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Chapter 11: Problem 27

At one instant, force \(\vec{F}=4.0 \hat{\mathrm{j}} \mathrm{N}\) acts on a \(0.25 \mathrm{~kg}\) object that has position vector \(\vec{r}=(2.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{k}}) \mathrm{m}\) and velocity vector \(\vec{v}=(-5.0 \hat{\mathrm{i}}+5.0 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}\). About the origin and in unit-vector notation, what are (a) the object's angular momentum and (b) the torque acting on the object?

Short Answer

Expert verified
The object's angular momentum is \( \vec{L} = 5.0 \hat{\mathrm{j}} \text{ kg}\cdot \text{m}^2/\text{s} \) and the torque is \( \vec{\tau} = -8.0 \hat{\mathrm{i}} + 8.0 \hat{\mathrm{k}} \text{ N}\cdot\text{m} \).

Step by step solution

01

Identify the relevant formula for angular momentum

Angular momentum \( \vec{L} \) is given by the cross product of the position vector \( \vec{r} \) and the linear momentum \( \vec{p} \), where \( \vec{p} = m\vec{v} \) and \( m \) is the mass. The formula for angular momentum is: \[ \vec{L} = \vec{r} \times (m\vec{v}) \]
02

Calculate linear momentum

Calculate the linear momentum \( \vec{p} \) by multiplying the mass \( m = 0.25 \text{ kg} \) with the velocity vector \( \vec{v} = (-5.0 \hat{\mathrm{i}} + 5.0 \hat{\mathrm{k}}) \text{ m/s} \): \[ \vec{p} = 0.25 \times (-5.0 \hat{\mathrm{i}} + 5.0 \hat{\mathrm{k}}) = (-1.25 \hat{\mathrm{i}} + 1.25 \hat{\mathrm{k}}) \text{ kg}\cdot\text{m/s} \]
03

Calculate the angular momentum

Substitute the position vector \( \vec{r} = (2.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{k}}) \text{ m} \) and linear momentum \( \vec{p} = (-1.25 \hat{\mathrm{i}} + 1.25 \hat{\mathrm{k}}) \text{ kg}\cdot\text{m/s} \) into the cross product formula: \[ \vec{L} = \vec{r} \times \vec{p} = (2.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{k}}) \times (-1.25 \hat{\mathrm{i}} + 1.25 \hat{\mathrm{k}}) \] Evaluate the cross product to find: \[ \vec{L} = [0.0 \hat{\mathrm{i}} + 0.0 \hat{\mathrm{j}} + (2.0)(1.25) - (-2.0)(-1.25)] \hat{\mathrm{j}} = 5.0 \hat{\mathrm{j}} \text{ kg}\cdot\text{m}^2/\text{s} \]
04

Identify the relevant formula for torque

Torque \( \vec{\tau} \) acting on an object is given by the cross product of the position vector \( \vec{r} \) and the force \( \vec{F} \): \[ \vec{\tau} = \vec{r} \times \vec{F} \]
05

Calculate the torque

Substitute the position vector \( \vec{r} = (2.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{k}}) \text{ m} \) and force vector \( \vec{F} = 4.0 \hat{\mathrm{j}} \text{ N} \) into the cross product formula: \[ \vec{\tau} = (2.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{k}}) \times 4.0 \hat{\mathrm{j}} \] Evaluate the cross product to find: \[ \vec{\tau} = [(-2.0)(4.0) \hat{\mathrm{i}} + 0.0 \hat{\mathrm{j}} + (2.0)(4.0)] \hat{\mathrm{k}} = -8.0 \hat{\mathrm{i}} + 8.0 \hat{\mathrm{k}} \text{ N}\cdot\text{m} \]

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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 is a vector quantity, meaning it has both magnitude and direction. You can think of torque as the force"twisting" an object around a pivot point. The SI unit for torque is the Newton-meter (N路m). Below are key points to help you understand torque better:

  • Torque ( \(\tau\) ) is calculated as the cross product of the position vector (\(\vec{r}\)) and the force vector (\(\vec{F}\)). This formula is expressed as: \[\vec{\tau} = \vec{r} \times \vec{F}\]

  • The direction of the torque vector is perpendicular to the plane formed by the position and force vectors. You can determine the direction using the right-hand rule.

