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At a certain time, a \(0.25 \mathrm{~kg}\) object has a position vector \(\vec{r}=(2.0 \mathrm{~m}) \hat{\mathrm{i}}+\) \((-2.0 \mathrm{~m}) \hat{\mathrm{y}}\) in meters. At that instant, its velocity in meters per second is \(\vec{v}=(-5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and the force in newtons acting on it is \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{j}} .\) (a) What is the rotational momentum of the object about the origin? (b) What torque acts on it?

Short Answer

Expert verified
The rotational momentum is \(0\). The torque is \(8 \hat{k} \text{N} \cdot \text{m}\).

Step by step solution

01

- Identify given values

Given: Mass of the object, \(m = 0.25 \, \text{kg}\)Position vector, \(\vec{r} = (2.0 \, \text{m}) \hat{i} - (2.0 \, \text{m}) \hat{j}\)Velocity vector, \(\vec{v} = (-5.0 \, \text{m/s}) \hat{i} + (5.0 \, \text{m/s}) \hat{j}\)Force vector, \(\vec{F} = (4.0 \, \text{N}) \hat{j}\)
02

- Calculate the angular momentum

The angular momentum \(\vec{L}\) of the object about the origin is given by \(\vec{L} = \vec{r} \times (m \vec{v})\). First, determine the momentum \(m \vec{v} = 0.25 \,( -5.0 \hat{i} + 5.0 \hat{j}) \,\text{kg m/s} = (-1.25 \,\hat{i} + 1.25 \,\hat{j})\, \text{kg m/s}\).Now, calculate the cross product \(\vec{r} \times m\vec{v}\):\(\vec{r} = 2 \hat{i} - 2 \hat{j}$$m \vec{v} = -1.25 \hat{i} + 1.25 \hat{j}$$\vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 0 \ -1.25 & 1.25 & 0 \end{vmatrix}\)= \(\hat{k} (2 \cdot 1.25 - (-2) \cdot (-1.25))\)= \(\hat{k} (2.5 - 2.5 )\)= 0
03

- Calculate the torque

The torque \(\vec{\tau}\) acting on the object is given by \(\vec{\tau} = \vec{r} \times \vec{F}\). Calculate the cross product \(\vec{r} \times \vec{F}\):\(\vec{r} = 2 \hat{i} - 2 \hat{j}$$\vec{F} = 0 \hat{i} + 4 \hat{j}$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 0 \ 0 & 4 & 0 \end{vmatrix}\)= \(\hat{k} (2 \cdot 4 - (-2) \cdot 0)\)= \(\hat{k} (8 - 0)\)= \(8 \hat{k}\)

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Key Concepts

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

Torque
Torque is a measure of how much a force acting on an object causes that object to rotate. It's like the rotational equivalent of force. The torque \(\vec{\tau}\) acting on an object is given by the cross product of the position vector \(\vec{r}\) and the force vector \(\vec{F}\).

This can be written as: \[ \vec{\tau} = \vec{r} \times \vec{F} \]

In simpler terms, torque is all about how a force applied at a certain distance from a pivot point can cause an object to start spinning. The unit of torque is Newton-meter (N·m). In this exercise, you calculated the torque using the provided position and force vectors. The resulting torque vector \(\vec{\tau} = 8 \hat{k}\) shows that the torque is acting in the direction of the z-axis.

Understanding torque is crucial in the study of rotational motion, as it directly influences the rotational acceleration of an object.
Cross Product
The cross product is a mathematical operation performed on two vectors that results in a third vector perpendicular to the plane formed by the original vectors. It is used extensively in physics to calculate quantities such as torque and angular momentum.

The cross product of two vectors \(\vec{A}\) and \(\vec{B}\) is denoted as \(\vec{A} \times \vec{B}\). It can be calculated using the determinant of a matrix:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix} \]

Where \(\hat{i}, \hat{j},\) and \(\hat{k}\) are unit vectors along the x, y, and z axes, respectively. In this exercise, you used the cross product to calculate both angular momentum and torque. This involved creating matrices from the position, velocity, and force vectors and finding their determinants. The concept is fundamental to many areas of physics, especially when dealing with 3D space and rotations.
Rotational Motion
Rotational motion refers to the movement of an object around a central axis. It is characterized by quantities such as angular velocity, angular acceleration, and moments of inertia. Just like linear motion has properties like velocity and acceleration, rotational motion has its own set of unique properties.

Some key concepts in rotational motion include:
  • Angular Momentum (\(\vec{L}\)): This is the rotational equivalent of linear momentum and is calculated as \(\vec{L} = \vec{r} \times m\vec{v}\).
  • Torque (\(\vec{\tau}\)): This is the measure of the force that causes an object to rotate. It is calculated using \(\vec{\tau} = \vec{r} \times \vec{F}\).
  • Moment of Inertia: This is a measure of an object's resistance to changes in its rotational motion.


In the exercise, you encountered rotational motion in the context of an object's angular momentum and the torque acting on it. Understanding these principles is vital for explaining how objects spin, rotate and respond to various forces.

By mastering these concepts, you can better understand the dynamics of anything that rotates, from simple wheels to the complex orbits of celestial bodies.

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Most popular questions from this chapter

What is the torque about the origin on a jar of jalapeño peppers located at coordinates \((3.0 \mathrm{~m},-2.0 \mathrm{~m}, 4.0 \mathrm{~m})\) due to (a) force \(\vec{F}_{A}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}+(5.0 \mathrm{~N}) \hat{\mathrm{k}},(\mathrm{b})\) force \(\vec{F}_{B}=\) \((-3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}-(5.0 \mathrm{~N}) \hat{\mathrm{k}}\), and \((\mathrm{c})\) the vector sum of \(\vec{F}_{A}\) and \(\vec{F}_{B} ?\) (d) Repeat part (c) about a point with coordinates \((3.0 \mathrm{~m},\), \(2.0 \mathrm{~m}, 4.0 \mathrm{~m}\) ) instead of about the origin.

(A) Show that \(\vec{a} \cdot(\vec{b} \times \vec{a})\) is zero for all vectors \(\vec{a}\) and \(\vec{b}\). (b) What is the magnitude of \(\vec{a} \times(\vec{b} \times \vec{a})\) if there is an angle \(\phi\) between the directions of \(\vec{a}\) and \(\vec{b}\) ?

A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g}\). Its axle radius is \(3.2 \mathrm{~mm}\), and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its translational acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) translational speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) rotational speed?

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of \(150 \mathrm{~kg}\), a radius of \(2.0 \mathrm{~m}\), and a rotational inertia of \(300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axis of rotation. A \(60 \mathrm{~kg}\) student walks slowly from the rim of the platform toward the center. If the rotational speed of the system is \(1.5 \mathrm{rad} / \mathrm{s}\) when the student starts at the rim, what is the rotational speed when she is \(0.50 \mathrm{~m}\) from the center?

A cockroach of mass \(m\) runs counterclockwise around the rim of a lazy Susan (a circular dish mounted on a vertical axle) of radius \(R\) and rotational inertia \(I\) and having frictionless bearings. The cockroach's speed (relative to the ground) is \(v\), whereas the lazy Susan turns clockwise with rotational speed \(\omega_{1}\). The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the rotational speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved?

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