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Projectile Motion and acceleration of particle

  • #1
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Homework Statement


If a particle moves in X-Y plane with acceleration non zero in X and Y , the particle will not move in a parabolic path
True or False ?

Homework Equations


X = UxT + 1/2(Ax)T^2
Y= UyT + 1/2 (Ay)T^2


The Attempt at a Solution


The equation of trajectory that i came up with involved Y^2 , X^1/2 and X
I am not able to draw conclusions with it
So i guessed it to be True as for a parabolic path you need constant acc in only one axis and zero acc in other axis
But the book says false
Please help !
 

Answers and Replies

  • #2
berkeman
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Welcome to the PF. :smile:

Think about how a projectile traces out a parabolic shape when you throw it up and out at a 45 degree angle. What forces and accelerations are acting on it in that situation (after it leaves your hand)?
 
  • #3
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Welcome to the PF. :smile:

Think about how a projectile traces out a parabolic shape when you throw it up and out at a 45 degree angle. What forces and accelerations are acting on it in that situation (after it leaves your hand)?
Welcome to the PF. :smile:

Think about how a projectile traces out a parabolic shape when you throw it up and out at a 45 degree angle. What forces and accelerations are acting on it in that situation (after it leaves your hand)?
Just the gravity pull and air friction
 
  • #4
berkeman
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Just the gravity pull and air friction
Good. So neglecting air friction (which will alter the path away from a pure parabola), in how many dimensions is the projectile accelerating while it traces out the parabola? :smile:
 
  • #5
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Good. So neglecting air friction (which will alter the path away from a pure parabola), in how many dimensions is the projectile accelerating while it traces out the parabola? :smile:
2 dimensional motion with downward acceleration (g)
 
  • #6
berkeman
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@AkshayM -- You marked this thread as solved -- does that mean you understand now?
 
  • #7
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@AkshayM -- You marked this thread as solved -- does that mean you understand now?
Sorry no bymistakely i did it
 
  • #8
berkeman
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If a particle moves in X-Y plane with acceleration non zero in X and Y , the particle will not move in a parabolic path
True or False ?
But the book says false
So the book is saying that the path is parabolic even when there is acceleration in 2 dimensions?
 
  • #9
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So the book is saying that the path is parabolic even when there is acceleration in 2 dimensions?
Yes it says parabolic path
Case of a misprint ?
 
  • #10
berkeman
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Maybe. Can you ask the professor or a TA about it? Do you know any other students in your class who are also working on the problem?
 
  • #11
haruspex
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Homework Statement


If a particle moves in X-Y plane with acceleration non zero in X and Y , the particle will not move in a parabolic path
True or False ?

Homework Equations


X = UxT + 1/2(Ax)T^2
Y= UyT + 1/2 (Ay)T^2


The Attempt at a Solution


The equation of trajectory that i came up with involved Y^2 , X^1/2 and X
I am not able to draw conclusions with it
So i guessed it to be True as for a parabolic path you need constant acc in only one axis and zero acc in other axis
But the book says false
Please help !
First, let us be clear that the question is asking whether a nonzero acceleration in both coordinates guarantees that it will not be a parabola. My point is nothing has been said about constancy of acceleration, so certainly it might not be a parabola.

So now suppose there is constant acceleration in each coordinate. What if you were to use different axes? Might there be a direction in which there is no acceleration? Will there always be such a direction?
 
  • #12
Ray Vickson
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Yes it says parabolic path
Case of a misprint ?
To re-express the answer of haruspex in post #11: if you take the path to be ##x = v_x t + \frac{1}{2} a_x t^2## and ##y = v_y t + \frac{1}{2} a_y t^2,## (with constants ##v_x, v_y, a_x,a_y##) can you find a constant angle ##\theta## and a rotated coordinate system
$$\begin{array}{rcl}
X&=& x \cos(\theta) - y \sin (\theta)\\
Y&=& x \sin (\theta) + y \cos (\theta)
\end{array} $$ in which the new path has the form ##X = U_x t, \; Y = U_y t + \frac{1}{2} W_y t^2,## (with constant ##U_x, U_y, W_y##) or is that not possible? If it is possible, that would give you a "rotated" parabola.
 
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