Find the velocity and acceleration vectors for uniform circular motion

In summary: This derivative is:a(t) = -{\omega}R{\sin}\left({\omega}t\right)\hat{i}+{\omega}R{\cos}\left({\omega}t\right)\hat{j}
  • #1
cdlegendary
15
0

Homework Statement


To find the velocity and acceleration vectors for uniform circular motion and to recognize that this acceleration is the centripetal acceleration.
Suppose that a particle's position is given by the following expression:
%7D%28t%29%3D+R%5B%5Ccos%28%5Comega+t%29%5Chat%7Bi%7D+%2B+%5Csin%28%5Comega+t%29%5Chat%7Bj%7D%5D.gif

2C%5C+%5Cqquad+%3D+R%5Ccos%28%5Comega+t%29%5Chat%7Bi%7D+%2B+R%5Csin%28%5Comega+t%29%5Chat%7Bj%7D.gif


A.) Choose the answer that best completes the following sentence:
The particle's motion at t=0 can be described by ____________.

B.) When does the particle first cross the negative x axis?
Express your answer in terms of some or all of the variables omega, R, and pi.

C.) 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).

D.) 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.

E.) 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).

F.) 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.

H.) 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.
Express your answer in terms of one or both of the variables R and v.

The Attempt at a Solution


I did parts A-D. But am stuck starting with Part E, and haven't really tried the rest
below is the particle's velocity as a function of time. I know to get acceleration you're supposed to take the derivative of that, but I seem to be getting it wrong. If anyone could help me out I would appreciate it.

t%29%5Chat%7Bi%7D%2B%7B%5Comega%7DR%7B%5Ccos%7D%5Cleft%28%7B%5Comega%7Dt%5Cright%29%5Chat%7Bj%7D.gif
 
Last edited:
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  • #2
The question is not complete. Post the complete question.
 
  • #3
Alright sorry, whole question is up.
 
  • #4
Your method sounds good. What did you get for the second derivative of r(t)?
 
  • #5
derivative of r(t) is:
t%29%5Chat%7Bi%7D%2B%7B%5Comega%7DR%7B%5Ccos%7D%5Cleft%28%7B%5Comega%7Dt%5Cright%29%5Chat%7Bj%7D.gif


My main issue is I can't find the derivative of a(t) from the v(t) eq. or at least, I seem to be doing it wrong.
 
  • #6
You found the derivative of r(t):

[tex]v(t) = \frac{dr}{dt} = -{\omega}R{\sin}\left({\omega}t\right)\hat{i}+{\omega}R{\cos}\left({\omega}t\right)\hat{j}[/tex]

So find a(t)=dv/dt in just the same way.
 
  • #7
would the derivative be:
ght%29%5Chat%7Bi%7D-R%7B%5Comega%7D%7B%5Csin%7D%5Cleft%28%7B%5Comega%7Dt%5Cright%29%5Chat%7Bj%7D.gif
 
  • #8
Almost. You forgot to bring out the omega. It might help to ignore the scalars (constants R and omega on the OUTSIDE of your functions) when taking the derivative. Try just

[tex]\frac{d}{dt}( -{\sin}\left({\omega}t\right)\hat{i}+{\cos}\left({\omega}t\right)\hat{j})[/tex]

then put the factors back in! (Hint: you should get omega squared!)
 

1. What is uniform circular motion?

Uniform circular motion is a type of motion in which an object moves in a circular path at a constant speed.

2. How do you find the velocity vector for uniform circular motion?

The velocity vector for uniform circular motion can be found by taking the derivative of the position vector with respect to time. This will give you the direction and magnitude of the velocity at any given point along the circular path.

3. What is the acceleration vector for uniform circular motion?

The acceleration vector for uniform circular motion is directed towards the center of the circle and has a magnitude equal to the square of the velocity divided by the radius of the circle.

4. How is uniform circular motion different from linear motion?

In uniform circular motion, the direction of the velocity is constantly changing, whereas in linear motion, the direction of the velocity remains constant. Additionally, in uniform circular motion, there is always an acceleration towards the center of the circle, while in linear motion, there may be no acceleration or an acceleration in a different direction.

5. Can uniform circular motion have a changing speed?

No, uniform circular motion by definition has a constant speed. However, the velocity vector is changing due to the changing direction of the object's motion.

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