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Chapter 6: Problem 16

A hawk flies in a horizontal are of radius \(12.0 \mathrm{m}\) at a constant speed of \(4.00 \mathrm{m} / \mathrm{s} .\) (a) Find its centripetal acceleration. (b) It continues to fly along the same horizontal arc but increases its speed at the rate of \(1.20 \mathrm{m} / \mathrm{s}^{2}\). Find the acceleration (magnitude and direction) under these conditions.

Short Answer

Expert verified
The centripetal acceleration of the hawk is \(1.33 \, m/s^2\). When it increases its speed at a rate of \(1.20 \, m/s^2\), the total acceleration is \(1.79 \, m/s^2\) toward the inside of the circle from its tangential path at an angle of about \(42.0掳\).

Step by step solution

01

Find centripetal acceleration

Centripetal acceleration can be found using the formula \(a_c = \frac{v^2}{r}\), where \(v\) is the speed and \(r\) is the radius. Substituting the given values, we get \(a_c = \frac{(4.00 \, m/s)^2}{12.0\, m}\)
02

Calculate the answer for part a

Upon simplifying the above equation we get the centripetal acceleration as \(1.33 \, m/s^2\) for part a of the exercise.
03

Find total acceleration

In part b, in addition to the centripetal acceleration, there is also tangential acceleration due to the increasing speed. The tangential acceleration can be calculated as \(a_t = \frac{鈭唙}{鈭唗}\). Given the rate of change of speed is \(1.20 \, m/s^2\), the tangential acceleration \(a_t\) is equal to this value.
04

Calculate the magnitude of total acceleration

Since the hawk is moving in a circle, the total acceleration will be the vector sum of the centripetal and tangential accelerations. The total acceleration can be found by using the Pythagorean theorem: \(a_{total} = \sqrt{a_c^2+a_t^2}\). Substituting the given values, we get \(a_{total} = \sqrt{(1.33 \, m/s^2)^2 + (1.20 \, m/s^2)^2}\)
05

Calculate the answer for part b (magnitude)

Upon simplifying the above equation we get the total acceleration as \(1.79 \, m/s^2\). This is the magnitude of the acceleration in part b.
06

Find the direction of acceleration

The direction of acceleration can be calculated using the formula, \(tan胃 = \frac{a_t}{a_c}\). Substituting the values found above, \(tan胃 = \frac{1.20 \, m/s^2}{1.33 \, m/s^2}\)
07

Calculate the answer for part b (direction)

After performing the inverse tangent operation, we will find that the angle 胃 is about \(42.0掳\). The acceleration is therefore in the direction \(42.0掳\) toward the inside of the circle formula its tangential path.

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Key Concepts

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

Circular Motion
When a hawk or any object moves in a circular path at a constant speed, it is undergoing circular motion. This type of motion means that there's a change in direction of motion happening continually, even if the speed stays the same. Key concepts in circular motion include:
  • Radius of the circle: Like in our exercise, the radius is 12 meters, defining the path's size.

  • Speed: This is how fast the object is moving along the path. In the case of the hawk, it's moving at a constant speed of 4 meters per second.

  • Centripetal acceleration: It is the acceleration directed towards the center of the circular path. It's responsible for changing the object's direction to keep it moving in a circle.

Understanding these components helps us see that even though the speed is constant, the object is not in a state of rest because its direction is continuously changing.
Tangential Acceleration
Tangential acceleration occurs when there's a change in the speed of an object moving along a circular path. It acts along the edge of the circle, or tangent to the curve at the object's position. This type of acceleration can be seen in our exercise when the hawk increases its speed.
  • Change in speed: The hawk starts to fly faster, increasing its speed by 1.20 meters per second squared. This increase in speed introduces tangential acceleration.

  • Direction: Unlike centripetal acceleration, which points towards the circle's center, tangential acceleration points in the direction of motion along the path.

When objects experience both centripetal and tangential acceleration, their total acceleration is a combination or sum of these different accelerations.
Vector Addition
To find the total acceleration of an object moving with both centripetal and tangential acceleration, we use vector addition. Vectors are quantities that have both magnitude and direction.
  • Centripetal and tangential vector: These vectors need to be combined to get the total acceleration. They act perpendicular to each other, making vector addition straightforward using the Pythagorean theorem.

  • Total acceleration magnitude: The total magnitude is found by calculating the square root of the sum of the squares of both accelerations. In our exercise, this results in around 1.79 meters per second squared.

  • Direction: Determining the direction requires finding the angle using tangent inverse operations. For our hawk, the angle comes out to be about 42掳 toward the inside of the circular path.

This combination can help us understand not just how quickly something is accelerating, but also in what direction it is accelerating.

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Most popular questions from this chapter

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of \(20.0^{\circ}\) with the horizontal. A piece of luggage having mass \(30.0 \mathrm{kg}\) is placed on the carousel, \(7.46 \mathrm{m}\) from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of static friction between the bag and the carousel. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, \(7.94 \mathrm{m}\) from the axis of rotation. Now going around once in every \(34.0 \mathrm{s}\), the bag is on the verge of slipping. Calculate the coefficient of static friction between the bag and the carousel.

A small block is at rest on the floor at the front of a rail. road boxcar that has length \(\ell .\) The coefficient of kinetic friction between the floor of the car and the block is \(\mu_{k}\) The car, originally at rest, begins to move with acceleration a. The block slides back horizontally until it hits the back wall of the car. At that moment, what is its speed (a) relative to the car? (b) relative to Earth?

A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius \(35.0 \mathrm{m}\). If the coefficient of static friction between crate and truck is \(0.600,\) how fast can the truck be moving without the crate sliding?

The pilot of an airplane executes a constant-speed loop-theloop maneuver in a vertical circle. The speed of the airplane is \(300 \mathrm{mi} / \mathrm{h},\) and the radius of the circle is \(1200 \mathrm{ft}\). (a) What is the pilot's apparent weight at the lowest point if his true weight is 160 lb? (b) What is his apparent weight at the highest point? (c) What If? Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. (Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body.)

In a cyclotron (one type of particle accelerator), a deuteron (of atomic mass \(2.00 \mathrm{u}\) ) reaches a final speed of \(10.0 \%\) of the speed of light while moving in a circular path of radius \(0.480 \mathrm{m}\). The deuteron is maintained in the circular path by a magnetic force. What magnitude of force is required?\(214.897 N\)
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