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Chapter 12: Problem 81

Suppose an object moves on the surface of a sphere with \(|\mathbf{r}(t)|\) constant for all \(t\) Show that \(\mathbf{r}(t)\) and \(\mathbf{a}(t)=\mathbf{r}^{\prime \prime}(t)\) satisfy \(\mathbf{r}(t) \cdot \mathbf{a}(t)=-|\mathbf{v}(t)|^{2}\)

Short Answer

Expert verified
Question: Prove the relationship between the position, velocity, and acceleration vectors for an object moving on the surface of a sphere with a constant magnitude of the position vector, \(\mathbf{r}(t) \cdot \mathbf{a}(t)=-|\mathbf{v}(t)|^{2}\). Answer: Following the steps of defining the given information, calculating the magnitude of the position vector, differentiating the equation, involving the acceleration vector by differentiating again, and establishing a relationship, we proved that \(\mathbf{r}(t) \cdot \mathbf{a}(t)=-|\mathbf{v}(t)|^{2}\).

Step by step solution

01

Define given information

We know that the object moves on the surface of the sphere and the magnitude of the position vector, \(|\mathbf{r}(t)|\), is constant for all \(t\). We are asked to prove the relationship \(\mathbf{r}(t) \cdot \mathbf{a}(t)=-|\mathbf{v}(t)|^{2}\).
02

Calculate the magnitude of the position vector

Given that \(|\mathbf{r}(t)|\) is constant, we can represent it as a constant value \(R\). Since \(\mathbf{r}(t)\) is the position vector, the sphere's equation can be written as: $$|\mathbf{r}(t)|^2 = R^2$$
03

Differentiate the equation of the sphere

To bring the velocity vector into the equation, differentiate the equation found in Step 2 with respect to time \(t\). As \(\mathbf{r}^{\prime}(t)=\mathbf{v}(t)\), the differentiation becomes: $$2\mathbf{r}(t)\cdot\mathbf{v}(t)=0$$
04

Differentiate the equation again to involve the acceleration vector

To involve the acceleration vector, differentiate the equation obtained in Step 3 with respect to time \(t\) again. As \(\mathbf{v}^{\prime}(t)=\mathbf{a}(t)\), the differentiation becomes: $$\mathbf{v}(t)\cdot\mathbf{v}(t) + \mathbf{r}(t)\cdot\mathbf{a}(t)=0$$
05

Rearrange and establish the relationship

Rearrange the equation from Step 4 to establish the required relationship between the position, velocity, and acceleration vectors: $$\mathbf{r}(t)\cdot\mathbf{a}(t) = -\mathbf{v}(t)\cdot\mathbf{v}(t)$$ Since \(|\mathbf{v}(t)|^2 = \mathbf{v}(t) \cdot \mathbf{v}(t)\), the final relationship is: $$\mathbf{r}(t) \cdot \mathbf{a}(t)=-|\mathbf{v}(t)|^{2}$$ Thus, the relationship between the position, velocity, and acceleration vectors for an object moving on the surface of a sphere with constant magnitude of the position vector is proven as stated in the exercise.

