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Chapter 9: Problem 26

We require the fact that \(d \mathbf{r} / d t=\mathbf{v} .\) Then $$\frac{d \mathbf{L}}{d t}=\frac{d}{d t}(\mathbf{r} \times \mathbf{p})=\mathbf{r} \times \frac{d \mathbf{p}}{d t}+\frac{d \mathbf{r}}{d t} \times \mathbf{p}=\tau+\mathbf{v} \times \mathbf{p}=\tau+\mathbf{v} \times m \mathbf{v}=\tau+m(\mathbf{v} \times \mathbf{v})=\tau+\mathbf{0}=\tau.$$

Short Answer

Expert verified
The rate of change of angular momentum is equal to the torque: \( \frac{d \mathbf{L}}{d t} = \tau \).

Step by step solution

01

Understand the given equation

The exercise presents a derivation involving the rate of change of angular momentum \( \frac{d \mathbf{L}}{d t} \). We need to show that \( \mathbf{v} \times \mathbf{v} = \mathbf{0} \) and thus simplify the expression to \( \tau \).
02

Differentiate the cross product

Firstly, differentiate the cross product \( \mathbf{r} \times \mathbf{p} \), where \( \mathbf{p} \) is the linear momentum, \( \mathbf{p} = m\mathbf{v} \). Using the product rule for cross products, we apply: \[ \frac{d}{d t} (\mathbf{r} \times \mathbf{p}) = \mathbf{r} \times \frac{d \mathbf{p}}{d t} + \frac{d \mathbf{r}}{d t} \times \mathbf{p}. \]
03

Substitute derivatives

Substitute \( \frac{d \mathbf{r}}{d t} = \mathbf{v} \) and rewrite \( \mathbf{p} = m\mathbf{v} \). This gives us: \[ \mathbf{r} \times \frac{d \mathbf{p}}{d t} + \mathbf{v} \times m\mathbf{v} = \tau + \mathbf{v} \times m\mathbf{v}. \]
04

Simplify using vector properties

The cross product of a vector with itself is zero, \( \mathbf{v} \times \mathbf{v} = \mathbf{0} \). Therefore: \[ \tau + m(\mathbf{v} \times \mathbf{v}) = \tau + \mathbf{0} = \tau. \]
05

Write final solution

The final simplification confirms that the rate of change of angular momentum simplifies to just the torque, \( \frac{d \mathbf{L}}{d t} = \tau \).

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Key Concepts

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

Cross Product
The cross product is a fundamental concept in vector mathematics, specifically used in three-dimensional space. It produces a vector that is perpendicular to the plane formed by the two input vectors. The result of the cross product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) is denoted as \( \mathbf{A} \times \mathbf{B} \).
  • The magnitude of the resulting vector equals the area of the parallelogram that the vectors span.
  • This magnitude is calculated as \( |\mathbf{A}||\mathbf{B}|\sin(\theta) \), where \( \theta \) is the angle between the two vectors.
In the context of the exercise, the cross product is vital for understanding how angular momentum behaves. When differentiating \( \mathbf{r} \times \mathbf{p} \), or position vector cross linear momentum vector, we apply the product rule for cross products. Each term in this product has a geometric interpretation, and the cross product helps describe rotational dynamics alongside linear ones. Utilizing the cross product ensures that when operations involve velocity, like \( \mathbf{v} \times \mathbf{v} \), we reach results such as the nullification of like directional vectors, yielding zero.
Torque
Torque is the rotational equivalent of force in physics. It signifies the tendency of a force to cause an object to rotate around an axis. Represented as \( \tau \), torque plays a crucial role in rotational mechanics.
  • Calculated as the cross product of the lever arm vector (\( \mathbf{r} \)) and force applied (\( \mathbf{F} \)), given by \( \tau = \mathbf{r} \times \mathbf{F} \).
  • The value of torque determines how effectively a force causes rotational motion.
In this exercise, torque is directly related to angular momentum (\( \mathbf{L} \)). When we take the time derivative of angular momentum, the torque emerges as the primary component of the equation, illustrating the dependency of rotational change on the exerted torque. This connection helps simplify complex rotational effects to more manageable terms and understand how forces affect objects in motion.
Linear Momentum
Linear momentum, symbolized by \( \mathbf{p} \), is a vector quantity representing the motion of an object. It is defined as the product of the mass \( m \) and velocity \( \mathbf{v} \) of an object.
  • The formula for linear momentum is given by \( \mathbf{p} = m \mathbf{v} \).
  • Linear momentum is conserved in closed systems, meaning it remains constant when no external forces are applied.
Understanding linear momentum is important for analyzing how objects move and interact. In the context of this exercise, linear momentum is used to break down rotational motion through the relationship \( \mathbf{r} \times \mathbf{p} \). As part of the equation for angular momentum, linear momentum's interaction with the position vector via the cross product provides insights into rotational effects. Such analysis helps demonstrate how linear and angular perspectives interrelate in physical phenomena.

