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Particle with gravity and electric field

  • Thread starter XanziBar
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  • #1
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Homework Statement



A particle of mass m carries an electric charge Q and is subject to the combined action of gravity and a uniform horizontal electric field of strength E. It is projected in the vertical plane parallel to the field at a positive angle [tex]\theta[/tex] to the horizontal. Show that the horizontal distance it has traveled will be maximum if tan (2 [tex]\theta[/tex])=-mg/(EQ)



Homework Equations


Kinematics, F=QE etc.


The Attempt at a Solution



There is constant acceleration in y and constant acceleration in x and so I wrote down those equations and eliminated the time variable. I got the range of the projectile to be R=(2*V^2/g)* sin (theta)*(EQ*sin(theta)/(mg)+cos(theta))

So to get the max range I set all the cos/sin equal to one:

Rmax=(2*V^2/g)(EQ/(mg)+1)

So for my next step I tried setting the two equal to each other and solving for theta but there must be some kind of trig identity I am not seeing because I can not solve for theta or even figure out how to plug in tan(2 theta) to check the condition that I am supposed to prove. Help?
 

Answers and Replies

  • #2
collinsmark
Homework Helper
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Hello XanziBar,
There is constant acceleration in y and constant acceleration in x and so I wrote down those equations and eliminated the time variable. I got the range of the projectile to be R=(2*V^2/g)* sin (theta)*(EQ*sin(theta)/(mg)+cos(theta))
You might want to show how you got to your range equation. I think there is a mistake somewhere (there are problems with your equation's dimensional units too, so there's definitely something not right) . I used the

[tex] s = v_0 t + \frac{1}{2}a t^2 [/tex]

kinematics equation for both x and y, and combined them to make a range equation. My range equation is different than yours however.
So to get the max range I set all the cos/sin equal to one:
Ummm, :uhh: I do not follow what you mean by that approach. :confused:

Have you tried taking the derivative and setting it equal to zero? (That's usually how you maximize or minimize something.)
Rmax=(2*V^2/g)(EQ/(mg)+1)

So for my next step I tried setting the two equal to each other and solving for theta but there must be some kind of trig identity I am not seeing because I can not solve for theta or even figure out how to plug in tan(2 theta) to check the condition that I am supposed to prove. Help?
I'm still not quite sure of what your approach is there. But whatever the case, you should find this trigonometric identity very useful.

[tex] \mathrm{sin}(2 \theta) = 2 \mathrm{cos} \theta \mathrm{sin} \theta [/tex]

Good luck! :smile:
 
  • #3
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Alrighty well for the y direction I have

0=vsin(theta)t-.5*g*t^2

so v*sin(theta)=.5*g*t and that means that time in flight is t=2v*sin(theta)/g

for the x direction I have that delta x (range)=vcos(theta)*t+.5*Q*E*t^2/m where I went ahead and replaced the x acceleration with a=F/m=QE/m

So I plugged that expression for t into the x acceleration and got Range=cos(theta)*(2v*sin(theta)/g)+.5*Q*E*(2v*sin(theta)/g)^2/m

So after I simplify that is where I got my range equation, could you tell me where the units are wrong?
R=(2*V^2/g)* sin (theta)*(EQ*sin(theta)/(mg)+cos(theta)) because I think I simplified that right.

Anyway my approach that I was describing is that the biggest sine and cosine can be is unity so I figured that I could figure out what angle produces the maximum range.

Yeah I tried using 2sin(x)cos(x)=sin(2x) I didn't get anywhere but maybe I don't have the right intuition.

And yeah I did just try taking the derivative of the range w.r.t. theta...made a HUGE mess. Sorry if I'm not describing things well but I'm still really stuck...
 
  • #4
collinsmark
Homework Helper
Gold Member
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Alrighty well for the y direction I have

0=vsin(theta)t-.5*g*t^2

so v*sin(theta)=.5*g*t and that means that time in flight is t=2v*sin(theta)/g

for the x direction I have that delta x (range)=vcos(theta)*t+.5*Q*E*t^2/m where I went ahead and replaced the x acceleration with a=F/m=QE/m
So far so good. Very nice. :approve:
So I plugged that expression for t into the x acceleration and got Range=cos(theta)*(2v*sin(theta)/g)+.5*Q*E*(2v*sin(theta)/g)^2/m
I'm assuming that by 'v' you mean 'v0'. This term needs to be squared in the first instance of your above equation.
So after I simplify that is where I got my range equation, could you tell me where the units are wrong?
R=(2*V^2/g)* sin (theta)*(EQ*sin(theta)/(mg)+cos(theta)) because I think I simplified that right.
Okay. I see now. Your equation is correct. It's just hard to follow if your equations are completely in text form. Sorry for the confusion. :blushing: For clarity let me retype your equation in LaTex:

Your equation (replacing v with v0 too):
[tex] R = \left(\frac{2 v_0^2}{g} \right) \mathrm{sin}\theta \left(\frac{EQ \mathrm{sin} \theta}{mg} + \mathrm{cos}\theta \right) [/tex]

Your equation is correct, but let me give you some advice. Don't try to simplify things too much yet. Leave things un-factored for now. Let's go back to your unsimplified version of the equation.

[tex] R = \frac{2 v_0^2 \mathrm{cos}\theta \mathrm{sin}\theta}{g}
+ \frac{2 v_0^2 EQ \mathrm{sin}^2 \theta}{mg^2} [/tex]

Now see if you can use that trigonometric identity somewhere,

[tex] \mathrm{sin}(2 \theta) = 2 \mathrm{cos} \theta \mathrm{sin} \theta [/tex]

Anyway my approach that I was describing is that the biggest sine and cosine can be is unity so I figured that I could figure out what angle produces the maximum range.
But that won't maximize anything. :frown: The sines and cosines are a necessary part of the equation, and it's not legal to just arbitrarily set them to 1 (well, at least not in this case, where there are a combination of both sines and cosines). By forcing the sines and cosines to be 1, you are essentially forcing θ to be 90o in parts of the equation, and 0o in other parts. But you don't necessarily want θ to be either 0o or 90o. You want it to be (generally) some other value -- whatever value that maximizes the range R.

Your goal is find the particular θ that maximizes the range R. To do this, find:

[tex] \frac{\partial R}{\partial \theta} = 0 [/tex]

(Take the derivative of your range equation with respect to θ, and set that equal to 0.)

Yeah I tried using 2sin(x)cos(x)=sin(2x) I didn't get anywhere but maybe I don't have the right intuition.
Try it again now, in your unfactored equation. And keep in in your back pocket too, because you'll use it again later.
And yeah I did just try taking the derivative of the range w.r.t. theta...made a HUGE mess. Sorry if I'm not describing things well but I'm still really stuck...
Try it again, after using the trig identity first. :wink:
 
  • #5
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Yes! Thank you so much!

I don't know what I was thinking just setting things to 1 for no reason...haha it seems so ridiculous in retrospect.

It worked out just a couple of more lines after the derivative. Thanks so much. You were really helpful and clear! :smile:
 
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