tossing a ball

[UrbanXrisis]

When tossing a ball, the ball has a constant negative net force because of the equation F=ma...in that acceleration is always negative so the net force would be negative. Is this correct?

[dav2008]

If you're saying that down is negative then yes.

It's all just a matter of convention really

[UrbanXrisis]

What if an object is moving in the xy plane vary with time according to the equation

x=-(5.00m) sin xt

and

y=(4.00m)-(5.00m)cos xt

where t is in seconds and x has units of seconds^-1

I need to find the componets of velocit and acceleration...
All I need to do is find the derivative of x for the velocity and the second derivative of x for the acceleration. Is my thought process correct?

[Pyrrhus]

The object has movement on y, too. So you need tyo derivate both equations and treat them as component of the velocity, find the magnitude of the velocity for the speed at t instant.

[UrbanXrisis]

I have never seen the notation of seconds^-1

So I have to find the derivative of x and the derivative of y then find the componet by x^2+y^2=c^2?
what is the deveivative of x having units of seconds^-1?

[Pyrrhus]

I've only seen s^-1 so far in Period, which is rad/s, but s^-1 refers to Hertz. The units are there to pretty much cancel each other, so you're left with meters.

[UrbanXrisis]

how would I take the derivative of x=-(5.00m) sin xt?

x'=-5*cosxt*x

btw, x is actually the omega symbol...which I replaced with x

[Pyrrhus]

Now it seems clearer to me, w oh, Simple Harmonic Motion :smile:

Well anyhow, use the chainrule.

Btw, It's not a good math practice to say

x = Acosxt, put w.

[UrbanXrisis]

hahaha whoops!

okay, I didnt know what to write for omega...

x'=-5*cos(wt)*w
x'=-5wcos(wt)

Is the derivative of wt...w?

[Pyrrhus]

It is, w is just a constant.

[dav2008]

Here's how you would display it on these forums: \omega

[UrbanXrisis]

I never learned how to use Latex...but Iwill give it a try...

I am still confused on the whole omega thing. If \omega is a constant, then the derivative of a constant is zero.

x=-5sin\omegat
applying chain rule
x'=-5cos(\omegat^1)*derivative of \omegat

for the derivative of \omegat, do I need to use product rule?

[Pyrrhus]

The derivative of a constant accompyniyn a variable is the constant times the derivative of the variable. or do the product rule, it will yield the same.

[UrbanXrisis]

so x'=-5wcos(wt) was correct?

[Pyrrhus]

It is correct yes.

[UrbanXrisis]

geez... :rolleyes:

Okay, so
x'=-5wcos(wt)
y'=5wsin(wt)

Then then componet of velocity is...x' and y'?

[Pyrrhus]

Yup. :approve:

[UrbanXrisis]

who would I get an expression for the vector of velocity?

[Pyrrhus]

\vec{v} = v_{x} \hat i + v_{y} \hat j

[UrbanXrisis]

what exactly is i and j?

[Pyrrhus]

Unit vectors. They are used to represent a vector. You know a scalar multiplied by the vector is equal another vector with magnitude of the original vector multiplied by the scalar, so because unit vectors has magnitude of 1, you can multiply it by the components of any vector and it will represent a vector in each of the unit directions.

i goes along x - axis
j goes along y - axis
k goes along z - axis

You could also represent your answer by

\vec{v} = (Magnitude, Direction)

[UrbanXrisis]

x=-(5.00m) sin ωt
x’= -(5.00m) ω cos ωt
x’’= -(5.00m) ω * -sin ωt * ω
x’’=(5.00m)ω^2 sin ωt
y= (4.00m) – (5.00 m) cos ωt
y’= 0-(5.00m) –sin ωt * ω
y’= (5.00m) ω sin ωt
y’’= (5.00m) ω^2 cos ωt

The componet for velocity is x’= -(5.00m) ω cos ωt and y’= (5.00m) ω sin ωt
The componet for acceleration is x’’=(5.00m)ω^2 sin ωt and y’’= (5.00m) ω^2 cos ωt

[UrbanXrisis]

velocity vector= vi+ vj
velocity vector= (5.00m) ω sin ωt-(5.00m) ω cos ωt
velocity vector= 5.00m ω (sin ωt- cos ωt)

like this?

[Pyrrhus]

no, like this

\vec{v} = v_{x} \hat i + v_{y} \hat j

\vec{v} = -(5.00m) \omega cos \omega t \hat i + (5.00m) \omega sin \omega t \hat j

[UrbanXrisis]

is the i and j necessary at then end of the equation?

[UrbanXrisis]

\vec{v} = -5.00m\omega (cos \omega t \hat i + sin \omega t \hat j)

is this correct?

[Pyrrhus]

Yes it is necessary, personally i've never seen a an answer in unit vector as a product, so don't factorize it.

[UrbanXrisis]

could I not include the i and j, would that still be valid? so dont factor the vector?

[Pyrrhus]

It wouldn't be valid without the unit vectors, and don't factorize it.

[UrbanXrisis]

accleleration vector= ai+ aj
accleleration vector=(5.00m)ω2 sin ωti+ (5.00m) ω2 cos ωtj

would that be it?

[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 :smile:

[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"?

[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