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Chapter 3: Problem 54

It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius \(8.00 \mathrm{~m}\) every \(5.00 \mathrm{~s}\) and rises vertically at a rate of \(3.00 \mathrm{~m} / \mathrm{s}\). Determine: (a) the speed of the bird relative to the ground; (b) the bird's acceleration (magnitude and direction); and (c) the angle between the bird's velocity vector and the horizontal.

Short Answer

Expert verified
Speed: 10.49 m/s; Acceleration: 12.61 m/s虏; Angle: 16.70掳

Step by step solution

01

Calculate the Circumference and Horizontal Speed

The bird is moving in a circle with a radius of \(8.00 \, \text{m}\). First, we calculate the circumference of the circle:\[C = 2\pi r = 2\pi (8.00) \approx 50.27 \, \text{m}\]The bird completes this circle in \(5.00 \, \text{s}\), so the horizontal speed (or tangential speed) \(v_H\) is:\[v_H = \frac{C}{T} = \frac{50.27}{5.00} = 10.05 \, \text{m/s}\]
02

Determine Total Speed of the Bird

The total speed of the bird relative to the ground is a combination of its horizontal speed and vertical speed. Given that the vertical speed \(v_V\) is \(3.00 \, \text{m/s}\), we use the Pythagorean theorem to find the total speed \(v\):\[v = \sqrt{v_H^2 + v_V^2} = \sqrt{(10.05)^2 + (3.00)^2} \approx 10.49 \, \text{m/s}\]
03

Calculate the Bird's Acceleration

For uniform circular motion, the bird experiences centripetal acceleration directed towards the center of its circular path. The formula for centripetal acceleration \(a_c\) is:\[a_c = \frac{v_H^2}{r} = \frac{(10.05)^2}{8.00} \approx 12.61 \, \text{m/s}^2\]There is no additional vertical acceleration since the vertical speed is constant.
04

Compute the Angle of the Velocity Vector

The angle \( \theta \) between the bird's velocity vector and the horizontal can be found using the tangent function:\[\tan \theta = \frac{v_V}{v_H} = \frac{3.00}{10.05}\]Solving for \( \theta \), we get:\[\theta = \tan^{-1} \left( \frac{3.00}{10.05} \right) \approx 16.70^\circ\]
05

Summary of Results

- Speed of the bird relative to the ground: \(10.49 \, \text{m/s}\)- Acceleration (magnitude and direction): \(12.61 \, \text{m/s}^2\) towards the center of the circle- Angle between the velocity vector and the horizontal: \(16.70^\circ\)

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Key Concepts

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

Understanding Uniform Circular Motion
When an object, like a bird, moves in a circle at a constant speed, it is said to be in uniform circular motion. This means the object travels equal distances along the circle's circumference within equal time intervals.
In the case of our bird, it moves in a spiral path that can be broken down into circular motion in the horizontal plane, with a constant vertical motion.
  • The radius of the circular path here is given as 8.00 m.
  • The time taken to complete one full circle is 5.00 seconds.
  • The bird's horizontal speed (tangential speed) is calculated by dividing the circumference of the circle by the time taken to complete the circle.
This type of motion is important because, even though the speed is constant, the bird is continuously changing direction, which implies a continuous change in velocity.
Centripetal Acceleration in Circular Motion
In uniform circular motion, there is a continuous change in the direction of the velocity vector.
This change is due to a specific type of acceleration known as centripetal acceleration, which points towards the center of the circle.
  • The relevant formula to calculate centripetal acceleration is \[a_c = \frac{v_H^2}{r} \]
  • The horizontal component of the bird's velocity (tangential speed) is necessary for this calculation.
  • In our bird's scenario, the tangential speed is \(10.05 \, \text{m/s}\) and the radius is 8.00 m.
This tells us how much the bird's direction changes each second, which in this case is \(12.61 \, \text{m/s}^2\).
Centripetal acceleration is essential for maintaining the bird in its circular path as it rises.
Deciphering Vector Analysis
In physics, vectors are crucial for understanding motion because they provide both a magnitude (size) and a direction.
For our flying bird, several vectors are involved:
  • The bird's velocity vector combines both horizontal and vertical components.
  • The horizontal speed (tangential speed) is directed along the circle's tangent.
  • The vertical speed is upward, perpendicular to the circular path.
By using vector analysis, we determine the bird's overall speed or resultant velocity vector.
This is achieved by combining the vector components using the Pythagorean theorem.
These vectors also help calculate the angle of elevation, that is, the angle between the resultant velocity and the horizontal plane.
Tangential Speed Calculation
Calculating tangential speed is essential in understanding circular motion.
It refers to the speed at which the bird travels along the circumference of the circle.
  • First, we find the circumference of the circle using \[C = 2\pi r\], which gives \(50.27 \, \text{m}\).
  • The tangential speed \(v_H\) is found by dividing this circumference by the time it takes to complete the circle, \[v_H = \frac{C}{T} = \frac{50.27}{5.00} = 10.05 \, \text{m/s}\].
Knowing the tangential speed is crucial because it provides insight into how quickly the bird moves through its horizontal circular path.
This speed, when coupled with the vertical speed, gives the comprehensive motion picture of spiral motion.

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Most popular questions from this chapter

An airplane is flying with a velocity of \(90.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(23.0^{\circ}\) above the horizontal. When the plane is \(114 \mathrm{~m}\) directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? You can ignore air resistance.

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of \(58.0^{\circ}\) above the horizontal, some of the tiny critters have reached a maximum height of \(58.7 \mathrm{~cm}\) above the level ground. (See Nature, Vol. 424,31 July \(2003,\) p. \(509 .\) ) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

A pool ball is rolling along a table with a constant velocity. The components of its velocity vector are \(v_{x}=0.5 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=0.8 \mathrm{~m} / \mathrm{s} .\) Calculate the distance it travels in \(0.4 \mathrm{~s}\).

A cart carrying a vertical missile launcher moves horizontally at a constant velocity of \(30.0 \mathrm{~m} / \mathrm{s}\) to the right (Figure 3.39 ). It launches a rocket vertically upward. The missile has an initial vertical velocity of \(40.0 \mathrm{~m} / \mathrm{s}\) relative to the cart. (a) How high does the rocket go? (b) How far does the cart travel while the rocket is in the air? (c) Where does the rocket land relative to the cart?

A particle starts from rest at the origin with an acceleration vector that has magnitude \(4 \mathrm{~m} / \mathrm{s}^{2}\) and direction \(30^{\circ}\) above the positive \(x\) axis. (a) What are the components of its velocity vector \(20 \mathrm{~s}\) later? (b) What is the particle's position at that time?
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