/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A hawk flies in a horizontal arc... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 16

A hawk flies in a horizontal arc of radius 12.0 \(\mathrm{m}\) at a con- stant 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 hawk's initial centripetal acceleration is 1.33 \(\mathrm{m/s^2}\). The magnitude of its total acceleration after increasing its speed can be found by calculating the vector sum of centripetal and tangential accelerations; its direction can be determined using the arctangent of the centripetal to tangential acceleration ratio.

Step by step solution

01

Determine the initial centripetal acceleration

Use the formula for centripetal acceleration, which is given by \( a_c = \frac{v^2}{r} \), where \(v\) is the speed and \(r\) is the radius of the circular path. Here, \(v = 4.00 \, \mathrm{m/s}\) and \(r = 12.0 \, \mathrm{m}\).
02

Calculate the initial centripetal acceleration

Plugging in the values into the centripetal acceleration formula gives: \( a_c = \frac{(4.00 \, \mathrm{m/s})^2}{12.0 \, \mathrm{m}} = \frac{16.0 \, \mathrm{m^2/s^2}}{12.0 \, \mathrm{m}} = 1.33 \, \mathrm{m/s^2} \).
03

Determine the tangential acceleration

The tangential acceleration is given as \( 1.20 \, \mathrm{m/s^2} \). This is the rate at which the hawk's speed is increasing.
04

Calculate the new centripetal acceleration at increased speed

The new speed after an infinitesimal time dt will be \( v + dv = 4.00 \, \mathrm{m/s} + 1.20 \, dt \, \mathrm{m/s^2} \). The corresponding new centripetal acceleration \( a_c' \) is given by \( a_c' = \frac{(v + dv)^2}{r} \). As dt approaches zero, this becomes \( a_c' = \frac{v^2}{r} + \frac{2v \cdot dv}{r} + \frac{(dv)^2}{r} \), or \( a_c' = a_c + \frac{2v \cdot a_t}{r} \) ignoring the \( (dv)^2 \) term as it is infinitesimally small.
05

Calculate the total acceleration as a vector sum

Total acceleration is the vector sum of the centripetal and tangential accelerations. As centripetal acceleration always points towards the center of the circle and tangential acceleration is in the direction of velocity, which is perpendicular to centripetal acceleration, we can use Pythagoras' theorem to find the total acceleration: \[|\mathbf{a}| = \sqrt{a_c'^2 + a_t^2} \] where \(a_c' = a_c + \frac{2v \cdot a_t}{r} \) and \(a_t = 1.20 \, \mathrm{m/s^2}\).
06

Calculate the magnitude of the total acceleration

Substitute the given values and the value from Step 4: \[|\mathbf{a}| = \sqrt{(1.33 \, \mathrm{m/s^2} + \frac{2 \cdot 4.00 \, \mathrm{m/s} \cdot 1.20 \, \mathrm{m/s^2}}{12.0 \, \mathrm{m}})^2 + (1.20 \, \mathrm{m/s^2})^2} \] Solving the above expression gives the magnitude of the total acceleration.
07

Calculate the direction of the total acceleration

The direction of the total acceleration can be determined by finding the angle \(\theta\) it makes with the tangential acceleration using the following tangent relation: \[\tan(\theta) = \frac{a_c'}{a_t} \] where \(a_c' = a_c + \frac{2v \cdot a_t}{r} = 1.33 \, \mathrm{m/s^2} + \frac{2 \cdot 4.00 \, \mathrm{m/s} \cdot 1.20 \, \mathrm{m/s^2}}{12.0 \, \mathrm{m}}\).
08

Determine the combined acceleration and its angle

Combine the results from Steps 6 and 7 to find the magnitude and direction of the hawk's total acceleration. The magnitude has been calculated, and the angle \(\theta\) can be calculated using an arctangent function from the ratio derived in Step 7.

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Key Concepts

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

Circular Motion
Circular motion refers to the movement of an object along a circular path. It's important to understand that even when the speed of an object in circular motion is constant, it is still undergoing acceleration due to its changing direction of velocity. This acceleration, called centripetal acceleration, always points toward the center of the circular path.

