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Calculating a ball of unit mass' velocity.

A ball of unit mass is constrained to roll along a straight track in such a way that overall
it experiences a resistive force which is inversely proportional to its velocity. Given that its
initial velocity c is positive, show that the velocity at some later time t has the form:
V(t)=c2κt V(t)=\sqrt{c^2-{\kappa}t}
for some constant κ>0{\kappa}> 0. Find an expression for the position of the particle at time t, and hence prove that the ball travels a distance
2c33κ\dfrac{2c^3}{3\kappa} before coming to rest, when the model breaks down.

The answer for the first part of the question kind of makes it look like I have to use the constant acceleration formula v2=u2+2asv^2=u^2+2as, however I'm not sure how I would get the acceleration from this :s-smilie:. Please help!
Reply 1
If the resistive force is inversely proportional to its velocity, it has the form -c/v for some positive constant c.

In which case, what is dv/dt? You get an easy differential equation. Solve it, and you'll get the first part out.
Reply 2
Original post by Happy2Guys1Hammer
A ball of unit mass is constrained to roll along a straight track in such a way that overall
it experiences a resistive force which is inversely proportional to its velocity. Given that its
initial velocity c is positive, show that the velocity at some later time t has the form:
V(t)=c2κt V(t)=\sqrt{c^2-{\kappa}t}
for some constant κ>0{\kappa}> 0. Find an expression for the position of the particle at time t, and hence prove that the ball travels a distance
2c33κ\dfrac{2c^3}{3\kappa} before coming to rest, when the model breaks down.

The answer for the first part of the question kind of makes it look like I have to use the constant acceleration formula v2=u2+2asv^2=u^2+2as, however I'm not sure how I would get the acceleration from this :s-smilie:. Please help!


Tne ac(dec)celeration is function of v with κ\kappa factor.

a=Fm=κ1va=-\frac{F}{m}=-\kappa \frac{1}{v}
so
dvdt=κv\frac{dv}{dt}=-\frac{\kappa}{v}
Solve for v
t=0 v=c

for the position
dsdt=v(t)\frac{ds}{dt}=v(t)
solve s with t=0 s=0
(edited 10 years ago)
Original post by ztibor
Tne ac(dec)celeration is function of v with κ\kappa factor.

a=Fm=κ1va=-\frac{F}{m}=-\kappa \frac{1}{v}
so
dvdt=κv\frac{dv}{dt}=-\frac{\kappa}{v}
Solve for v
t=0 v=c

for the position
dsdt=v(t)\frac{ds}{dt}=v(t)
solve s with t=0 s=0


Okay so I've done:
dvdt=κvvdv=κdt12v2=κt\frac{dv}{dt}=-\frac{\kappa}{v} \Rightarrow \int v dv = \int -{\kappa} dt \Rightarrow \frac{1}{2}v^2=-{\kappa}t. Is this right? I can't see where c comes into this.

Also, could you please explain why "Fm=κ1v-\frac{F}{m}=-\kappa \frac{1}{v}" as I can't see the transition there.
Reply 4
Original post by Happy2Guys1Hammer
Okay so I've done:
dvdt=κvvdv=κdt12v2=κt\frac{dv}{dt}=-\frac{\kappa}{v} \Rightarrow \int v dv = \int -{\kappa} dt \Rightarrow \frac{1}{2}v^2=-{\kappa}t. Is this right? I can't see where c comes into this.

Also, could you please explain why "Fm=κ1v-\frac{F}{m}=-\kappa \frac{1}{v}" as I can't see the transition there.


You haven't included a constant of integration!

You're told that resistive force is inversely proportional to speed i.e. Fres = -B/v for some constant B. Also use the general Newton's 2nd law F = ma where a is acceleration and F is the total force acting.
Original post by davros
You haven't included a constant of integration!

You're told that resistive force is inversely proportional to speed i.e. Fres = -B/v for some constant B. Also use the general Newton's 2nd law F = ma where a is acceleration and F is the total force acting.


Okay but if I use a constant of integration, I will just get +c or whatever, but that's not the same c that v is equal to surely. Also I need it to be c^2. but it'll just be c. There's also a 1/2 in there which isn't in the final answer and there's no constant in the final answer. Are you sure this is the correct method because It doesn't seem to take shape to me.
Reply 6
Original post by Happy2Guys1Hammer
Okay but if I use a constant of integration, I will just get +c or whatever, but that's not the same c that v is equal to surely. Also I need it to be c^2. but it'll just be c. There's also a 1/2 in there which isn't in the final answer and there's no constant in the final answer. Are you sure this is the correct method because It doesn't seem to take shape to me.


You can always rescale your constants so k -> 2k (or k -> k/2). The c^2 is inside a square root sign, so when you put in t = 0 you get v = c whjch is the required initial condition.
Original post by davros
You can always rescale your constants so k -> 2k (or k -> k/2). The c^2 is inside a square root sign, so when you put in t = 0 you get v = c whjch is the required initial condition.


Oh okay I understand that now. Thank you very much :smile:.
Original post by davros
You can always rescale your constants so k -> 2k (or k -> k/2). The c^2 is inside a square root sign, so when you put in t = 0 you get v = c whjch is the required initial condition.


Okay so I've looked back through my answer and I really don't think I did the method correctly.

Here's what I've got:
dvdt=κvvdv=κdt12v2=κt+c\frac{dv}{dt}=-\frac{\kappa}{v} \Rightarrow \int v dv = \int -{\kappa} dt \Rightarrow \frac{1}{2}v^2=-{\kappa}t+c. As v(0)=cv=2c2κtv(0)=c \Rightarrow v=\sqrt{2c^2-{\kappa}t}.

Letting 2c2=c2,2κ=κv=c2κt 2c^2=c^2, 2{\kappa}=\kappa \Rightarrow v=\sqrt{c^2-{\kappa}t}.

This doesn't quite seem right to me, is it correct?
Reply 9
Original post by Happy2Guys1Hammer
Okay so I've looked back through my answer and I really don't think I did the method correctly.

Here's what I've got:
dvdt=κvvdv=κdt12v2=κt+c\frac{dv}{dt}=-\frac{\kappa}{v} \Rightarrow \int v dv = \int -{\kappa} dt \Rightarrow \frac{1}{2}v^2=-{\kappa}t+c. As v(0)=cv=2c2κtv(0)=c \Rightarrow v=\sqrt{2c^2-{\kappa}t}.

Letting 2c2=c2,2κ=κv=c2κt 2c^2=c^2, 2{\kappa}=\kappa \Rightarrow v=\sqrt{c^2-{\kappa}t}.

This doesn't quite seem right to me, is it correct?


It looks about right, but to make it more "convincing" I'd use something like A and B for your constant of proportionality and your constant of integration respectively, and then use the boundary conditions and the factor of 2 to get c and κ\kappa, rather than starting with the constants stated in the question and then making it look like you're fudging things by changing their values :smile:

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