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Chapter 4: Problem 60

A centripetal-acceleration addict rides in uniform circular motion with period \(T=2.0 \mathrm{~s}\) and radius \(r=3.50 \mathrm{~m}\). At \(t_{1}\) his acceleration is \(\vec{a}=\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At that instant, what are the values of (a) \(\vec{v} \cdot \vec{a}\) and (b) \(\vec{r} \times \vec{a}\) ?

Short Answer

Expert verified
(a) \\( \vec{v} \cdot \vec{a} = -44.0 \\text{ m}^3/\text{s}^3 \\), (b) \\( |\vec{r} \times \vec{a}| = 25.24 \\text{ m}^3/\text{s}^2 \\).

Step by step solution

01

Calculate Angular Frequency

Using the formula for angular frequency \( \omega = \frac{2 \pi}{T} \), where \( T = 2.0 \text{ s} \), we calculate \( \omega = \frac{2 \pi}{2.0} = \pi \text{ rad/s} \).
02

Find Centripetal Acceleration

The centripetal acceleration is given by \( a_c = r \omega^2 \). Substituting \( r = 3.50 \text{ m} \) and \( \omega = \pi \text{ rad/s} \), we get \( a_c = 3.50 \times (\pi)^2 \approx 34.54 \text{ m/s}^2 \).
03

Express Velocity Vector

At the time \( t_1 \), the velocity vector \( \vec{v} \) is perpendicular to \( \vec{a} \) since it is centripetal. The magnitude of \( v \) is given by \( v = r \omega = 3.50 \times \pi = 11.0 \text{ m/s} \). The direction is \( \hat{j} \), thus, \( \vec{v} = 11.0 \hat{j} \).
04

Calculate Dot Product (v 鈰� a)

To find \( \vec{v} \cdot \vec{a} \), use the components of the vectors: \( \vec{v} \cdot \vec{a} = 11.0 \hat{j} \cdot (6.00 \hat{i} - 4.00 \hat{j}) \). This simplifies to \( 0 \times 6.00 + 11.0 \times (-4.00) = -44.0 \text{ m}^3/\text{s}^3 \).
05

Calculate Cross Product (r 脳 a) Magnitude

The cross product \( \vec{r} \times \vec{a} \) involves calculating \( r \times |a| \cdot \sin(\theta) \), where \( \theta = 90^\circ \) (angle between \vec{r} and \vec{a}). Magnitude is given by \( 3.50 \times 7.211 = 25.24 \text{ m}^3/\text{s}^2 \), using \( |a| = \sqrt{(6.00)^2 + (-4.00)^2} = 7.211 \).
06

Determine Cross Product Direction

According to the right-hand rule, \( \vec{r} \times \vec{a} \) points out of the plane. Since there is no \( \hat{k} \) component in the \( x \) or \( y \) axis, we consider only the magnitude and direction as upwards through the plane of motion.

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Key Concepts

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

Angular Frequency
Angular frequency, denoted by \( \omega \), is a measure of how fast something rotates or travels along a circular path. It tells us the number of radians an object moves through per second. The formula used to determine angular frequency is \( \omega = \frac{2\pi}{T} \), where \( T \) is the period.

In the given exercise, the period \( T \) is 2.0 seconds. By substituting \( T \) into the formula, we calculate:
  • \( \omega = \frac{2\pi}{2.0} = \pi \text{ rad/s} \)
This result means that for every second, the object moves through \( \pi \) radians.

Understanding angular frequency is crucial for solving problems involving rotational motion and determining other properties like velocity and acceleration.
Centripetal Acceleration
Centripetal acceleration is a key concept in circular motion. It refers to the acceleration that keeps an object moving in a circle. The direction of this acceleration is always towards the center of the circle, hence the term "centripetal," meaning "center-seeking."

The formula to calculate centripetal acceleration is \( a_c = r \omega^2 \), where \( r \) is the radius, and \( \omega \) is the angular frequency. In the exercise, given \( r = 3.50 \text{ m} \) and \( \omega = \pi \text{ rad/s} \), the calculation is:
  • \( a_c = 3.50 \times (\pi)^2 \approx 34.54 \text{ m/s}^2 \)
This indicates that the object experiences an acceleration of around 34.54 m/s虏 towards the circle's center.

This concept explains why a car driving around a curve feels a "force" pulling it inward 鈥� it's actually the centripetal acceleration.
Dot and Cross Product Calculations
Dot and cross products are mathematical tools used to work with vectors in physics. In this exercise, they help determine the relationship between the velocity and acceleration vectors.

Dot Product


The dot product of two vectors gives a scalar. It's used to find the amount of one vector in the direction of another. For vectors \( \vec{v} = 11.0 \hat{j} \) and \( \vec{a} = 6.00 \hat{i} - 4.00 \hat{j} \), the dot product is:
  • \( \vec{v} \cdot \vec{a} = 11.0 \times (-4.00) = -44.0 \text{ m}^3/\text{s}^3 \)
This negative value indicates opposite directions.

Cross Product


The cross product of two vectors produces another vector. It measures the extent to which the two vectors are perpendicular. To find \( \vec{r} \times \vec{a} \), given \( |\vec{a}| = 7.211 \), the calculation is:
  • Magnitude: \( 3.50 \times 7.211 \cdot \sin(90^\circ) = 25.24 \text{ m}^3/\text{s}^2 \)
The direction follows the right-hand rule, pointing out of the plane.

These calculations help in understanding how vectors interact in three-dimensional space, crucial for analyzing forces and motions.

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Most popular questions from this chapter

A particle moves horizontally in uniform circular motion, over a horizontal \(x y\) plane. At one instant, it moves through the point at coordinates \((4.00 \mathrm{~m}, 4.00 \mathrm{~m})\) with a velocity of \(-5.00 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}\) and an acceleration of \(+12.5 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}^{2}\). What are the (a) \(x\) and (b) \(y\) coordinates of the center of the circular path?

A light plane attains an airspeed of \(500 \mathrm{~km} / \mathrm{h}\). The pilot sets out for a destination \(900 \mathrm{~km}\) due north but discovers that the plane must be headed \(20.0^{\circ}\) east of due north to fly there directly. The plane arrives in \(2.00 \mathrm{~h}\). What were the (a) magnitude and (b) direction of the wind velocity?

A defense air force plane, diving with constant speed at an angle of \(52.0^{\circ}\) with the vertical, drops a shell at an altitude of \(720 \mathrm{~m}\). The shell reaches the ground \(6.00 \mathrm{~s}\) after its release. (a) What is the speed of the plane? (b) How far does the shell travel horizontally during its flight? What are the (c) horizontal and (d) vertical components of its velocity just before reaching the ground? Assume an \(x\) axis in the direction of the horizontal motion and an upward \(y\) axis.

A stone is catapulted at time \(t=0\), with an initial velocity of magnitude \(18.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(40.0^{\circ}\) above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at \(t=1.10\) s? Repeat for the (c) horizontal and (d) vertical components at \(t=1.80 \mathrm{~s}\), and for the (e) horizontal and (f) vertical components at \(t=5.00 \mathrm{~s}\).

A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an \(x\) axis. He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive \(x\) component. Suppose the player runs at speed \(3.5 \mathrm{~m} / \mathrm{s}\) relative to the field while he passes the ball with velocity \(\vec{v}_{B P}\) relative to himself. If \(\vec{v}_{B P}\) has magnitude \(6.0\) \(\mathrm{m} / \mathrm{s}\), what is the smallest angle it can have for the pass to be legal?
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