Classical Mechanics Dynamic Drag Resistance

[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.

 

[PhanthomJay]

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?

 

[Biest]

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

 

[cameo_demon]

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?

 

[PhanthomJay]

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.

 

[cameo_demon]

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...

 

[PhanthomJay]

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.

 

[cameo_demon]

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.

 

[PhanthomJay]

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.

 

[cameo_demon]

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?

 

[PhanthomJay]

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).

 

[cameo_demon]

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!