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At time \(t=0\), a \(2.0 \mathrm{~kg}\) particle has the position vector \(\vec{r}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}}\) relative to the origin. Its velocity is given by \(\vec{v}=\left(-6.0 t^{2} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}\) for \(t \geq 0\) in seconds. About the origin, what are (a) the particle's angular momentum \(\vec{L}\) and \((\mathrm{b})\) the torque \(\vec{\tau}\) acting on the particle, both in unit-vector notation and for \(t>0 ?\) About the point \((-2.0 \mathrm{~m},-3.0 \mathrm{~m}, 0)\), what are (c) \(\vec{L}\) and (d) \(\vec{\tau}\) for \(t>0 ?\)

Short Answer

Expert verified
(a) \( \vec{L} = 24.0t^2 \hat{\mathrm{k}} \text{ kg m}^2/\text{s}\) (b) \( \vec{\tau} = 0 \hat{\mathrm{k}} \text{ N·m}\) (c) \( \vec{L'} = -12.0t^2 \hat{\mathrm{k}} \text{ kg m}^2/\text{s}\) (d) \( \vec{\tau'} = 0 \hat{\mathrm{k}} \text{ N·m}\).

Step by step solution

01

Calculate Angular Momentum About the Origin

To determine the angular momentum \( \vec{L} \) about the origin, we use the formula \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{p} = m \vec{v} \) is the linear momentum. At time \( t \), the velocity is \( \vec{v} = (-6.0t^2) \hat{\mathrm{i}} \). Thus, the linear momentum \( \vec{p} = (2.0 \text{ kg})(-6.0t^2) \hat{\mathrm{i}} = -12.0t^2 \hat{\mathrm{i}} \text{ kg m/s} \).The position vector is constant: \( \vec{r}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}} \). The cross product is:\[\vec{L} = (4.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{j}}) \times (-12.0t^2 \hat{\mathrm{i}}) = (4.0 \times 0 + (-2.0) \times (-12.0t^2)) \hat{\mathrm{k}} = 24.0t^2 \hat{\mathrm{k}} \text{ kg m}^2/\text{s}\].
02

Calculate Torque About the Origin

Torque \( \vec{\tau} \) about the origin is given by \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{F} = m\frac{d\vec{v}}{dt} \). Taking the derivative of \( \vec{v} = (-6.0t^2) \hat{\mathrm{i}} \), we get : \( \frac{d\vec{v}}{dt} = (-12.0t) \hat{\mathrm{i}} \). Thus, \( \vec{F} = (2.0)(-12.0t) \hat{\mathrm{i}} = -24.0t \hat{\mathrm{i}} \text{ N} \).The position vector is \( \vec{r} = 4.0 \hat{\mathrm{i}} -2.0 \hat{\mathrm{j}} \).Thus, \( \vec{\tau} = (4.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{j}}) \times (-24.0t \hat{\mathrm{i}}) = 0 \hat{\mathrm{k}} \text{ N·m}\).
03

Calculate Angular Momentum About Point (-2.0 m, -3.0 m)

Now, consider the new position vector \( \vec{r'} \) relative to the point \((-2.0 \mathrm{~m}, -3.0 \mathrm{~m})\). The vector is:\[\vec{r'} = (4.0 - (-2.0)) \hat{\mathrm{i}} + (-2.0 - (-3.0)) \hat{\mathrm{j}} = 6.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}} \text{ m}\]. Recall \( \vec{p} = -12.0t^2 \hat{\mathrm{i}} \text{ kg m/s}\).The cross product \( \vec{L'} \) is:\[\vec{L'} = (6.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}}) \times (-12.0t^2 \hat{\mathrm{i}}) = (6.0 \times 0 + 1.0 \times (-12.0t^2)) \hat{\mathrm{k}} = -12.0t^2 \hat{\mathrm{k}} \text{ kg m}^2/\text{s}.\]
04

Calculate Torque About Point (-2.0 m, -3.0 m)

