# Relative motion in a circle

1. Sep 10, 2016

### nysnacc

1. The problem statement, all variables and given/known data

2. Relevant equations
a= omega^2*r

3. The attempt at a solution
a_A/B = a_A - a_B

a_A = 10^2*r (-î)
a_B = 10^2*r (+ĵ)

a_A/B = 10^2*r (-î) - 10^2*r (+ĵ)
=> -200 î -200 ĵ

why is the answer 200 î -200 ĵ (accelerate in +î direction) ???

2. Sep 10, 2016

### Student100

The question is asking for the acceleration of A relative to B. What's the acceleration at A and at B independently?

Hint: Check the direction of acceleration vectors.

Last edited: Sep 10, 2016
3. Sep 10, 2016

### nysnacc

Err... isn't my expression correct..?

a_A = 10^2*r (-î). A is going -x, -y direction
a_B = 10^2*r (+ĵ). B is going -x, +y direction

4. Sep 10, 2016

### Student100

No. Should your acceleration be tangential or radial to the disk? Remeber the disk has constant angular velocity.

5. Sep 12, 2016

### nysnacc

should be tangential cuz constant radius??

6. Sep 12, 2016

### Student100

Actually it should only have radial acceleration. Can you see why?

7. Sep 12, 2016

### nysnacc

because constant rotation?

8. Sep 12, 2016

### nysnacc

so the a_r for B is pointing from point B to Point O
and for A it is pointing from Point A to Point O?

9. Sep 13, 2016

### Student100

Exactly, so what's your relative acceleration now?

10. Sep 13, 2016

### nysnacc

so the a_r for B is omega^2*r (-i)
a_r for A is omega^2*r (-j)

a_A/B = a_A - a_B = -omega^2*r (j) - -omega^2*r (i)
rearrange = omega^2*r (i) - omega^2*r (j)

Got it!! Thanks!