- #1
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- Homework Statement
- Show That Newton's law for the arrow can be writteb in the form
- Relevant Equations
- F=ma
I don't understand what the gamma letter in the equation is supposed to represent.
1=-gamma v^2?kuruman said:What do you get when you multiply both sides of the equation with ##-\gamma v^2##? Can you interpret what it says?
Nope. You get $$\frac{dv}{dt}=-\gamma~v^2.$$If I tell you that ##\gamma## is just a positive constant, can you interpret what this equation is saying to you regarding the motion of the arrow?Some1WhoNeedsHelp said:1=-gamma v^2?
it's a function of speed relative to the time it representskuruman said:Nope. You get $$\frac{dv}{dt}=-\gamma~v^2.$$If I tell you that ##\gamma## is just a constant, can you interpret what this equation is saying to you regarding the motion of the arrow?
Nope. What you wrote down is the position of an object as a function of time under constant acceleration. Is the acceleration constant in this case? As a matter of fact look at the equation. Can you figure out an expression for the acceleration? Hint: How is the acceleration defined in terms of the velocity?Some1WhoNeedsHelp said:it's a function of speed relative to the time it represents
The change in speed divided by the change in time
Is that got something to do with the function:
X=X_{0}+VT+(1/2)AT^{2}
?
a = dv/dtkuruman said:Nope. What you wrote down is the position of an object as a function of time under constant acceleration. Is the acceleration constant in this case? As a matter of fact look at the equation. Can you figure out an expression for the acceleration? Hint: How is the acceleration defined in terms of the velocity?
kuruman said:Gamma is a positive constant. The motion of the arrow is opposed by air resistance (it's like friction). So in the equation ##\dfrac{dv}{dt}=-\gamma~v^2##, the left-hand side is a negative number because the speed ##v## decreases with time. Can you answer the questions posed by the problem?
Because the force is opposite to direction of motion, the equation needs to be negative?kuruman said:Look
dv/dt = -γV^{2}
a = dv/dt = -γV^{2}
F = ma = -mγV^{2}
The force is in the same direction as the acceleration and both of them are opposite to the direction of motion. The constant γ is positive. You don't need the absolute value if you handle the signs correctly.
No, it is not correct because it is self contradictory.Some1WhoNeedsHelp said:F=(Cd*ρ*A*V^{2})/2=-mγV^{2}?
Is that correct?
sokuruman said:No, it is not correct because it is self contradictory.
The magnitude |F| of the drag force is positive and one can write
|F|=(Cd*ρ*A*V^{2})/2=mγV^{2}.
If one wants to write Newton's second law, one must write
F= m dv/dt = -(Cd*ρ*A*V^{2})/2 = -mγV^{2}.
yes!kuruman said:Yes.
∫^{v0}_{v}(1/γV^{2}) dv=∫^{0}_{t}dtkuruman said:Time is not negative. You forgot the limits of integration. Assume that at ##t=0## the arrow has velocity ##v_0## and some later time ##t## it has some other velocity ##v##. Set the limits of integration, then integrate.
I have no idea how you get that result from the integration. What is the integral (1/x^{2}).dx?Some1WhoNeedsHelp said:∫^{v0}_{v}(1/γV^{2}) dv=∫^{0}_{t}dt
(v_{0}/γ)*(-1/v_{0})-(v/γ)*(-1/v)=-t
(-1/γ)-(-1/γ)=t
0=t
correct?
The integral of (1/x^{2}).dx is (-1/x)haruspex said:I have no idea how you get that result from the integration. What is the integral (1/x^{2}).dx?
Try again, taking one small step at a time so that we can see where your error is.
That looks better.Some1WhoNeedsHelp said:The integral of (1/x^{2}).dx is (-1/x)
so
(-1/γv_{0})-(-1/γv)=-t
-1/(γv_{t})+1/(γv_{0})=tharuspex said:That looks better.
But please use parentheses appropriately. What you have written means (1/γ)v, where you mean 1/(γv). And it would have been more natural to have set the bounds the other way up, ##\int_{v_0}^v, \int_0^t##.
No, it asks you to find the value of t.Some1WhoNeedsHelp said:but the question asks me to calculate the value of Vt
haruspex said:No, it asks you to find the value of t.
You did that in post #21Some1WhoNeedsHelp said:Is that mean I need to find an expression of V depending on t?
That's for you to figure out. In fact it asks you to find how long it will take:Some1WhoNeedsHelp said:So am I right about the drag force will stop the arrow eventually despite being the only force?
Alright so according to the equation I created from post #21haruspex said:You did that in post #21
That's for you to figure out. In fact it asks you to find how long it will take:
https://www.physicsforums.com/attachments/help2-png.318235/
How long will it take to stop, according to the formula?Some1WhoNeedsHelp said:v_{t} will decrease to the point the speed of the arrow will become neglectable, so the arrow will stop
We too exist in the "real life" plane. Have we not been helping you understand?Some1WhoNeedsHelp said:I found someone in real life who help me to better understand the subject.
I realize now that the arrow won't stop as long as only drag force is active and gamma is 1/meter
Still thanks to both of you.
you help me a lot to understand too, even with this guy help I wouldn't be able to understand what he tries to explain to me if not for you twokuruman said:We too exist in the "real life" plane. Have we not been helping you understand?
You can find an answer quite easily using power (energy loss per unit time) considerations. The integral of power over time is energy loss in this case. If you can find an expression for energy loss as a function of time, you can find how long it takes to lose ##\frac{1}{2}mv_0^2.##
Ok, we're here to help you if you get stuck.Some1WhoNeedsHelp said:you help me a lot to understand too, even with this guy help I wouldn't be able to understand what he tries to explain to me if not for you two
That is why I am thanking you
Let me just upload my homework and I will solve this