by step solution

01

Understanding Torque and Angular Momentum Relationship

Torque \(\mathbf{\tau}\) and angular momentum \(\mathbf{L}\) are related through the derivative \(\mathbf{\tau} = \frac{d\mathbf{L}}{dt}\). This means that the torque acting on a particle is equal to the time rate of change of its angular momentum. We will use this relationship to solve parts (a) through (d) of the problem.
02

Solve for Torque in Part (a)

For part (a), the angular momentum \(L = 4.0 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}\), which is constant over time. Therefore, \(\frac{dL}{dt} = 0\). So the torque \(\tau = \frac{dL}{dt} = 0\).
03

Solve for Torque in Part (b)

For part (b), the angular momentum \(L = 4.0 t^2 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}\). We differentiate with respect to time: \(\frac{dL}{dt} = \frac{d}{dt}(4.0 t^2) = 8.0 t\). Thus, the torque \(\tau = 8.0 t\).
04

Solve for Torque in Part (c)

For part (c), the angular momentum \(L = 4.0 \sqrt{t} \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}\). We differentiate with respect to time: \(\frac{dL}{dt} = \frac{d}{dt}(4.0 \sqrt{t}) = \frac{4.0}{2\sqrt{t}} = \frac{2.0}{\sqrt{t}}\). Thus, the torque \(\tau = \frac{2.0}{\sqrt{t}}\).
05

Solve for Torque in Part (d)

For part (d), the angular momentum \(L = \frac{4.0}{t^2} \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}\). We differentiate with respect to time: \(\frac{dL}{dt} = \frac{d}{dt}(\frac{4.0}{t^2}) = -\frac{8.0}{t^3}\). Thus, the torque \(\tau = -\frac{8.0}{t^3}\).

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Key Concepts

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

Angular Momentum
Angular momentum is a fundamental concept in physics that helps describe the rotational motion of objects. It is analogous to linear momentum, but for rotating bodies. The key difference is that while linear momentum deals with objects moving in straight lines, angular momentum deals with objects rotating around an axis. For a particle moving in a plane around a point (often the origin), the angular momentum is given by the product of its position vector (from the origin) and its linear momentum.
  • The symbol for angular momentum is usually denoted by \( \mathbf{L} \).
  • Measured in units of \( \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s} \).
  • It is a vector quantity, meaning it has both magnitude and direction.
For our exercise, angular momentum is expressed in ways that change over time, as seen in parts (b), (c), and (d). These changes affect torque, which represents how effectively forces cause an object to rotate. Understanding the dynamic nature of angular momentum through time helps us to grasp how systems behave under rotational influences.
Differentiation
Differentiation is a mathematical tool used to find the rate at which one quantity changes with respect to another. It is extensively utilized in physics to understand how dynamic processes evolve over time, especially when dealing with motion.In the context of our problem, differentiation is used to calculate the torque from angular momentum, as torque \( \mathbf{\tau} \) is the time derivative of angular momentum \( \mathbf{L} \). In simpler terms, torque is how fast angular momentum changes over time.
  • For a constant angular momentum, the derivative is zero, resulting in zero torque, as seen in part (a).
  • For time-dependent angular momentum, using differentiation provides expressions for torque, like \( 8.0 t \) for \( L = 4.0 t^2 \) in part (b).
  • Other functional forms of \( L \) also help demonstrate different responding torques, such as \( \frac{2.0}{\sqrt{t}} \) and \(-\frac{8.0}{t^3} \) in parts (c) and (d).
Differentiation thus acts as an intermediary between angular momentum and torque, letting us see the relationship between them clearly and calculating exactly how changes in momentum produce changes in rotational force.
Unit-Vector Notation
Unit-vector notation is a way to express vector quantities, which have both magnitude and direction, in a clear and standardized form. In three-dimensional space, unit vectors are typically used along the Cartesian coordinates: the \( x \), \( y \), and \( z \) axes.
  • In unit-vector notation, any vector can be broken down into its components along these axes.
  • The standard unit vectors are denoted as \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) for the \( x \), \( y \), and \( z \) directions respectively.
This notation is particularly handy when dealing with physical quantities that may change in multiple directions simultaneously. For instance, in our exercise, although the torque is not explicitly written in unit-vector notation, the understanding of vectors is critical.Expressing torque and angular momentum in this form allows physicists and engineers to easily analyze and visualize the effect of striking forces and movements in each independent direction. By breaking them down, we simplify the complex relationships in rotational dynamics, making the calculations and conceptual understanding more manageable.

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Most popular questions from this chapter

A body of radius \(R\) and mass \(m\) is rolling smoothly with speed \(v\) on a horizontal surface. It then rolls up a hill to a maximum height \(h\). (a) If \(h=3 v^{2} / 4 g\), what is the body's rotational inertia about the rotational axis through its center of mass? (b) What might the body be?

At the instant the displacement of a \(2.00 \mathrm{~kg}\) object relative to the origin is \(\vec{d}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}-(3.00 \mathrm{~m}) \hat{\mathrm{k}}\), its velocity is \(\vec{v}=-(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\) and it is subject to a force \(\vec{F}=(6.00 \mathrm{~N}) \hat{\mathrm{i}}-(8.00 \mathrm{~N}) \hat{\mathrm{j}}+(4.00 \mathrm{~N}) \hat{\mathrm{k}}\). Find (a) the acceleration of the object, (b) the angular momentum of the object about the origin, (c) the torque about the origin acting on the object, and (d) the angle between the velocity of the object and the force acting on the object.

A \(2.50 \mathrm{~kg}\) particle that is moving horizontally over a floor with velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) undergoes a completely inelastic collision with a \(4.00 \mathrm{~kg}\) particle that is moving horizontally over the floor with velocity \((4.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\). The collision occurs at \(x y\) coordinates \((-0.500 \mathrm{~m},-0.100 \mathrm{~m})\). After the collision and in unit- vector notation, what is the angular momentum of the stuck-together particles with respect to the origin?

A uniform disk of mass \(10 \mathrm{~m}\) and radius \(3.0 r\) can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass \(m\) and radius \(r\) lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\). Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their angular velocity about the center of the larger disk? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the two-disk system to the system's initial kinetic energy?

A wheel of radius \(0.250 \mathrm{~m}\), moving initially at \(43.0 \mathrm{~m} / \mathrm{s}\), rolls to a stop in \(225 \mathrm{~m}\). Calculate the magnitudes of its (a) linear acceleration and (b) angular acceleration. (c) Its rotational inertia is \(0.155\) \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its central axis. Find the magnitude of the torque about the central axis due to friction on the wheel.

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