# Classical Mechanics Dynamic Drag Resistance

• cameo_demon
In summary, the conversation discusses a problem involving a baseball being released in air and finding its terminal velocity. The forces acting on the ball and the equations used to solve for the velocity and position are also mentioned. The conversation continues to discuss a follow-up question and how to solve for velocity and position at a specific time.
cameo_demon
i'm not even sure where to get started with this one because the thorton and marion classical dynamics book is really awful.

BEGIN PROBLEM

Suppose a baseball, which has a mass of 150 g and a diameter D of 7 cm is released from rest. For a sphere in air, the dynamic drag is F_d=.25 D^2 V^2

let F_d=kmv^2, find the terminal velocity of a baseball.

END PROBLEM

the question seems really vague given that we're not told if its being released in space, off of something, thrown or whatever. any suggestions? I am lost as to how to approach this problem.

The sphere is released from rest in air. In what direction does it move? What are the forces acting on it as it moves? What is the net force acting on it when it reaches terminal velocity ? (At terminal velocity, its speed is constant). I assume the F_d force is given in Newtons?

I am guessing it is Classical Dynamics of Particles and Systems right? Which question and chapter?

this problem is one just made up by my professor, and isn't in the thorton and marion book. is terminal velocity at a=0?

cameo_demon said:
this problem is one just made up by my professor, and isn't in the thorton and marion book. is terminal velocity at a=0?
For all practical purposes, yes.

a diagram he attached just shoes it moving in the downward y direction, without a given height. as it falls gravity is acting on the ball so we have mg and then with drag acting against it, given here as kmv^2, we have mg - kmv^2 = ma = 0 for the terminal velocity. then setting mg = kmv^2. solving for v we have v = sqrt{g / k}. setting kmv^2 = .25D^2v^2, cancelling v^2's and solving for k i get 8.2*10^-3. plugging k back into my formula for v=34.6 m/s which is approx 77mph and that seems like a reasonable answer for a baseball don't you think?

maybe i just had a brainfart earlier and needed to look at this one again...
...or maybe i had just been away from mechanics for too long...

cameo_demon said:
a diagram he attached just shoes it moving in the downward y direction, without a given height. as it falls gravity is acting on the ball so we have mg and then with drag acting against it, given here as kmv^2, we have mg - kmv^2 = ma = 0 for the terminal velocity. then setting mg = kmv^2. solving for v we have v = sqrt{g / k}. setting kmv^2 = .25D^2v^2, cancelling v^2's and solving for k i get 8.2*10^-3. plugging k back into my formula for v=34.6 m/s which is approx 77mph and that seems like a reasonable answer for a baseball don't you think?

maybe i just had a brainfart earlier and needed to look at this one again...
...or maybe i had just been away from mechanics for too long...
Yes, looks good. Faster than a Wakefield knuckler, but not near as fast as Clemens heater.

well see the follow up to that question is that major leaguers throw around 90 to 100mph often enough, and so the professor asks if there is a contradiction. my conjecture is that the spin often applied to the ball would give it additional momentum, and that a dead spinless ball would only top out at 77mph.

cameo_demon said:
well see the follow up to that question is that major leaguers throw around 90 to 100mph often enough, and so the professor asks if there is a contradiction. my conjecture is that the spin often applied to the ball would give it additional momentum, and that a dead spinless ball would only top out at 77mph.
No, it's not the spin. When thrown from pitcher to catcher, it has an initial speed, and there is not enough time for it to slow down appreciably. If you stood atop a very high building and threw the ball vertically downward with an initial speed of 100mph, it would still ultimately reach terminal speed of 77mph or so, as long as the building was tall enough.

for the next part of the problem, we're asked:

at t=5 sec, find v and x

again, mg - kmv^2 = ma, but this time a does not equal zero, solving for a, a=g-kv^2
to find x and v, can i plug in the terminal velocity as v_final and solve using standard kinematic equations, or do i need to solve a differential equation?

cameo_demon said:
for the next part of the problem, we're asked:

at t=5 sec, find v and x

again, mg - kmv^2 = ma, but this time a does not equal zero, solving for a, a=g-kv^2
to find x and v, can i plug in the terminal velocity as v_final and solve using standard kinematic equations, or do i need to solve a differential equation?
Terminal velocity occurs when the acceleration approaches zero, so you can't just plug it in. Besides, the standard kinematic equations are developed based on constant acceleration. Note that a is not constant (a=dv/dt).

ah so using the equation for a i can sub it into a = dv/dt and just solve the differential equation, same deal for position once i find the velocity equation. easy enough, thanks!

## 1. What is classical mechanics dynamic drag resistance?

Classical mechanics dynamic drag resistance, also known as air resistance, is a force that opposes the motion of an object through a fluid (such as air or water). It is caused by the interaction between the object and the molecules of the fluid, and depends on factors such as the object's speed, shape, and the density of the fluid.

## 2. How does drag affect the motion of an object?

Drag force acts in the opposite direction of an object's motion, slowing it down and reducing its acceleration. As an object moves faster, the drag force increases, reaching a point where it balances out the force of gravity and the object reaches a constant speed, known as terminal velocity.

## 3. How is drag force calculated?

The drag force on an object can be calculated using the equation FD = 1/2 * ρ * v2 * CD * A, where ρ is the density of the fluid, v is the speed of the object, A is the cross-sectional area of the object, and CD is the drag coefficient which depends on the object's shape and the fluid's viscosity.

## 4. What factors affect the amount of drag force on an object?

The amount of drag force on an object depends on several factors, including the object's speed, shape, and size, as well as the density and viscosity of the fluid. For example, a larger and more streamlined object will experience less drag than a smaller and more irregularly shaped object at the same speed.

## 5. How can drag force be reduced?

There are several ways to reduce drag force on an object, including changing its shape to be more streamlined, reducing its speed, and altering the properties of the fluid it is moving through (e.g. using a less dense or less viscous fluid). Additionally, adding features such as fins or dimples can also help to reduce drag force.

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