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You throw a baseball straight up. The drag force is proportional to \(v^{2} .\) In terms of \(g,\) what is the \(y\) -component of the ball's acceleration when its speed is half its terminal speed and (a) it is moving up? (b) It is moving back down?

Short Answer

Expert verified
(a) \(-\frac{5g}{4}\) for upward motion; (b) \(\frac{3g}{4}\) for downward motion.

Step by step solution

01

Understand the Forces Involved

When a baseball is thrown upwards, it experiences two main forces: gravity and air drag. The gravitational force is constant and acts downward with a magnitude of \(mg\), where \(m\) is the mass of the baseball and \(g\) is the acceleration due to gravity. The drag force is given by \(D = kv^2\) and acts opposite to the direction of motion, where \(k\) is a constant and \(v\) is the velocity.
02

Define Terminal Velocity

Terminal velocity \(v_t\) is the constant speed that occurs when the drag force equals the gravitational force \(mg\). At terminal velocity, \(kv_t^2 = mg\). Solving for \(k\), we find \(k = \frac{mg}{v_t^2}\), allowing us to express the drag force in terms of the terminal velocity.
03

Determine Drag Force at Half Terminal Speed

When the speed \(v = \frac{v_t}{2}\), the drag force becomes \(D = k\left(\frac{v_t}{2}\right)^2 = \frac{kv_t^2}{4}\). Substituting \(k = \frac{mg}{v_t^2}\), we get \(D = \frac{mg}{v_t^2} \times \frac{v_t^2}{4} = \frac{mg}{4}\).
04

Equation of Motion for Upward Movement

When the ball is moving up, the gravitational force and drag force both act downward. The net force is \(ma = -(mg + D)\). Substituting \(D = \frac{mg}{4}\), the equation becomes \(ma = -(mg + \frac{mg}{4})\). The net force is \(ma = -\frac{5mg}{4}\). Thus, the acceleration is \(a = -\frac{5g}{4}\), indicating downward acceleration.
05

Equation of Motion for Downward Movement

When the ball is moving down, gravity acts downward and drag force acts upward. The net force becomes \(ma = mg - D\). Using \(D = \frac{mg}{4}\), the equation is \(ma = mg - \frac{mg}{4} = \frac{3mg}{4}\). Therefore, the acceleration is \(a = \frac{3g}{4}\), indicating downward motion but with less resistance compared to going up.

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Key Concepts

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

Drag Force
When an object moves through a fluid, like air, it encounters a resistance known as drag force. Drag force can be a little tricky because it depends on the velocity of the object. The faster you go, the greater the drag force you experience. In our exercise with the baseball, the drag force is directly proportional to the square of the velocity, expressed as:
  • \(D = k v^2\)
where \(D\) is the drag force, \(k\) is a constant that depends on factors like the shape of the object and air density, and \(v\) is the velocity of the object.
The drag force acts in the opposite direction to the motion of the object. So, when the baseball is thrown upwards, the drag force is downward. Conversely, when the baseball is coming down, the drag force is upwards, opposing gravity.
Understanding how drag force works helps in predicting how quickly an object will slow down as it moves through the air.
Terminal Velocity
Terminal velocity is a fascinating concept! It represents the highest speed an object can reach as it falls through a fluid. At terminal velocity, an object stops accelerating and moves at a constant speed. This happens when the drag force and the gravitational force are equal in magnitude but opposite in direction.
For the baseball scenario, terminal velocity happens when:
  • \(k v_t^2 = mg\)
Here, \(v_t\) is the terminal velocity, \(m\) is the mass of the baseball, and \(g\) is the acceleration due to gravity. Solving for \(k\) provides:
  • \(k = \frac{mg}{v_t^2}\)
This formula ensures that terminal velocity is maintained, balancing drag and weight. When the baseball reaches its terminal speed, it glides down at a steady pace, no longer speeding up.
Acceleration Due to Gravity
Gravity is the force that pulls objects towards each other. On Earth, every object experiences a gravitational pull toward the center of the planet. This pull is characterized by what we call the acceleration due to gravity, usually denoted as \(g\). On Earth, \(g\) is approximately \(9.81 \, m/s^2\), acting downward.
When dealing with physics problems, especially in mechanics, it's important to remember that gravitational force depends on the mass of the object and the gravitational constant. For the baseball, gravity acts as a constant force downwards, causing it to accelerate when it's thrown up or dropped.
When the baseball is thrown upwards, gravity slows it down until it stops and then pulls it back to Earth. During its downward journey, gravity accelerates it until other forces, like drag, modify this acceleration, eventually reaching a terminal velocity where the two forces balance each other out.

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Most popular questions from this chapter

A \(1125-k g\) car and a \(2250-k g\) pickup truck aproach a curve on the expressway that has a radius of 225 \(\mathrm{m}\) . (a) At what angle should the highway engineer bank this curve so that vehicles traveling at 65.0 \(\mathrm{mi} / \mathrm{h}\) can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? (b) As the car and truck round the curve at 65.0 \(\mathrm{mi} / \mathrm{h}\) , find the normal force on each one due to the highway surface.

The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 \(\mathrm{m}\) . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 \(\mathrm{s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 \(\mathrm{N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

Stopping Distance. (a) If the coefficient of kinctic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

Falling Baseball. You drop a bascball from the roof of a tall building. As the ball falls, the air exerts a drag force proportional to the square of the ball's speed \(\left(f=D v^{2}\right) .\) (a) In a diagram, show the direction of motion and indicate, with the aid of vectors, all the forces acting on the ball. (b) Apply Newton's second law and infer from the resulting equation the general properties of the motion. (c) Show that the ball acquires a terminal speed that is as given in Eq. (5.13). (d) Derive the equation for the speed at any time. (Note: $$ \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{a} \operatorname{arctanh}\left(\frac{x}{a}\right) $$ where $$ \tanh (x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\frac{e^{2 x}-1}{e^{2 x}+1} $$ defines the hyperbolic tangent.)

Maximum Safe Speed. As you travel every day to campus, the road makes a large turn that is approximately an are of a circle. You notice the warning sign at the start of the turm, asking for a maximum speed of 55 \(\mathrm{mi} / \mathrm{h}\) . You also notice that in the curved portion the road is level - that is, not banked at all. On a dry day with very little traffic, you enter the turn at a constant speed of 80 \(\mathrm{mi} / \mathrm{h}\) and feel that the car may skid if you do not slow down quickly. You conclude that your speed is at the limit of safety for this curve and you slow down. However, you remember reading that on dry pavement new tires have an average coefficient of static friction of about 0.76 .while under the worst winter driving conditions, you may encounter wet ice for which the coefficient of static friction can be as low as \(0.20 .\) Wet ice is not unheard of on this road, so you ask yourself whether the speed limit for the turn on the roadside warning sign is for the worst-case scenario. (a) Estimate the radius of the curve from your \(80-\) mi/h experience in the dry turn. (b) Use this estimate to find the maximum speed limit in the turn under the worst wet-ice conditions. How does this compare with the speed limit on the sign? Is the sign misleading drivers?(c) On a rainy day, the coefficient of static friction would be about \(0.37 .\) What is the maximum safe speed of for the turn when the road is wet? Does your answer help you understand the maximum-speed sign?

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