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Chapter 11: Problem 30

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 veloc- ity 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 accel- eration 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.

Short Answer

Expert verified
a) \(\vec{a} = (3.00) \hat{\mathrm{i}} - (4.00) \hat{\mathrm{j}} + (2.00) \hat{\mathrm{k}}\,\mathrm{m/s}^2\); b) \(\vec{L} = (42.00) \hat{\mathrm{i}} + (12.00)\hat{\mathrm{j}} + (30.00)\hat{\mathrm{k}}\,\mathrm{kg}\cdot\mathrm{m}^2/\mathrm{s}\); c) \(\vec{\tau} = (-8.00) \hat{\mathrm{i}} - (26.00) \hat{\mathrm{j}} - (40.00) \hat{\mathrm{k}}\,\mathrm{N}\cdot\mathrm{m}\); d) \(\theta \approx 138.59^\circ \).

Step by step solution

01

Calculate Acceleration

To find the acceleration, use Newton's second law, \( \vec{F} = m \vec{a} \). Given \( \vec{F} = (6.00 \mathrm{~N}) \hat{\mathrm{i}} - (8.00 \mathrm{~N}) \hat{\mathrm{j}} + (4.00 \mathrm{~N}) \hat{\mathrm{k}} \) and \( m = 2.00 \mathrm{~kg} \), solve for \( \vec{a} \) by dividing each component of the force by the mass:\[\vec{a} = \left(\frac{6.00}{2.00}\right) \hat{\mathrm{i}} - \left(\frac{8.00}{2.00}\right) \hat{\mathrm{j}} + \left(\frac{4.00}{2.00}\right) \hat{\mathrm{k}} = (3.00) \hat{\mathrm{i}} - (4.00) \hat{\mathrm{j}} + (2.00) \hat{\mathrm{k}} \, \mathrm{m/s}^2\]
02

Calculate Angular Momentum

Angular momentum \( \vec{L} \) is given by \( \vec{L} = \vec{r} \times m\vec{v} \), where \( \vec{r} = (2.00 \mathrm{~m}) \hat{\mathrm{i}} + (4.00 \mathrm{~m}) \hat{\mathrm{j}} - (3.00 \mathrm{~m}) \hat{\mathrm{k}} \) and \( \vec{v} = -(6.00 \mathrm{~m/s}) \hat{\mathrm{i}} + (3.00 \mathrm{~m/s}) \hat{\mathrm{j}} + (3.00 \mathrm{~m/s}) \hat{\mathrm{k}} \). Use the cross product:\[\vec{L} = m \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 2.00 & 4.00 & -3.00 \ -6.00 & 3.00 & 3.00 \end{vmatrix}\]Calculating the determinant:\[\vec{L} = 2.00 \left((4.00 \times 3.00) - (-3.00 \times 3.00)\right)\hat{\mathrm{i}} - \left((2.00 \times 3.00) - (-3.00 \times -6.00)\right)\hat{\mathrm{j}} + \left((2.00 \times 3.00) - (4.00 \times -6.00)\right)\hat{\mathrm{k}}\]\[= 2.00 \left(12 + 9\right)\hat{\mathrm{i}} - \left(6 - 18\right)\hat{\mathrm{j}} + \left(6 + 24\right)\hat{\mathrm{k}} \]\[= (42.00) \hat{\mathrm{i}} + (12.00)\hat{\mathrm{j}} + (30.00)\hat{\mathrm{k}} \, \mathrm{kg}\cdot\mathrm{m}^2/\mathrm{s}\]
03

