- #1

Some1WhoNeedsHelp

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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.

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- Thread starter Some1WhoNeedsHelp
- Start date

- #1

Some1WhoNeedsHelp

- 16

- 0

- 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.

- #2

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- #3

Some1WhoNeedsHelp

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1=-gamma v^2?

I also need to find gamma, So I need to realize what the letter supposed to represent

I looked at the formula sheet and didn't find it there

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- #4

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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?1=-gamma v^2?

- #5

Some1WhoNeedsHelp

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it's a function of speed relative to the time it representsNope. 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?

The change in speed is divided by the change in time.

So does it represent the location of the arrow?

Is that got something to do with the function:

X=X

?

- #6

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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?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}

?

- #7

Some1WhoNeedsHelp

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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?

acceleration is meter/(seconds)

and v

So gamma must be meter

and using Newton's law

ma=(Cd*ρ*A*V

and since

Gamma=-(Cd*ρ*A)/2m

Is that correct?

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- #8

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- #9

Some1WhoNeedsHelp

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So should I do Absolute value?

F=ma

F=(Cd*ρ*A*V

ma=(Cd*ρ*A*V

a=(Cd*ρ*A*V

dv/dt = -γV

-γV

γ=|-

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- #10

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dv/dt = -γV

a = dv/dt = -γV

F = ma = -mγV

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.

- #11

Some1WhoNeedsHelp

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Because the force is opposite to direction of motion, the equation needs to be negative?

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.

F=-(Cd*ρ*A*V

so

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- #12

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No, it is not correct because it is self contradictory.F=(Cd*ρ*A*V^{2})/2=-mγV^{2}?

Is that correct?

The magnitude |F| of the drag force is positive and one can write

|F|=(Cd*ρ*A*V

If one wants to write Newton's second law, one must write

F= m dv/dt = -(Cd*ρ*A*V

- #13

Some1WhoNeedsHelp

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soNo, 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}.

?

- #14

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Yes.

- #15

Some1WhoNeedsHelp

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yes!Yes.

But since

I am trying to calculate the V

so I multiply by "dt" and do integral

∫(1/

(v/

(-1/

So time is negative

- #16

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- #17

Some1WhoNeedsHelp

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∫

(v

(-1/

0=t

correct?

- #18

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I have no idea how you get that result from the integration. What is the integral (1/x∫^{v0}_{v}(1/γV^{2}) dv=∫^{0}_{t}dt

(v_{0}/γ)*(-1/v_{0})-(v/=-tγ)*(-1/v)

(-1/γ)-=t(-1/γ)

0=t

correct?

Try again, taking one small step at a time so that we can see where your error is.

- #19

Some1WhoNeedsHelp

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The integral of (1/xI 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.

so

- #20

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That looks better.The integral of (1/x^{2}).dx is (-1/x)

so

(-1/γv_{0})-=-t(-1/γv)

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##.

- #21

Some1WhoNeedsHelp

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-1/(γvThat 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##.

I know

γ,V

so I get - 1 equation with 2 unknown things (V

I would assume V

I also know η, Cd , A, ρ and the length of the arrow, if that helps.

- #22

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No, it asks you to find the value of t.but the question asks me to calculate the value of Vt

- #23

Some1WhoNeedsHelp

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No, it asks you to find the value of t.

Oh wait, Is that mean I need to find an expression of V depending on t?

So am I right about the drag force will stop the arrow eventually despite being the only force?

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- #24

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You did that in post #21Is 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:So am I right about the drag force will stop the arrow eventually despite being the only force?

https://www.physicsforums.com/attachments/help2-png.318235/

- #25

Some1WhoNeedsHelp

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Alright so according to the equation I created from post #21You 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/

-1/(γv

-v

v

v

v

So as time keeps increasing the v

- #26

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How long will it take to stop, according to the formula?v_{t}will decrease to the point the speed of the arrow will become neglectable, so the arrow will stop

- #27

Some1WhoNeedsHelp

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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.

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.

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- #28

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We too exist in the "real life" plane. Have we not been helping you understand?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 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.##

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- #29

Some1WhoNeedsHelp

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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 twoWe 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.##

That is why I am thanking you

Let me just upload my homework and I will solve this

- #30

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Ok, we're here to help you if you get stuck.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

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