Particle's Motion in XY Coordinate System

[OcaliptusP]

Homework Statement

Coordinates of a particle which moves on a xy coordinate system given with:
x=-(5m)sinωt
y=(4m)-(5m)cosωt
In these equlations t's unit given as second, and ω's unit
second^-1. A-) Found velocity and acceleration components when t=0 B-) Write equlations for position and acceleration vektors when t>0C-)Define particle's way
on a xy graph.

Homework Equations

V=dx/dt a=dV/dt

The Attempt at a Solution

I tried to take deridatives of equlations, but I couldn't.
Can someone explain deridativation method clearly:D

 

[person123]

Can someone explain deridativation method clearly:D

Are you sure you didn't mean derivation? Derivation can allow you to find the motion of a particle at a specific point in time.

 

[OcaliptusP]

Yes my English is not well so there can be some writing mistakes. I mean I actually cannot derivate the equlations which given in that problem

 

[person123]

Do you know how to find the derivatives of trig functions? Also, where is $t$ in the functions, and what does $m$ indicate?

 

[OcaliptusP]

Okay I am so sorry this is my first post. In the equlations y's at right side are actually t I know how to derivate trig functions

 

[OcaliptusP]

Do you know how to find the derivatives of trig functions? Also, where is $t$ in the functions, and what does $m$ indicate?

I think I fixed it

 

[person123]

So is $x=-5m\sin(\frac 1 t) t$? and $y=4m-5m\cos(\frac 1 t) t$? I'm guessing $m$ is a constant. By the way, if you want to write equations more clearly here's the page which shows you how: https://www.physicsforums.com/help/latexhelp/. I found it pretty simple.

 

[OcaliptusP]

So is $x=-5m\sin(\frac 1 t) t$? and $y=4m-5m\cos(\frac 1 t) t$? I'm guess $m$ is a constant.

Yes m is meter unit. Equlations are correct

 

[person123]

Alright. So if you find the derivative with respect to time you will get the velocity. Do it again and you'll get acceleration. Knowing the derivative of trig functions and the chain rule should make it manageable.

 

[OcaliptusP]

Alright. So if you find the derivative with respect to time you will get the velocity. Do it again and you'll get acceleration. Knowing the derivative of trig functions and the chain rule should make it manageable.

I cannot derivate it still :/ can you explain me step by step

 

[person123]

I cannot derivate it still :/ can you explain me step by step

Do you know the chain rule?

 

[OcaliptusP]

Do you know the chain rule?

Yes it is df(g(x))/dx=g'(x)f'(g(x)) am I wrong?

 

[scottdave]

Originally you have this:
x=-(5m)sinωt
y=(4m)-(5m)cosωt

In these equations t's unit given as second, and ω's unit
second^-1.

Then somehow you have x = (-5 meters)sin(1/t)t, but where did the extra t come from? You need to keep ω in there, which has dimension of (1 / time) which cancels out the time [the argument of sine or cosine need to be dimensionless].

 

[OcaliptusP]

Originally you have this:
x=-(5m)sinωt
y=(4m)-(5m)cosωt

In these equations t's unit given as second, and ω's unit
second^-1.

Then somehow you have x = (-5 meters)sin(1/t)t, but where did the extra t come from? You need to keep ω in there, which has dimension of (1 / time) which cancels out the time [the argument of sine or cosine need to be dimensionless].

Yes when you dimensionally analyze the problem this problem occurs. The problem is about problem then ?

 

[scottdave]

Let's say you have f(θ), and you want to take the derivative with respect to t, and θ is a function of t, then you have df/dt = (df/dθ)*(dθ/dt).
So for sin(θ) you have (d/dt) of sin(θ) = cos(θ)*(dθ/dt). Since θ = ωt, then dθ/dt = ω

 

[scottdave]

Looking at x(t) = (-5 meters)*sin(ωt), taking the first derivative gives (ω)*(-5 meters)*cos(ωt) {trig function and chain rule}. Note that this now has the dimension of {length / time} which is velocity. Taking the derivative of velocity {second derivative of x} gives a = x'' = (ω²)*(5 meters)*sin(ωt), which has the dimension of {length / time²} which is acceleration. Do similar for the y component.

 

[OcaliptusP]

Let's say you have f(θ), and you want to take the derivative with respect to t, and θ is a function of t, then you have df/dt = (df/dθ)*(dθ/dt).
So for sin(θ) you have (d/dt) of sin(θ) = cos(θ)*(dθ/dt). Since θ = ωt, then dθ/dt = ω

I unterstood until last equlation but why θ=ωt? Did I missunderstood something in question?

 

[scottdave]

I unterstood until last equlation but why θ=ωt? Did I missunderstood something in question?

This was given in the problem statement: x=-(5m)sinωt. It probably should have been clarified as x=-(5m)sin(ωt). Omega is the angular frequency, and multiply that by time gives you a dimensionless quantity as the argument for sine.

 

[OcaliptusP]

Okay now I understand that I completely miss understood that question I assumed that x= -5m sin(ω)t
not x= -5sin(ωt) Thanks for all your help

 

[Ray Vickson]

I cannot derivate it still :/ can you explain me step by step

PF helpers are not allowed to do that; read the posting rules, which state that YOU must first do some work on the problem before receiving help.