Is Gravity Always Negatively Affecting a Ball?

[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"?

 

[Pyrrhus]

Yea, it seems.

 

[UrbanXrisis]

Could you check if I did this correct...

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]

Looks ok to me.

 

[UrbanXrisis]

I wasnt sure if the omega symbol needed to be included in the answer or not, just checking