Circular motion of a particle

1. Sep 22, 2007

HeLLz aNgeL

Suppose that a particle's position is given by the expression in the attachment

Explain:
1. When does the particle first cross the negative x axis?

2. Find the particle's velocity as a function of time. Express your answer using unit vectors.(e.g., A i_unit+ B j_unit, where A and B are functions of omega, R, t, and pi).

3. Find the speed of the particle at time t. Express your answer in terms of some or all of the variables.

4. Find the speed of the particle at time t. Express your answer in terms of some or all of the variables omega, R, and pi.

5. Now find the acceleration of the particle. Express your answer using unit vectors (e.g., A i_unit+ B j_unit, where A and B are functions of omega, R, t, and pi).

6. Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r_vec(t).
Express your answer in terms of some or all of the variables r_vec(t) and omega.

ok, i know that omega gives you the angular velocity, but i dont understand how that factors into the "i" and "j" equations... are they the x, and y axis components ??? if they are do i have to plug in values one-by-one to check when it crosses the negetive x-asis ???

i'm confused !

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2. Sep 22, 2007

FedEx

I am sorry,but i am not able to see the picture.

3. Sep 22, 2007

Vidatu

It'll cross the x axis when omega*time is equal to pi, if that helps

4. Sep 22, 2007

HeLLz aNgeL

there was no picture, just the equation i attached in the first post..

5. Sep 22, 2007

FedEx

I know that there is just an equation but i am not even able to see that.And by the way for posting equations try using latex.

6. Sep 23, 2007

fubag

i am seeing the picture,

and have similar questions to ask, mostly what type of circle would this represent at t= 0?

An ellipse, or circle, in positive y or x direction?

Also if no one can see the formula its:

r(->) = Rcos(omega*time)(i^) + Rsin(omega*time)(j^)

7. Sep 23, 2007

FedEx

Ok,I am sure that you people would be knowing differentiation.

For the first question the answer has already been given by someone.

For the second, the answer would be obtained by taking the time derivative of the equation which you already have.

For the fifth take the time derivative of the equation you are getting after solving the second question.

For the third and the fourth i think that there is some problem but still you can solve it by the scalar relations given as follows

$$v$$ = $$r\omega$$

and

$$t\omega$$ = $$\theta$$.

Moreover while differentiating the equations you can treat $$t\omega$$ = $$\theta$$

8. Sep 23, 2007

HeLLz aNgeL

but what would r be ???

9. Sep 23, 2007

FedEx

Sorry. Its the radius. I must have written it as R and not r.

10. Sep 23, 2007

HeLLz aNgeL

but we dont have the radius anyhwere in the question ??

11. Sep 23, 2007

FedEx

The question is asking to define in terms of omega and radius and all.And in the equation which you have written what is R? It is the radius.

12. Sep 23, 2007

fubag

does this then mean:

v = sqrt. ( (-Rwsinx)^2 + (Rwcosx)^2)

where x is equal to theta?

i also noticed that they told us a hint to look for an important trig identity.

im assuming it will most likely be (sinx)^2 + (cosx)^2= 1.

Sorry, I got it, never mind.

Last edited: Sep 23, 2007
13. Sep 23, 2007

fubag

having difficulty with one more part:

Finally, express the magnitude of the particle's acceleration in terms of R and v using the expression you obtained for the speed of the particle.

I am getting a = v*omega...but this isn't right it seems.

It says to express in terms of R and v (speed function of the particle)

I know that a(t) = Rw^2
and that v(t) = Rw

now if I relate them to each other shouldn't I get = v*w...?

this doesn't work however.

Never mind, I got it... sorry.

Last edited: Sep 23, 2007