Basic angular motion problem

1. Feb 24, 2009

KD-jay

1. The problem statement, all variables and given/known data
A small object with mass 3.90 kg moves counterclockwise with constant speed 5.40 m/s in a circle of radius 5.00 m centered at the origin. It starts at the point with position vector (5.00i + 0 j) m. Then it undergoes an angular displacement of 8.00 rad.

(a) What it its position vector?
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
(c) What is its velocity? (in components)
(d) In what direction is it moving? (clockwise/anti?)
(e) What is its acceleration? (also in components)
(f) What total force is exerted on the object? (again in components)

2. Relevant equations
a=v^2/r
F=ma
s=θr
v=ωr
a=αr

3. The attempt at a solution
a) Find the position
<-0.739, 4.94>

Quadrant 2, 98.5 degrees from the horizontal.

c) Velocity?
This is the one I'm having trouble with. First of all, would that 5.40 m/s be angular? If it was angular I would have to multiply it by the radius (5 meters) and find its components. If it's already linear, I would just find the components of that? I don't understand how angular velocity can have components though. Is it like acceleration where one component is towards the center of the circle and the other component is tangent to it?

d) Direction? (Clockwise/Anti)
I think it is going anti-clockwise.

e) Both components of acceleration?
The acceleration would be given by a=v^2/r, but again I don't get how to break this up into its components.

f) Total force on object? (In components)
After I get components of acceleration I'm assuming I can just apply F=ma in both directions separately and get this answer?

Any help would be appreciated.

2. Feb 24, 2009

Staff: Mentor

c) Yes, I believe they are just asking for the vector components of the speed they have given you. They are distinguishing between scalar speed and vector velocity.

e) For uniform circular motion, what is the direction of the acceleration of the object? Since you know this direction, and you have correctly stated the magnitude, you can find the components....

3. Feb 26, 2009

KD-jay

Ok, I've got it. I used the fact that the velocity vector is 90 degrees from the position vector since it's tangential. Finding the acceleration is similar, except the acceleration vector is 180 degrees (i.e. oppositely directed) from the position vector.