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Most popular questions from this chapter

Note that \(z=1+x^{2}+y^{2}\) is always positive. Then \(V=\int_{0}^{1} \int_{0}^{3-3 x}\left(1+x^{2}+y^{2}\right) d y d x=\left.\int_{0}^{1}\left(y+x^{2} y+\frac{1}{3} y^{3}\right)\right|_{0} ^{3-3 x} d x\) \(=\int_{0}^{1}\left[(3-3 x)+x^{2}(3-3 x)+9(1-x)^{3}\right] d x=\int_{0}^{1}\left(12-30 x+30 x^{2}-12 x^{3}\right) d x\) \(=\left.\left(12 x-15 x^{2}+10 x^{3}-3 x^{4}\right)\right|_{0} ^{1}=4\).

From Problem \(40, m=k \pi\) and \(d s=d t\) $$\begin{array}{l} M_{x}=\int_{C} y \rho d s=\int_{C} k x y d s=k \int_{0}^{\pi}(1+\cos t) \sin t d t=\left.k\left(-\cos t+\frac{1}{2} \sin ^{2} t\right)\right|_{0} ^{\pi}=2 k \\\ M_{y}=\int_{C} x \rho d s=\int_{C} k x^{2} d s=k \int_{0}^{\pi}(1+\cos t)^{2} d t=k \int_{0}^{\pi}\left(1+2 \cos t+\cos ^{2} t\right) d t \end{array}$$ $$=\left.k\left(t+2 \sin t+\frac{1}{2} t+\frac{1}{4} \sin 2 t\right)\right|_{0} ^{\pi}=\frac{3}{2} k \pi$$ \(\bar{x}=M_{y} / m=\frac{3 k \pi / 2}{k \pi}=\frac{3}{2} ; \quad \bar{y}=M_{x} / m=\frac{2 k}{k \pi}=\frac{2}{\pi} .\) The center of mass is \((3 / 2,2 / \pi)\)

The other five integrals are \(\int_{0}^{3} \int_{0}^{\sqrt{36-4 y^{2}} / 3} \int_{1}^{3} F(x, y, z) d z d x d y\), $$\begin{aligned} &\int_{1}^{3} \int_{0}^{2} \int_{0}^{\sqrt{36-9 x^{2} / 2}} F(x, y, z) d y d x d z, \int_{1}^{3} \int_{0}^{3} \int_{0}^{\sqrt{36-4 y^{2}} / 3} F(x, y, z) d x d y d z,\\\ &\int_{0}^{3} \int_{1}^{3} \int_{0}^{\sqrt{36-4 y^{2}}} F(x, y, z) d x d z d y, \int_{0}^{2} \int_{1}^{3} \int_{0}^{\sqrt{36-9 x^{2}} / 2} F(x, y, z) d y d z d x. \end{aligned}$$

We let \(u=y-x\) and \(v=y+x\) \(R 1: y=0 \Longrightarrow u=-x, v=x \Longrightarrow v=-u\) \(R 2: x+y=1 \Longrightarrow v=1\) \(R 3: x=0 \Longrightarrow u=y, v=y \Longrightarrow v=u\) \(\frac{\partial(u, v)}{\partial(x, y)}=\left|\begin{array}{cc}-1 & 1 \\ 1 & 1\end{array}\right|=-2 \Rightarrow \frac{\partial(x, y)}{\partial(u, v)}=-\frac{1}{2}\) $$\begin{aligned} \iint_{R} e^{(y-x) /(y+x)} d A &=\iint_{S} e^{u / v}\left|-\frac{1}{2}\right| d A^{\prime}=\frac{1}{2} \int_{0}^{1} \int_{-v}^{v} e^{u / v} d u d v=\left.\frac{1}{2} \int_{0}^{1} v e^{u / v}\right|_{-v} ^{v} d v \\ &=\frac{1}{2} \int_{0}^{1} v\left(e-e^{-1}\right) d v=\left.\frac{1}{2}\left(e-e^{-1}\right) \frac{1}{2} v^{2}\right|_{0} ^{1}=\frac{1}{4}\left(e-e^{-1}\right) \end{aligned}$$

\(\int_{0}^{4} \int_{\sqrt{y}}^{2} \sqrt{x^{3}+1} d x d y=\int_{0}^{2} \int_{0}^{x^{2}} \sqrt{x^{3}+1} d y d x=\left.\int_{0}^{2} y \sqrt{x^{3}+1}\right|_{0} ^{x^{2}} d x\) \(=\int_{0}^{2} x^{2} \sqrt{x^{3}+1} d x=\left.\frac{2}{9}\left(x^{3}+1\right)^{3 / 2}\right|_{0} ^{2}=\frac{2}{9}\left(9^{3 / 2}-1^{3 / 2}\right)=\frac{52}{9}\)
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