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Resistance - fluid resistance to ball motion

  • #1
Resistance -- fluid resistance to ball motion

A ball of mass m = 3 kg is falling from an initial height y = 3 m in an area where the acceleration due to gravity can be approximated as g = 10 m/s2. As it falls the ball is subjected to a fluid resistance of magnitude FR = kv where k = 6 kg/s. How many seconds will it take for the acceleration of the ball to decrease to a value a = 9 m/s2? Type the numerical value only, not the unit. Round off your answer to 3 decimal place.


The Attempt at a Solution


I did FR=kv
FR=mg
3(10)=30
30=6v
v=5m/s

I don't know if this step is right...
v=v0+at
v=at
5=9t
5/9=t

PART B: I have no clue on this one.

Consider the problem of an object falling from rest in a fluid where the resistance is FR = kv. How many time constants will it take for the object to reach a velocity that represents 72% of its terminal velocity? The time constant is the ratio m/k where m is the mass of the falling object and k is the coefficient of the resistance force. Round off your answer to 1 decimal place.

Any help would be appreciated! Thanks
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
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A ball of mass m = 3 kg is falling from an initial height y = 3 m in an area where the acceleration due to gravity can be approximated as g = 10 m/s2. As it falls the ball is subjected to a fluid resistance of magnitude FR = kv where k = 6 kg/s. How many seconds will it take for the acceleration of the ball to decrease to a value a = 9 m/s2? Type the numerical value only, not the unit. Round off your answer to 3 decimal place.


The Attempt at a Solution


I did FR=kv
FR=mg
3(10)=30
30=6v
v=5m/s

I don't know if this step is right...
v=v0+at
v=at
5=9t
5/9=t

PART B: I have no clue on this one.

Consider the problem of an object falling from rest in a fluid where the resistance is FR = kv. How many time constants will it take for the object to reach a velocity that represents 72% of its terminal velocity? The time constant is the ratio m/k where m is the mass of the falling object and k is the coefficient of the resistance force. Round off your answer to 1 decimal place.

Any help would be appreciated! Thanks
No, that's not going well. In the first attempt you solved for the terminal velocity, which is not what you want. What you did from there is also wrong those equations only apply to motion at uniform acceleration. The way to do it is considerably more complicated. You need to get a differential equation for dv/dt and solve it. Have seen or done anything like that?
 
  • #3
I have in calculus but not in physics before. Was your first reply correct? The F=mg-FR=mg-kv? If so, I got v=1/2 m/s.
F=mg-kv
ma=mg-kv
3(9)=3(10)-6v
-3=-6v
1/2=v

I guess I need a full on explanation on what's going on in this problem :/ sorry
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
I have in calculus but not in physics before. Was your first reply correct? The F=mg-FR=mg-kv? If so, I got v=1/2 m/s.
F=mg-kv
ma=mg-kv
3(9)=3(10)-6v
-3=-6v
1/2=v

I guess I need a full on explanation on what's going on in this problem :/ sorry
If you've done it in calculus, you should be able to pull it off in physics. F=ma means F=m*dv/dt, right? Since a=dv/dt. And yes, F=mg-kv. It's the difference between the gravitational force and the resistive force. So m*dv/dt=mg-kv. Try and solve that differential equation for v as a function of time v(t). You can't just put a=9. a depends on time, a(t)=10 at t=0 and a(t)=9 only at the time you are looking for. Remember differential equations?
 
  • #5
Alright I went ahead and tried to differentiate equation.

m*dv/dt=mg-kv
dv/dt=g-(kv/m)
dv=(g-(kv/m))*dt
dt=(1/(g-(kv/m))*dv
Let:
u=g-(kv/m)
du=(-k/m)dv
dv=(-m/k)du
(-m/k)du/u=dt
(1/u)du=dt(-k/m)
ln(u)=(-k/m)t+C
e^ln(u)=e^[(-kt/m)(kC/m)]
g-(kv/m)=e^[(-kt/m)(kC/m)]
v=(m*e^[(-kt/m)(kC/m)]-g)/-k

Is this right? I'm not very confident it is but I gave it my best shot.
 

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