The formula for centripetal acceleration is expressed as: \[ a_c = \frac{v^2}{r} \] where \(v\) represents the constant speed of the object and \(r\) is the radius of the circle. In our exercise, a hawk flying in a circular arc is a practical example of circular motion, with the centripetal acceleration keeping it on the curved path.
Tangential Acceleration
In contrast to centripetal acceleration which points towards the center of a circle, tangential acceleration occurs along the edge of the circle, in the direction of motion, and is associated with the change in the magnitude of velocity - in other words, how fast an object's speed increases or decreases.

In the given problem, the hawk increases its speed at a rate of \(1.20 \, \mathrm{m/s^2}\), which is the tangential acceleration. This acceleration is a result of an applied force and indicates that the hawk is not only moving in a circle but also speeding up. Tangential acceleration is often denoted as \(a_t\) and can be added or subtracted from the speed of the object to find the new speed at any given point as: \[ v_{new} = v + a_t \cdot dt \] where \(dt\) represents a small time interval.
Vector Sum of Accelerations
When an object in circular motion is also changing its speed, it experiences both centripetal and tangential accelerations simultaneously. To determine the overall acceleration of the object, we take the vector sum of these two accelerations. Since they act perpendicular to each other, we can use the Pythagorean theorem to find the total acceleration as follows:

\[|\mathbf{a}| = \sqrt{a_c'^2 + a_t^2} \]
Here, \(a_c'\) is the modified centripetal acceleration at the new speed, calculated after considering any tangential acceleration, and \(a_t\) remains the tangential acceleration. Calculating this vector sum provides us with the total acceleration's magnitude. Meanwhile, the direction of this total acceleration can be found using the arctangent function, giving us a comprehensive understanding of the object's motion at that moment.

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Most popular questions from this chapter

A roller coaster at the Six Flags Great America amusement park in Gurnee, IL, incorporates some clever design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop (Fig. P6.20). The cars ride on the inside of the loop at the top, and the speeds are high enough to ensure that the cars remain on the track. The biggest loop is 40.0 m high, with a maximum speed of 31.0 \(\mathrm{m} / \mathrm{s}\) (nearly 70 \(\mathrm{mi} / \mathrm{h} )\) at the bottom. Suppose the speed at the top is 13.0 \(\mathrm{m} / \mathrm{s}\) and the corresponding centripetal acceleration is 2 \(\mathrm{g}\) . (a) What is the radius of the arcof the teardrop at the top? (b) If the total mass of a car plus the riders is \(M,\) what force does the rail exert on the car at the top? (c) Suppose the roller coaster had a circular loop of radius 20.0 \(\mathrm{m}\) . If the cars have the same speed, 13.0 \(\mathrm{m} / \mathrm{s}\) at the top, what is the centripetal acceleration at the top? Comment on the normal force at the top in this situation.

A light string can support a stationary hanging load of 25.0 \(\mathrm{kg}\) before breaking. A 3.00 -kg object attached to the string rotates on a horizontal, frictionless table in a circle of radius \(0.800 \mathrm{m},\) while the other end of the string is held fixed. What range of speeds can the object have before the string breaks?

A 3.00 -g leaf is dropped from a height of 2.00 \(\mathrm{m}\) above the ground. Assume the net downward force exerted on the leaf is \(F=m g-b v,\) where the drag factor is \(b=0.0300 \mathrm{kg} / \mathrm{s}\) (a) Calculate the terminal speed of the leaf. (b) Use Euler's method of numerical analysis to find the speed and position of the leaf, as functions of time, from the instant it is released until 99\(\%\) of terminal speed is reached. (Suggestion: Try \(\Delta t=0.005 \mathrm{s} . )\)

Consider an object on which the net force is a resistive force proportional to the square of its speed. For example, assume that the resistive force acting on a speed skater is \(f=-k m v^{2},\) where \(k\) is a constant and \(m\) is the skater's mass. The skater crosses the finish line of a straight-line race with speed \(v_{0}\) and then slows down by coasting on his skates. Show that the skater's speed at any time \(t\) after crossing the finish line is \(v(t)=v_{0} /\left(1+k t v_{0}\right) .\) This problem also provides the background for the two following problems.

(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.
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