Don't know where to start-Uniform Circular Motion

  • Thread starter prelic
  • Start date
  • #1
14
0
At t[tex]_{}1[/tex]=2.00s, the acceleration of a particle in counter-clockwise circular motion is (6.00m/s[tex]^{}2[/tex])i + (4.00m/s[tex]^{}2[/tex])j. It moves at constant speed. At time t[tex]_{}2[/tex]=5.00s, its acceleration is (4.00m/s[tex]^{}2[/tex])i + (-6m/s[tex]^{}2[/tex])j. What is the radius of the path taken by the particle if t[tex]_{}2[/tex] - t[tex]_{}1[/tex] is less than one period?

I really don't even know where to start. I know that the speed is constant, and that I'm probably going to have to use a=v[tex]^{}2[/tex]/r, but I really don't know where to even begin with this.
 

Answers and Replies

  • #2
Doc Al
Mentor
45,184
1,509
Hint: Since the motion is uniform circular motion, where must those acceleration vectors point?
 
  • #3
14
0
the acceleration vectors always point towards the center, but I still don't understand what that means...does that mean something nets to 0?
 
  • #4
one way to tackle this is draw yourself a simple diagram: you have two acceleration vectors at two points of the circular path. Can you work out the angle between the vectors?

You know how long it takes to traverse this angle... so from that you can work out something which will lead you straight to radius given constant speed.
 
  • #5
14
0
ok im starting to understand...before I do all the work for no reason, should i use a dot product with these 2 vectors to figure out the angle?
 
  • #6
Doc Al
Mentor
45,184
1,509
ok im starting to understand...before I do all the work for no reason, should i use a dot product with these 2 vectors to figure out the angle?
Sure. Good idea.
 

Related Threads on Don't know where to start-Uniform Circular Motion

Replies
6
Views
2K
  • Last Post
Replies
9
Views
5K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
4
Views
2K
Replies
3
Views
3K
Replies
5
Views
869
Replies
0
Views
3K
Replies
2
Views
964
  • Last Post
Replies
2
Views
2K
Replies
1
Views
796
Top