Using the new position vector \( \vec{r'} = 6.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}} \) and force \( \vec{F} = -24.0t \hat{\mathrm{i}} \text{ N}\), we can now find torque:\[\vec{\tau'} = (6.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}}) \times (-24.0t \hat{\mathrm{i}}) = (6.0 \times 0 + 1.0 \times 0) \hat{\mathrm{k}} = 0 \hat{\mathrm{k}} \text{ N·m}.\]

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Key Concepts

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

Torque
Torque is a concept used to describe how force affects the rotation of an object. It is calculated as the cross product of the position vector and the force vector acting on the object. One of the fascinating aspects of torque is that it takes both the magnitude of the force and the distance from the pivot point into account, making it essential in determining how effectively a force can cause an object to rotate.
To formulaically express torque:
  • We use: \( \vec{\tau} = \vec{r} \times \vec{F} \)
  • Where \( \vec{r} \) is the position vector and \( \vec{F} \) is the force vector.
To find the torque about a specific point, one must not only know the magnitude and direction of these vectors but also how they interact through the cross product.
In the problem described, the torque about the origin for a moving particle with position vector \( \vec{r} = 4.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{j}} \) and force \( \vec{F} = -24.0t \hat{\mathrm{i}} \) was calculated as zero, \( 0 \hat{\mathrm{k}} \), indicating that the force applied won't cause any rotational effect at the origin.
Cross Product
The cross product is a mathematical operation used extensively in physics to find a vector that is perpendicular to two given vectors. This is crucial in applications such as determining angular momentum and torque, where direction matters just as much as magnitude.
To calculate the cross product of two vectors \( \vec{a} \) and \( \vec{b} \), we follow the right-hand rule and use the formula:
  • \( \vec{a} \times \vec{b} = (a_y b_z - a_z b_y)\hat{\mathrm{i}} + (a_z b_x - a_x b_z)\hat{\mathrm{j}} + (a_x b_y - a_y b_x)\hat{\mathrm{k}} \)
  • It's important to note that the cross product is anti-commutative, meaning that \( \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} \).
In the angular momentum calculation for the problem, the cross product of the position vector \( \vec{r} = 4.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{j}} \) and the momentum \( \vec{p} = -12t^2 \hat{\mathrm{i}} \) resulted in a vector that points in the \( \hat{\mathrm{k}} \) direction, indicating the direction of the resultant angular momentum relative to the origin.
Unit Vector Notation
Unit vector notation is a crucial concept in physics, representing vectors in terms of their components along each axis. Using unit vectors, we can express complex vectors in a digestible format. Each unit vector is a vector with a magnitude of one and points along one of the principal axes of a coordinate system.
  • In a 3-dimensional space, we primarily use \( \hat{\mathrm{i}} \), \( \hat{\mathrm{j}} \), and \( \hat{\mathrm{k}} \).
  • These represent the x, y, and z-axis directions respectively.
For example, a position vector \( \vec{r} = 4.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{j}} \) clearly specifies that the object is located 4 meters along the x-axis and -2 meters along the y-axis from the origin.
This notation not only aids in visualizing vectors but also simplifies calculations like those in cross products and torque evaluations. By decomposing vectors into their unit components, complex problems become more manageable and intuitive.

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Most popular questions from this chapter

Shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800\) \(\mathrm{m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2} .\) By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1} ?\)

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is \(6.00 \mathrm{~m}\) long, weighs \(10.0 \mathrm{~N}\), and rotates at 240 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.

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.

\- 52 (co A cockroach of mass \(m\) lies on the rim of a uniform disk of mass \(4.00 m\) that can rotate freely about its center like a merrygo-round. Initially the cockroach and disk rotate together with an angular velocity of \(0.260 \mathrm{rad} / \mathrm{s}\). Then the cockroach walks halfway to the center of the disk. (a) What then is the angular velocity of the cockroach- disk system? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the system to its initial kinetic energy? (c) What accounts for the change in the ki- \(\quad\) netic energy?

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

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