mechanics question (1 Viewer)

[sasquatch]

Im having some trouble with the second question in the exericse 7.2 from cambridge.

I can get the distance, but not the time (can't seem to be able to integrate the acceleration in terms of t...)

2. A particle is moving vertically downward in a medium which exerts a resistence to the motion which is proportional to the square of the speed of the particle. It is released from rest at O and its terminal velocity V. Find the distance it has fallen below O and the time taken when its velocity is one half of its terminal velocity.

TO make it simpler, the acceleration equation is

a = g[V² - v² / V²]

Thanks.

 

[Riviet]

dv/dt = g/V².(V² - v²)

Rearranging and integrating,

t=V²/g.∫dv/(V²-v²)

=V²/2Vg.∫{1/(V-v) + 1/(V+v}dv by partial fractions

=V/2g.ln|(V+v)/(V-v)| (the constant happens to be zero)

When v=V/2,

t=V/2g.ln|(3V/2)/(V/2)|

=V/2g.ln3

 

[sasquatch]

oh god im so stupid.. forgot about partial fractions.. :|

 

[sasquatch]

I have another question that i hope someone can help me out with,

A particle of mass m is projected vertically upward under gravity in a medium in which the resistance is mk times the square of the pssed, where k is a positive constant. If its speed of projection is equal to the terminal velocity V in the medium, show that when it returns to the point of projection its speed is V/root2.

THe part i dont get, i say the force

F = -mg - mkv²

hence a = -g - kv²

Then as a -> 0, v -> V

so 0 = -g - kV²

so k = -g/V²

but the question says k > 0, so im confused, g > 0 and V² >= 0, so that doesnt make sense, meaning that the forces must be wrong.

But if the particle is projected vertically upward, it means its resistance is opposite to its original motion right? so both the forces acting upon the particle should be negative..

Could somebody explain it, cuz for another question:

A particle of mass m is projected vertically upward under gravity with speed u in a medium in which the resistance is mk times the speed, where k is a positve constant.

a) if the particle reaches its greatest height H in time T, show that u =gT + kH

and i did it the same way and got it correct.

Thanks for any help.

 

[sasquatch]

nevermind me figured it out..thanks anyway