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Key Concepts

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

Position Vector
In vector calculus, a position vector is a directed quantity that indicates the position of an object in space relative to an origin. When discussing motion on a sphere, the position vector, represented as \(\mathbf{r}(t)\), is crucial. It not only tells us where the object is located but also provides a framework for understanding how the object moves in relation to the sphere.
  • This vector is characterized by its magnitude and direction.
  • The magnitude \(|\mathbf{r}(t)|\) is often constant for an object constrained to move on a spherical surface, signifying a fixed distance from the center.
Understanding the concept of a position vector is fundamental, particularly in interpreting how an object in motion responds to forces and how its trajectory is confined to a spherical surface.
Velocity Vector
The velocity vector, represented as \(\mathbf{v}(t)\), is a measure of the rate and direction of change of an object's position. In the context of our exercise, this vector is found by differentiating the position vector \(\mathbf{r}(t)\) with respect to time \(t\).
  • It indicates how fast the object is moving and in what direction.
  • When the object moves on a sphere, the velocity vector must always be tangent to the sphere's surface.
  • This is because, on the sphere, the position vector maintains a constant magnitude \(|\mathbf{r}(t)|\).
Since the velocity vector is tangent to the sphere, it plays a vital role in showcasing the dynamics of movement on a spherical path.
Acceleration Vector
The acceleration vector \(\mathbf{a}(t)\) is concerned with how an object's velocity changes over time. This vector is derived from the derivative of the velocity vector or, equivalently, the second derivative of the position vector with respect to time \(t\).
  • It can provide insights into how the object speeds up, slows down, or changes direction.
  • In problems involving a sphere, the acceleration vector often interacts with the object's position and velocity vectors, forming essential relationships that govern motion.
In our exercise, \(\mathbf{r}(t) \cdot \mathbf{a}(t)\) being equal to \(-|\mathbf{v}(t)|^2\) illustrates such a relationship, ensuring that the vectors align in ways that respect the constraints of the spherical surface.
Differentiation
Differentiation is a fundamental operation in calculus, which involves finding the rate at which a function changes at any point. In the context of our exercise, differentiation is used to reveal the velocity and acceleration vectors by operating on the position vector.
  • The first differentiation of \(\mathbf{r}(t)\) gives the velocity vector \(\mathbf{v}(t)\).
  • Further differentiation provides the acceleration vector \(\mathbf{a}(t)\).
  • Understanding differentiation allows us to derive key relationships and dynamics of motion on the sphere.
Differentiation, hence, is pivotal in forming expressions and equations that describe how forces and motion interact over time, especially on constrained surfaces like spheres.
Sphere
A sphere is a three-dimensional shape that is perfectly symmetrical around its center, with every point on its surface at an equal distance from this center. In motion problems, the sphere introduces a constant magnitude for the position vector, simplifying calculations by confining motion to its surface.
  • Understanding the geometric constraints of the sphere helps in analyzing vector magnitudes and directions.
  • The equation of the sphere \(|\mathbf{r}(t)|^2 = R^2\) encapsulates the key property of movement on its surface - constant distance from the center.
  • This relationship directly influences how position, velocity, and acceleration vectors interact with each other, as shown in our exercise.
Motion on a sphere requires a deep grasp of these geometric principles to understand the complex interplay of vectors and how they create dynamic equations like \(\mathbf{r}(t) \cdot \mathbf{a}(t) = -|\mathbf{v}(t)|^2\).

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Most popular questions from this chapter

Suppose water flows in a thin sheet over the \(x y\) -plane with a uniform velocity given by the vector \(\mathbf{v}=\langle 1,2\rangle ;\) this means that at all points of the plane, the velocity of the water has components \(1 \mathrm{m} / \mathrm{s}\) in the \(x\) -direction and \(2 \mathrm{m} / \mathrm{s}\) in the \(y\) -direction (see figure). Let \(C\) be an imaginary unit circle (that does not interfere with the flow). a. Show that at the point \((x, y)\) on the circle \(C\) the outwardpointing unit vector normal to \(C\) is \(\mathbf{n}=\langle x, y\rangle\) b. Show that at the point \((\cos \theta, \sin \theta)\) on the circle \(C\) the outward-pointing unit vector normal to \(C\) is also $$ \mathbf{n}=\langle\cos \theta, \sin \theta\rangle $$ c. Find all points on \(C\) at which the velocity is normal to \(C\). d. Find all points on \(C\) at which the velocity is tangential to \(C\). e. At each point on \(C\) find the component of \(v\) normal to \(C\) Express the answer as a function of \((x, y)\) and as a function of \(\theta\) f. What is the net flow through the circle? That is, does water accumulate inside the circle?

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Verify that the Cauchy-Schwarz Inequality holds for \(\mathbf{u}=\langle 3,-5,6\rangle\) and \(\mathbf{v}=\langle-8,3,1\rangle\).

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An object on an inclined plane does not slide provided the component of the object's weight parallel to the plane \(\left|\mathbf{W}_{\text {par }}\right|\) is less than or equal to the magnitude of the opposing frictional force \(\left|\mathbf{F}_{\mathrm{f}}\right|\). The magnitude of the frictional force, in turn, is proportional to the component of the object's weight perpendicular to the plane \(\left|\mathbf{W}_{\text {perp }}\right|\) (see figure). The constant of proportionality is the coefficient of static friction, \(\mu\) a. Suppose a 100 -lb block rests on a plane that is tilted at an angle of \(\theta=20^{\circ}\) to the horizontal. Find \(\left|\mathbf{W}_{\text {parl }}\right|\) and \(\left|\mathbf{W}_{\text {perp }}\right|\) b. The condition for the block not sliding is \(\left|\mathbf{W}_{\mathrm{par}}\right| \leq \mu\left|\mathbf{W}_{\text {perp }}\right| .\) If \(\mu=0.65,\) does the block slide? c. What is the critical angle above which the block slides with \(\mu=0.65 ?\)

A race Two people travel from \(P(4,0)\) to \(Q(-4,0)\) along the paths given by $$ \begin{aligned} \mathbf{r}(t) &=(4 \cos (\pi t / 8), 4 \sin (\pi t / 8)\rangle \text { and } \\\ \mathbf{R}(t) &=\left(4-t,(4-t)^{2}-16\right) \end{aligned} $$ a. Graph both paths between \(P\) and \(Q\) b. Graph the speeds of both people between \(P\) and \(Q\) c. Who arrives at \(Q\) first?
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