  • Torque causes an angular acceleration, which is the rate at which the angular velocity of an object changes over time.

Understanding torque is crucial for solving rotational motion problems, as it helps us predict how objects will move when forces are applied.
Cross Product
The cross product is a mathematical operation used to find a vector perpendicular to two original vectors. In physics, it is used to calculate quantities like angular momentum and torque. It's particularly important when dealing with three-dimensional vectors.

  • The cross product of two vectors \(\vec{A}\) and \(\vec{B}\) is denoted by \(\vec{A} \times \vec{B}\), which results in a new vector.

  • The magnitude of the cross product is calculated as \(\|\vec{A} \times \vec{B}\| = \|\vec{A}\| \|\vec{B}\| \sin(\theta)\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

  • The resulting vector is perpendicular to both \(\vec{A}\) and \(\vec{B}\).

  • The direction of the cross product can be found using the right-hand rule: Point your index finger in the direction of the first vector, and your middle finger in the direction of the second vector. Your thumb will then point in the direction of the resulting vector.

The cross product helps determine the direction and magnitude of rotational quantities in physics problems.
Linear Momentum
Linear momentum is a measure of an object's motion and is a fundamental concept in physics. It provides insight into how an object's velocity and mass influence its behavior.

  • Linear momentum \(\vec{p}\) is calculated using the formula \(\vec{p} = m\vec{v}\), where \(m\) is the mass and \(\vec{v}\) is the velocity of the object.

  • This vector quantity has the same direction as the velocity vector.

  • The principle of conservation of momentum states that in a closed system, the total linear momentum remains constant if no external forces act on it.

  • Linear momentum plays an essential role in collision and impulse-related problems, helping to analyze and predict outcomes of interactions between objects.

Understanding linear momentum allows us to solve problems involving both linear and rotational motion.
Physics Problem Solving
Physics problem solving is a skill that involves applying physical laws and concepts to find solutions to various scientific problems. It can seem daunting, but with practice and strategy, it becomes much easier.

  • Begin by understanding the problem thoroughly, identifying what is known and what needs to be determined.

  • Select relevant physics principles that apply to the problem, such as Newton's laws of motion or the conservation of energy.

  • Use appropriate formulas and mathematical tools, like vectors, cross products, and algebra, to set up the problem.

  • Carry out calculations carefully, keeping track of units, directions, and ensuring all values are recorded accurately.

  • After finding a solution, review your work to ensure it makes sense in the context of the problem, and double-check any calculations.

Becoming proficient in physics problem solving takes practice, so keep working through examples and verify with reliable sources.

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Most popular questions from this chapter

The rotor of an electric motor has rotational inertia \(I_{m}=\) \(2.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia \(I_{p}=12 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about this axis. Calculate the number of revolutions of the rotor required to turn the probe through \(30^{\circ}\) about its central axis.

A thin-walled pipe rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length?

A wheel rotates clockwise about its central axis with an angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). At time \(t=0\), a torque of magnitude \(50 \mathrm{~N} \cdot \mathrm{m}\) is applied to the wheel to reverse the rotation. At what time \(t\) is the angular speed zero?

A particle is acted on by two torques about the origin: \(\vec{\tau}_{1}\) has a magnitude of \(2.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the positive direction of the \(x\) axis, and \(\vec{\tau}_{2}\) has a magnitude of \(4.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the negative direction of the \(y\) axis. In unit-vector notation, find \(d \vec{\ell} / d t\), where \(\vec{\ell}\) is the angular momentum of the particle about the origin.

In a playground, there is a small merry-go-round of radius \(1.20 \mathrm{~m}\) and mass \(180 \mathrm{~kg} .\) Its radius of gyration (see Problem 79 of Chapter 10 ) is \(91.0 \mathrm{~cm}\). A child of mass \(44.0 \mathrm{~kg}\) runs at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate (a) the rotational inertia of the merry-go-round about its axis of rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c) the angular speed of the merry-go- round and child after the child has jumped onto the merry-go-round.
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