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Chapter 3: Problem 224

A particle with a mass of \(4 \mathrm{kg}\) has a position vector in meters given by \(\mathbf{r}=3 t^{2} \mathbf{i}-2 t \mathbf{j}-3 \mathbf{k},\) where \(t\) is the time in seconds. For \(t=5\) s determine the angular momentum of the particle and the moment of all forces on the particle, both about the origin \(O\) of coordinates.

Short Answer

Expert verified
Angular momentum: \(-24\mathbf{i} - 360\mathbf{j} + 680\mathbf{k} \, \mathrm{kg \cdot m^2/s}\). Moment of forces: \(0\mathbf{i} - 72\mathbf{j} + 240\mathbf{k} \, \mathrm{N \cdot m}\).

Step by step solution

01

Find the Position Vector at t = 5s

To find the position vector \( \mathbf{r} \) at \( t = 5 \), substitute \( t = 5 \) into the given position vector equation \( \mathbf{r} = 3t^2 \mathbf{i} - 2t \mathbf{j} - 3\mathbf{k} \). This gives: \[ \mathbf{r} = 3(5)^2 \mathbf{i} - 2(5) \mathbf{j} - 3 \mathbf{k} = 75 \mathbf{i} - 10 \mathbf{j} - 3 \mathbf{k}. \]
02

Calculate the Velocity Vector

The velocity vector \( \mathbf{v} \) is the derivative of the position vector \( \mathbf{r} \) with respect to time \( t \). Differentiating \( \mathbf{r} = 3t^2 \mathbf{i} - 2t \mathbf{j} - 3\mathbf{k} \), we get \[ \mathbf{v} = \frac{d}{dt}(3t^2 \mathbf{i} - 2t \mathbf{j} - 3\mathbf{k}) = 6t \mathbf{i} - 2 \mathbf{j}. \]
03

Find the Velocity Vector at t = 5s

Substitute \( t = 5 \) into the velocity vector equation \( \mathbf{v} = 6t \mathbf{i} - 2 \mathbf{j} \). This gives: \[ \mathbf{v} = 6(5) \mathbf{i} - 2 \mathbf{j} = 30 \mathbf{i} - 2 \mathbf{j}. \]
04

Calculate the Angular Momentum

Angular momentum \( \mathbf{L} \) of the particle about the origin is given by \( \mathbf{L} = \mathbf{r} \times (m \cdot \mathbf{v}) \). Here, \( m = 4 \mathrm{kg} \). So, \[ \mathbf{L} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 75 & -10 & -3 \ 120 & -8 & 0 \end{vmatrix}. \] Simplifying this determinant, we find: \[ \mathbf{L} = -24\mathbf{i} - 360\mathbf{j} + 680\mathbf{k} \, \mathrm{kg \cdot m^2/s}. \]
05

Determine the Moment of Forces (Torque)

The force \( \mathbf{F} \) on the particle is the derivative of momentum, \( \mathbf{F} = \frac{d}{dt}(m \cdot \mathbf{v}) \). Since \( \mathbf{v} = 6t \mathbf{i} - 2 \mathbf{j} \), \( \mathbf{F} = m \cdot \frac{d\mathbf{v}}{dt} = 4 \cdot (6 \mathbf{i}) = 24\mathbf{i} \). To find the moment of forces (or torque) \( \tau \), use \( \tau = \mathbf{r} \times \mathbf{F} \), where \( \mathbf{r} = 75 \mathbf{i} - 10 \mathbf{j} - 3 \mathbf{k} \). Then, \[ \tau = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 75 & -10 & -3 \ 24 & 0 & 0 \end{vmatrix}. \] Simplifying this determinant, we find: \[ \tau = 0\mathbf{i} - 72\mathbf{j} + 240\mathbf{k} \, \mathrm{N \cdot m}. \]