Calculate Torque

Torque \( \vec{\tau} \) is given by \( \vec{\tau} = \vec{r} \times \vec{F} \). Use the given position \( \vec{r} = (2.00 \mathrm{~m}) \hat{\mathrm{i}} + (4.00 \mathrm{~m}) \hat{\mathrm{j}} - (3.00 \mathrm{~m}) \hat{\mathrm{k}} \) and force \( \vec{F} = (6.00 \mathrm{~N}) \hat{\mathrm{i}} - (8.00 \mathrm{~N}) \hat{\mathrm{j}} + (4.00 \mathrm{~N}) \hat{\mathrm{k}} \):\[\vec{\tau} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 2.00 & 4.00 & -3.00 \ 6.00 & -8.00 & 4.00 \end{vmatrix}\]Calculate the determinant:\[\vec{\tau} = \left((4.00 \times 4.00) - (-3.00 \times -8.00)\right)\hat{\mathrm{i}} - \left((2.00 \times 4.00) - (-3.00 \times 6.00)\right)\hat{\mathrm{j}} + \left((2.00 \times -8.00) - (4.00 \times 6.00)\right)\hat{\mathrm{k}}\]\[= \left(16 - 24\right)\hat{\mathrm{i}} - \left(8 + 18\right)\hat{\mathrm{j}} + \left(-16 - 24\right)\hat{\mathrm{k}}\]\[= (-8.00) \hat{\mathrm{i}} - (26.00) \hat{\mathrm{j}} - (40.00) \hat{\mathrm{k}} \, \mathrm{N}\cdot\mathrm{m}\]
04

Calculate Angle Between Velocity and Force

To find the angle \( \theta \) between the velocity \( \vec{v} \) and force \( \vec{F} \), use the dot product formula: \[ \vec{v} \cdot \vec{F} = |\vec{v}| |\vec{F}| \cos(\theta) \]. First, calculate the magnitudes:\[ |\vec{v}| = \sqrt{(-6.00)^2 + 3.00^2 + 3.00^2} = \sqrt{36 + 9 + 9} = \sqrt{54} \]\[ |\vec{F}| = \sqrt{6.00^2 + (-8.00)^2 + 4.00^2} = \sqrt{36 + 64 + 16} = \sqrt{116} \]Compute the dot product:\[\vec{v} \cdot \vec{F} = (-6.00 \times 6.00) + (3.00 \times -8.00) + (3.00 \times 4.00) = -36 - 24 + 12 = -48\]Then, find \( \cos(\theta) \) and \( \theta \):\[\cos(\theta) = \frac{-48}{\sqrt{54} \times \sqrt{116}}\]Solve for \( \theta \):\[\theta = \cos^{-1}\left(\frac{-48}{\sqrt{54} \times \sqrt{116}}\right)\]Use a calculator to find \( \theta \), resulting in approximately \( 138.59^\circ \).

