Is Gravity Always Negatively Affecting a Ball?

[UrbanXrisis]

whoops... I mean have w^2

accleleration vector=(5.00m)ω^2 sin ωti+ (5.00m) ω^2 cos ωtj

 

[Pyrrhus]

Yes, like that.

 

[UrbanXrisis]

I am asked to describe the path of the object on an xy graph, how would I figure this out? graph it?

 

[Pyrrhus]

By path maybe they mean direction of the velocity (probably displacement), I'm not sure about the question. Could be make a graph...

 

[UrbanXrisis]

well, it first asked for the componets of velocity and acceleration at t=0
then the expression for the velocity and acceleration vector at t>0
then decribe the path of the object on an xy graph, not quite sure what they are getting at for the graph

 

[Pyrrhus]

Well graph for the trajectory, use y and x. assign values to t. then explain how it moves... should look like a sinusoidal or cosinusoidal.

 

[UrbanXrisis]

I think they want me to describe the movement of the object...which is just like a sine curve right? A bouncing ball basically

 

[Pyrrhus]

Yes, basically

 

[UrbanXrisis]

how would I know if it was sinusoidal or cosinusoidal?

 

[Pyrrhus]

At t=0 sinusoidal starts at 0, and cosinusoidal starts at Amplitude. In this case 1 multiplied by the 5w.

 

[UrbanXrisis]

Okay, I have one more problem. You're a great help, seriously. I can't get though physics without PF...anyways. There's a fish that is swimming in a horzontal plane. It has an initial velocity of vi=(4.00i+1.00j)m/s at a point in the ocean whose displacement from a certain rock is r=(10.0i-4.00j)m. After the fish swims with constant acceleration for 20s, it has a v=(20.0i-5.00j) m/s. The componet of acceleration would be...
x=(4.00i-20.0i)/20s
y=(1.00j-(-5.00j))/20s

something like this? acceleration is the change in velocity over change in time...am I heading in the right direction?

 

[Pyrrhus]

It's Final Speed - Initial Speed.

 

[UrbanXrisis]

x=(20.0i-4.00i)/20s
y=(-5.00j-1.00j)/20s

ai=0.8m/s^2
aj=-0.3m/s^2

is this correct?

 

[Pyrrhus]

It is correct.

 

[UrbanXrisis]

so...if the fish swam for 25 seconds...to find its position relative to the rock @ r=(10.0i-4.00j) is to use

di=vit+.5at+xi
di=4.00(25)+(.5)(0.8m/s^2)(25s)+10
di=120

so do the same for dj

is this correct?

 

[Pyrrhus]

Incorrect, the time accompanying the acceleration goes squared.

 

[UrbanXrisis]

di=vit+.5at^2+xi
di=4.00(25)+(.5)(0.8m/s^2)(25s)^2+10
di=360m

 

[Pyrrhus]

It is correct.

 

[UrbanXrisis]

The componet for velocity is x’= -(5.00m) ω cos ωt and y’= (5.00m) ω sin ωt
At t=0 seconds, x’= -(5.00m) ω and y’=0m/s
The componet for acceleration is x’’=(5.00m)ω^2 sin ωt and y’’= (5.00m) ω^2 cos ωt
At t=0 seconds, x’’=0m/s^2 and y’’=(5.00m)ω^2

 

[Pyrrhus]

I was going to suggest working these parametric equations into one whole equation and examine that in the cartesian coordinate system.

 

[UrbanXrisis]

I just started this chapter and do not understand a lot of this...like the cartesian coordinate system

 

[Pyrrhus]

$$x=-(5.00m) sin \omega t$$

$$y=(4.00m)-(5.00m)cos \omega t$$

We could square the x expression so

$$x^2=(25.00m) sin^2 \omega t$$

then apply the pythogoras identities

$$x^2=(25.00m) (1 - cos^2 \omega t)$$

$$x^2=(25.00m) - (25.00m) cos^2 \omega t$$

$$(25.00m) - x^2 = (25.00m) cos^2 \omega t$$

$$\frac{(25.00m) - x^2}{(25.00m)} = cos^2 \omega t$$

$$\sqrt{\frac{(25.00m) - x^2}{(25.00m)}} = cos \omega t$$

Substitute in the other

$$y=(4.00m)-(5.00m)\sqrt{\frac{(25.00m) - x^2}{(25.00m)}}$$

Evaluate this graph, i think it will be better.

 

[Pyrrhus]

Well i graphed it in my calculator, i guess i was wrong, looks like a circle, i was thinking about it too, looked like an elipse expression for a polar coordinate, I'm not sure about the path. In the cartesian it forms an arc.

 

[UrbanXrisis]

I'm trying to understand what you did, is this relating my first problem or my second problem?

 

[Pyrrhus]

I asked a friend (mathematician) he says it will form a cardiode in a polar, and half a circle in cartesian, that should be the path, i don't know how to explain it to you.. but i think the question asked maybe requires a simpler solution, and i must had misunderstood it.

 

[UrbanXrisis]

wow, that is WAY beyond me. What question were you trying to answer?

 

[Pyrrhus]

The trajectory, if you want put an asterisk next to it, see what your teacher says.

 

[UrbanXrisis]

This is all I did:

x=-(5.00m) sin ωt
x=-(5.00m) sin ω(0)
x=0m

y=(4.00m) – (5.00m) cos ωt
y=(4.00m) – (5.00m) cos ω(0)
y= (4.00m) – (5.00m) = 1.00m

The path of the object would represent a sine cure. The x position would start at zero while the y position would start at 1m.

 

[Pyrrhus]

Its path is half a circle actually... Like my friend said. I was wrong about the sine curve.

 

[UrbanXrisis]

so it's like a parabolic arc? Not quite sure how I could describe this.

"The ball will bounce like a parabolic arc"?