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Key Concepts

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

Position Vector
In physics, the position vector is a fundamental concept that describes the location of a point particle in space relative to a chosen origin. For any given time, a position vector can be represented in the form of coordinate vectors:
  • The vector is expressed with components along the coordinate axes, typically denoted as \( \,\mathbf{i}, \,\mathbf{j}, \,\mathbf{k} \, \) for the x, y, and z directions, respectively.
  • Each component represents how far and in what direction the particle is from the origin along a particular axis.
The position vector in our example describes the motion of a particle over time \( \, t \). It's given by \( \,\mathbf{r} = 3t^2 \mathbf{i} - 2t \mathbf{j} - 3 \mathbf{k} \, \).
Here, substituting \( \, t = 5 \, \) seconds, we calculate:
  • \( \,75 \mathbf{i}, \, -10 \mathbf{j}, \, -3 \mathbf{k} \, \) which shows how the particle is placed 75 units in the x-direction, -10 units in the y-direction, and -3 units in the z-direction at 5 seconds.
These coordinate components allow us to describe not just the position but also any subsequent analysis on motion or force.
Velocity Calculation
Velocity is the rate at which the position of the particle changes with respect to time, and it is a vector quantity with both magnitude and direction. To find the velocity vector from a given position vector, you need to calculate the derivative with respect to time:
  • In mathematical terms, this means differentiating the components of the position vector individually.
  • The derivative of a component along a direction yields the velocity in that direction.
For our example:
  • Starting from \( \,\mathbf{r} = 3t^2 \mathbf{i} - 2t \mathbf{j} \), taking the derivative with respect to \( \, t \, \), we find \( \,\mathbf{v} = 6t \mathbf{i} - 2 \mathbf{j} \, \).
  • Substituting \( \, t = 5 \, \), the velocity vector is \( \, 30 \mathbf{i} - 2 \mathbf{j} \, \), denoting 30 units/s in the x-direction and -2 units/s in the y-direction.
This calculation provides insight into how fast and in which direction the particle is moving at a given time.
Torque (Moment of Forces)
Torque, also known as the moment of forces, measures the tendency of a force to rotate an object about an axis. It combines principles of force and distance from an axis, and is foundational in understanding rotational dynamics.
  • In our context, we calculate torque using the cross product of the position vector \( \, \mathbf{r} \, \) and the force vector \( \, \mathbf{F} \, \).
  • The force vector itself comes from the change in momentum, derived by differentiating the velocity vector according to time.
With mass \( \, m = 4 \mathrm{kg} \, \) and a derived force \( \,\mathbf{F} = 24 \mathbf{i} \, \), our torque \( \, \tau \, \) is computed to \( \, 0 \mathbf{i} - 72 \mathbf{j} + 240 \mathbf{k} \, \mathrm{N \cdot m} \, \).
This indicates how force will influence rotational movement along the coordinate axes. The larger the torque, the stronger its tendency to produce angular displacement.
Cross Product in Physics
The cross product is a mathematical operation used in physics to determine a vector perpendicular to two given vectors. It is vital for calculating physical quantities like torque and angular momentum.
  • The result of a cross product is a vector orthogonal to the original vectors, with a magnitude that is a product of their magnitudes and the sine of the angle between them.
  • In physics, cross products aid in defining quantities that involve perpendicular relationships.
Using our example determines angular momentum and torque.
  • For angular momentum \( \, \mathbf{L} \), the cross product of the position \( \, \mathbf{r} \, \) and the momentum \(( m\mathbf{v}) \) yields \( \, -24\mathbf{i} - 360\mathbf{j} + 680\mathbf{k} \, \mathrm{kg \cdot m^2/s} \, \).
  • The operation involves computing a determinant with these vectors' components, elegantly handling multi-dimensional interactions.
Understanding cross product application is crucial for evaluating rotational effects and forces in three-dimensional space.

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Most popular questions from this chapter

Two 425,000 -lb locomotives pull fifty \(200,000-1 b\) coal hoppers. The train starts from rest and accelerates uniformly to a speed of \(40 \mathrm{mi} / \mathrm{hr}\) over a distance of \(8000 \mathrm{ft}\) on a level track. The constant rolling resistance of each car is 0.005 times its weight. Neglect all other retarding forces and assume that each locomotive contributes equally to the tractive force. Determine \((a)\) the tractive force exerted by each locomotive at \(20 \mathrm{mi} / \mathrm{hr}\) (b) the power required from each locomotive at \(20 \mathrm{mi} / \mathrm{hr},(c)\) the power required from each locomotive as the train speed approaches \(40 \mathrm{mi} / \mathrm{hr},\) and \((d)\) the power required from each locomotive if the train cruises at a steady \(40 \mathrm{mi} / \mathrm{hr}\).

Car \(B\) is initially stationary and is struck by car \(A\) which is moving with speed \(v\). The mass of car \(B\) is \(p m,\) where \(m\) is the mass of car \(A\) and \(p\) is a positive constant. If the coefficient of restitution is \(e=\) \(0.1,\) express the speeds \(v_{A}^{\prime}\) and \(v_{B}^{\prime}\) of the two cars at the end of the impact in terms of \(p\) and \(v .\) Evaluate your expressions for \(p=0.5\).

The position vector of a particle is given by \(\mathbf{r}=\) \(8 t \mathbf{i}+1.2 t^{2} \mathbf{j}-0.5\left(t^{3}-1\right) \mathbf{k},\) where \(t\) is the time in seconds from the start of the motion and where \(\mathbf{r}\) is expressed in meters. For the condition when \(t=\) 4 s, determine the power \(P\) developed by the force \(\mathbf{F}=40 \mathbf{i}-20 \mathbf{j}-36 \mathbf{k} \mathrm{N}\) which acts on the particle.

A boy of mass \(m\) is standing initially at rest relative to the moving walkway inclined at the angle \(\theta\) and moving with a constant speed \(u .\) He decides to accelerate his progress and starts to walk from point \(A\) with a steadily increasing speed and reaches point \(B\) with a speed \(v_{r}\) relative to the walkway. During his acceleration he generates a constant average force \(F\) tangent to the walkway between his shoes and the walkway surface. Write the work-energy equations for the motion between \(A\) and \(B\) for his absolute motion and his relative motion and explain the meaning of the term \(m u v_{r}\) If the boy weighs 150 lb and if \(u=2 \mathrm{ft} / \mathrm{sec}, s=\) \(30 \mathrm{ft},\) and \(\theta=10^{\circ},\) calculate the power \(P_{\mathrm{rel}}\) developed by the boy as he reaches the speed of \(2.5 \mathrm{ft} / \mathrm{sec}\) relative to the walkway.

Beginning from rest when \(\theta=20^{\circ},\) a 35 -kg child slides with negligible friction down the sliding board which is in the shape of a 2.5 -m circular arc. Determine the tangential acceleration and speed of the child, and the normal force exerted on her \((a)\) when \(\theta=30^{\circ}\) and \((b)\) when \(\theta=90^{\circ}\).
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