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Key Concepts

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

Acceleration
Understanding acceleration is crucial to exploring Newton's second law of motion. Acceleration (\( \vec{a} \)) is how we measure how quickly an object changes its velocity.
According to Newton's second law, acceleration is produced when a force \( \vec{F} \) acts on a mass \( m \).
This is summarized by the formula:
  • \( \vec{a} = \frac{\vec{F}}{m} \).
In this specific problem, the force acting on the object is given as \( \vec{F} = (6.00 \, \mathrm{N}) \hat{\mathrm{i}} - (8.00 \, \mathrm{N}) \hat{\mathrm{j}} + (4.00 \, \mathrm{N}) \hat{\mathrm{k}}\), and the object's mass is \( 2.00 \, \mathrm{kg}\).
To find the acceleration, divide each component of the force by the mass:
  • i-component: \( 3.00 \, \mathrm{m/s}^2 \)
  • j-component: \(-4.00 \, \mathrm{m/s}^2 \)
  • k-component: \(2.00 \, \mathrm{m/s}^2 \)
Putting these together, the full acceleration vector is \( (3.00) \hat{\mathrm{i}} - (4.00) \hat{\mathrm{j}} + (2.00) \hat{\mathrm{k}} \, \mathrm{m/s}^2\).
Angular Momentum
Angular momentum (\( \vec{L} \)) is a fundamental concept in physics, often described as rotational equivalent of linear momentum.
For an object moving in a straight line, the angular momentum about a point is calculated using:
  • \( \vec{L} = \vec{r} \times m\vec{v} \),
where \( \vec{r} \) is the position vector, \( m \) is the mass, and \( \vec{v} \) is the velocity.
In the exercise, the position vector \( \vec{r} \) is \( (2.00 \, \mathrm{m}) \hat{\mathrm{i}} + (4.00 \, \mathrm{m}) \hat{\mathrm{j}} - (3.00 \, \mathrm{m}) \hat{\mathrm{k}}\), and the velocity \( \vec{v} \) is \( -(6.00 \, \mathrm{m/s}) \hat{\mathrm{i}} + (3.00 \, \mathrm{m/s}) \hat{\mathrm{j}} + (3.00 \, \mathrm{m/s}) \hat{\mathrm{k}}\).
To find the angular momentum, calculate the cross product:
\[\begin{vmatrix}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \2.00 & 4.00 & -3.00 \-6.00 & 3.00 & 3.00\end{vmatrix}\]By solving the determinant, the angular momentum vector results in:
  • \( (42.00) \hat{\mathrm{i}} + (12.00) \hat{\mathrm{j}} + (30.00) \hat{\mathrm{k}} \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\).
Torque
Torque (\( \vec{\tau} \)) describes how much a force acting on an object causes that object to rotate around a point or axis.
It is calculated using the formula:
  • \( \vec{\tau} = \vec{r} \times \vec{F} \),
where \( \vec{r} \) is the position vector and \( \vec{F} \) is the force vector.
From the exercise, the position vector \( \vec{r} \) is \( (2.00 \, \mathrm{m}) \hat{\mathrm{i}} + (4.00 \, \mathrm{m}) \hat{\mathrm{j}} - (3.00 \, \mathrm{m}) \hat{\mathrm{k}}\) and the force vector \( \vec{F} \) is \( (6.00 \, \mathrm{N}) \hat{\mathrm{i}} - (8.00 \, \mathrm{N}) \hat{\mathrm{j}} + (4.00 \, \mathrm{N}) \hat{\mathrm{k}}\).
To compute the torque, use the determinant:
\[\begin{vmatrix}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \2.00 & 4.00 & -3.00 \6.00 & -8.00 & 4.00\end{vmatrix}\]After simplification, the torque vector is:
  • \(-8.00 \hat{\mathrm{i}} - 26.00 \hat{\mathrm{j}} - 40.00 \hat{\mathrm{k}} \, \mathrm{N} \cdot \mathrm{m}\).
Vector Calculations
Dealing with vectors is essential in physics as they allow describing quantities that have both magnitude and direction.
In this problem, vectors such as displacement, velocity, and forces are represented in component form with the unit vectors \( \hat{\mathrm{i}} \), \( \hat{\mathrm{j}} \), and \( \hat{\mathrm{k}} \).
One key vector operation is the dot product, used to find the angle between two vectors \( \vec{a} \) and \( \vec{b} \).
The dot product formula is:
  • \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) \),
where \( \theta \) is the angle between them.
In this case, the velocity \( \vec{v} \) and the force \( \vec{F} \) allow finding \( \theta \) using their magnitudes and dot product.
  • Magnitude of \( \vec{v} \): \( \sqrt{54} \)
  • Magnitude of \( \vec{F} \): \( \sqrt{116} \)
  • Dot product: \( -48 \)
  • Calculate \( \cos(\theta) = \frac{-48}{\sqrt{54} \times \sqrt{116}} \)
Finally, solve for \( \theta \) using the inverse cosine function, resulting in approximately \( 138.59^\circ \).

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Most popular questions from this chapter

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

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?

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 angular speed of the system is \(1.5 \mathrm{rad} / \mathrm{s}\) when the student starts at the rim, what is the angular speed when she is \(0.50 \mathrm{~m}\) from the center?

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 \(t=3.00 \mathrm{~s}\) ?

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved \(1.0 \mathrm{~m}\) up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?
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