Spiraling Up

  • Thread starter Toranc3
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  • #1
Toranc3
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Homework Statement


It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 6.00m every 5.00s and rises vertically at a rate of 3.00m/s.

Part C: Find the direction of the bird's acceleration.

Part D: Find the angle between the bird's velocity vector and the horizontal

Homework Equations



Acceleration radius = v^(2)/R

V=(2pi*R)/T

Acceleration radius = (4*pi^(2)*R)/T^(2)


The Attempt at a Solution



I already found the speed of the bird relative to the ground

V=(2pi*R)/T= 2*pi*6.00m/5.00 sec = 7.54m/s

Vmagnitude= sqrt[(7.54m/s)^(2)+(3.00m/s)^(2)]

v=8.11m/s

I also found the magnitude of the birds acceleration

Arad.=v^(2)/R =(7.54m/s)^(2)/6.00m

a=9.48 m/s^(s)

Not sure how to go about the direction of acceleration though.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
SHISHKABOB
537
1
think about the acceleration in terms of of components

is the bird accelerating up or down? Is it accelerating in a direction parallel to the ground?
 
  • #3
Toranc3
189
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think about the acceleration in terms of of components

is the bird accelerating up or down? Is it accelerating in a direction parallel to the ground?

Well the bird is going at a constant 3.00m/s. Wouldn't that mean that it would have zero acceleration along that path?
 
  • #4
SHISHKABOB
537
1
if there's no change in velocity then there can't be any acceleration, that's by definition
 
  • #5
Toranc3
189
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Sorry but I want to start from the beginning of this problem to fully understand it because I do not know what is going on.

A: Find the speed of the bird relative to the ground.

Can somedody tell me if I have my picture correctly? this is what I have.

[url=http://www.freeimagehosting.net/y9ewq][PLAIN]http://www.freeimagehosting.net/t/y9ewq.jpg[/url][/PLAIN]

So in my z axis that is my 3.00 m/s right? How do I find my x velocity and my y velocity?
Do I use the equation V=2piR/T....if so what velocity does that give me? The y one or the x one? I am not understanding what is going on in this problem. Any help would be nice.
 
  • #6
SHISHKABOB
537
1
basically, in physics, a key strategy for solving problems like this is to separate the components

the way you separate them is dependent on your coordinate system, and in this case we want to separate them into x, y and z because that's the way you've set it up

the reason why we separate the components, is because *a component that is entirely in one direction does not affect what happens to things in the other direction*

Have you done projectile motion problems? I'm assuming you probably have. In those problems, you want to separate the components of the object's velocity into the y and x directions. Vertical and horizontal, respectively. This is because the force of gravity *only* affects the object in the vertical direction. It does not affect the object in the horizontal direction.

Think about it this way: if you have a crate, and you tie a rope around it and then pull that rope so that the rope is parallel with the ground, will any amount of force applied ever end up lifting the crate off the ground?

So in this problem how would you separate the components? Your picture is on the right track.
 
  • #7
Toranc3
189
0
basically, in physics, a key strategy for solving problems like this is to separate the components

the way you separate them is dependent on your coordinate system, and in this case we want to separate them into x, y and z because that's the way you've set it up

the reason why we separate the components, is because *a component that is entirely in one direction does not affect what happens to things in the other direction*

Have you done projectile motion problems? I'm assuming you probably have. In those problems, you want to separate the components of the object's velocity into the y and x directions. Vertical and horizontal, respectively. This is because the force of gravity *only* affects the object in the vertical direction. It does not affect the object in the horizontal direction.

Think about it this way: if you have a crate, and you tie a rope around it and then pull that rope so that the rope is parallel with the ground, will any amount of force applied ever end up lifting the crate off the ground?

So in this problem how would you separate the components? Your picture is on the right track.

Well I know in general you would seperate the componenets by using the formulas

Vx=Vcos(theta)
Vy=Vsin(theta)

I do not have the angle though. I also know that velocity is tangent to the circle and is always perpendicular to acceleration (Arad) of the circle. I am not sure how to apply it to this problem though.
 
  • #8
SHISHKABOB
537
1
you had the right idea with the uniform circular motion. The circular motion that the bird is making is independent from the upward rising motion.

If you can figure out the acceleration that the bird is undergoing from the circular motion, then you can compare it to any other accelerations that it is also undergoing, if it is undergoing any other accelerations at all.
 
  • #9
Toranc3
189
0
you had the right idea with the uniform circular motion. The circular motion that the bird is making is independent from the upward rising motion.

If you can figure out the acceleration that the bird is undergoing from the circular motion, then you can compare it to any other accelerations that it is also undergoing, if it is undergoing any other accelerations at all.

Ok so I have Vz=3.00m/s that is given. NOw I need to find Vx and Vy. When the bird travels in uniform circular motion once around the whole circle I get a velocity during that whole time right? which goes back to v=2piR/t. Would that be my total velocity?
 
  • #10
SHISHKABOB
537
1
the velocity determined from the uniform circular motion would be the component of the bird's velocity that is parallel to the x-y plane
 
  • #11
Toranc3
189
0
the velocity determined from the uniform circular motion would be the component of the bird's velocity that is parallel to the x-y plane

So the velocity determined from the uniform circular motion is one of my components then right? Then I have my other componenet in the Z direction. If this is so then wouldn't it have been easier for me to make only a y and x coordinate system? I thought I had to find Vx and Vy seperately. Or am I wrong?
 
  • #12
SHISHKABOB
537
1
So the velocity determined from the uniform circular motion is one of my components then right? Then I have my other componenet in the Z direction. If this is so then wouldn't it have been easier for me to make only a y and x coordinate system? I thought I had to find Vx and Vy seperately. Or am I wrong?

well, you could do that, but in this problem it asks for the *speed* of the bird relative to the ground. The velocity is a vector, and so has a direction and magnitude. The speed, then, is just the magnitude of the velocity.
 
  • #13
Toranc3
189
0
well, you could do that, but in this problem it asks for the *speed* of the bird relative to the ground. The velocity is a vector, and so has a direction and magnitude. The speed, then, is just the magnitude of the velocity.

So the velocity of uniform circular motion is my Vxy and then I have my Vz. Square those and take the square root of them to get My speed with relative to the ground. Could I find vx or vy seperately in this problem? Or does the uniform circular motion just solve for my vxy?
 
  • #14
SHISHKABOB
537
1
you could find vx and vy, but I think that would require vector notation and that doesn't seem to be what the problem is asking for
 

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