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Chapter 3: Problem 46

Birds of prey typically rise upward on thermals. The paths these birds take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume that a bird completes a circle of radius 6.00 m every 5.00 s and rises vertically at a constant rate of 3.00 m/s. Determine (a) the bird's speed 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
(a) Speed: 8.15 m/s; (b) Acceleration: 9.48 m/s虏, centripetal; (c) Angle: 21.8掳.

Step by step solution

01

Determine the Circular Speed

First, calculate the speed of the bird as it moves in a circular path. The speed of an object in uniform circular motion is given by \( v_c = \frac{2\pi r}{T} \), where \( r \) is the radius and \( T \) is the period of motion. For this problem, \( r = 6.00 \) m and \( T = 5.00 \) s. Thus, \( v_c = \frac{2\pi \times 6.00}{5.00} \approx 7.54 \) m/s.
02

Calculate the Total Speed Relative to the Ground

The bird's overall speed relative to the ground includes both its upward velocity and its circular speed. Given the upward velocity \( v_u = 3.00 \) m/s, use the Pythagorean theorem to find the total speed: \( v = \sqrt{v_c^2 + v_u^2} = \sqrt{(7.54)^2 + (3.00)^2} \approx 8.15 \) m/s.
03

Determine the Bird's Acceleration

For uniform circular motion, the centripetal acceleration, which points towards the center of the circle, is calculated by \( a_c = \frac{v_c^2}{r} \). Substituting the values, we get \( a_c = \frac{(7.54)^2}{6.00} \approx 9.48 \) m/s虏.
04

Calculate the Angle with the Horizontal

The angle \( \theta \) between the velocity vector and the horizontal can be found using \( \tan \theta = \frac{v_u}{v_c} \). Thus, \( \theta = \tan^{-1} \left( \frac{3.00}{7.54} \right) \approx 21.8^\circ \).

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Key Concepts

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

Spiral Motion
Birds of prey often engage in spiral motion as they ascend in the sky using thermals, or warm air currents. This allows them to rise with minimal energy expenditure. The spiral motion can be visualized as a combination of circular motion along a horizontal path, with a constant upward movement.

The key attributes of spiral motion include:
  • **Circular Component**: The bird moves in a circle of radius 6.00 meters, completing one revolution every 5.00 seconds. This is a perfect example of uniform circular motion, where the speed around the path remains constant.
  • **Vertical Ascent**: The bird additionally moves upward at a steady rate of 3.00 meters per second, which adds a vertical component to its overall motion.
  • **Resulting Motion**: When combined, these motions create a spiral path that rises upward, characterized by both rotational and linear movement simultaneously.
Understanding spiral motion helps us get a clear picture of how complex movements can be reduced to simpler, simultaneous motions. This is particularly useful in physics for dissecting compounded vectors into comprehensible parts.
Centripetal Acceleration
Centripetal acceleration is a crucial concept when an object moves in a circular path. It is defined as the acceleration directed towards the center of the circle, necessary for maintaining the object on its curved trajectory.

In the case of our bird, the centripetal acceleration keeps it firm on its circular path, despite its intention to rise vertically.
  • **Calculation**: Centripetal acceleration is calculated using the formula: \( a_c = \frac{v_c^2}{r} \), where \( v_c \) is the circular speed and \( r \) is the radius of the circle. For our bird example, this results in an acceleration of approximately 9.48 m/s虏 towards the center of its circular path.
  • **Physical Meaning**: Without this inward acceleration, the bird would fly off in a straight line due to inertia. It acts like a "centripetal force" that pulls the bird towards the center, maintaining its circular path.
  • **Distinction from Other Accelerations**: Unlike other forms of acceleration that might change speed, centripetal acceleration alters the direction of velocity but not its magnitude. This distinction helps in understanding why the bird's speed along its path remains constant, while its direction can continuously change.
Realizing the role of centripetal acceleration in circular motion gives us insights into how forces are balanced to maintain paths and stability in dynamic systems.
Velocity Vector Angle
When an object moves in a plane, the velocity vector describes not only the speed but also the direction of motion. The angle of this velocity vector relative to a reference line, such as the horizontal, provides insights into the overall direction of movement.

For our bird in spiral motion:
  • **Components of Velocity**: The velocity vector has both horizontal (circular movement) and vertical (upward movement) components, which need to be considered to determine the overall angle.
  • **Calculating the Angle**: This angle \( \theta \) can be computed using trigonometry, specifically: \( \tan \theta = \frac{v_u}{v_c} \). With the bird, the calculation gives an angle of approximately 21.8掳 from the horizontal.
  • **Real-world Interpretation**: The angle indicates how steeply the bird is rising as it circles. A smaller angle would suggest a flatter, more lateral path, while a larger angle indicates a steeper ascent.
Understanding the concept of the velocity vector angle aids in visualizing how direction changes contribute to overall motion, and is vital in fields ranging from physics to engineering.

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Most popular questions from this chapter

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of your rich uncle. The top of the fence is 5.00 m above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of 1.60 m above the ground and at an angle of 56.0\(^\circ\) above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance beyond the fence will the rock land on the ground?

On level ground a shell is fired with an initial velocity of 40.0 m/s at 60.0\(^\circ\) above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell's initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.

A baseball thrown at an angle of 60.0\(^{\circ}\) above the horizontal strikes a building 18.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.

At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravity") on test pilots and astronauts. In this device, an arm 8.84 m long rotates about one end in a horizontal plane, and an astronaut is strapped in at the other end. Suppose that he is aligned along the centrifuge's arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this device is typically 12.5\(g\). (a) How fast must the astronaut's head be moving to experience this maximum acceleration? (b) What is the \(difference\) between the acceleration of his head and feet if the astronaut is 2.00 m tall? (c) How fast in rpm (rev/min) is the arm turning to produce the maximum sustained acceleration?

A student sits atop a platform a distance \(h\) above the ground. He throws a large firecracker horizontally with a speed \(v\). However, a wind blowing parallel to the ground gives the firecracker a constant horizontal acceleration with magnitude \(a\). As a result, the firecracker reaches the ground directly below the student. Determine the height \(h\) in terms of \(v\), \(a\), and \(g\). Ignore the effect of air resistance on the vertical motion.
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