# Acceleration of particle along the curve

1. May 27, 2015

### Tanya Sharma

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

2. Relevant equations

3. The attempt at a solution

Since the tangent moves at uniform angular speed ,the particle also moves with constant angular speed . Is it correct ?

If it is right then the particle has only radial acceleration given by $ω^2r$ ,where 'r' is the radius of curvature .

r = sec2x

r ( x= πa/4) = 1/√2 .

Hence resultant acceleration at x= πa/4 = 2√2 .

I don't have the answer key . Have I interpreted the question correctly and done it right ?

I would be grateful if somebody could verify my work .

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2. May 27, 2015

### SammyS

Staff Emeritus
It may be true that the particle moves with constant angular speed (not sure how you would define that), but not for the reason you give.

I think that the problem statement is saying that the angle the tangent line makes with the x-axis increases at a constant rate.

3. May 27, 2015

### Tanya Sharma

Ok . So what should be the correct approach ?

4. May 27, 2015

### SammyS

Staff Emeritus
Get dy/dx.

Slope of tangent line = tan(θ), where θ is the angle the tangent line makes w.r.t the x-axis. Right ?

Start there.

5. May 27, 2015

### D H

Staff Emeritus
No, that isn't correct.

Angular velocity is not the way to solve this problem. As Sammy mentioned, determining dy/dx is the first step.

6. May 27, 2015

### D H

Staff Emeritus
The problem statement says the tangent to the curve rotates with a constant angular velocity. I would take that as meaning the unit tangent.

HINT: The magnitude of the time derivative of any unit vector is the rate at which that unit vector is rotating.

7. May 27, 2015

### Tanya Sharma

Ok .

dy/dx = tan2x i.e tanθ = tan2x .

θ = 2x .

dx/dt = 1

Is it correct ? If yes , what next ?

8. May 27, 2015

### SammyS

Staff Emeritus
Where did the $\ a\$ go ?

9. May 27, 2015

### Tanya Sharma

Sorry .

dy/dx = tan(x/a) i.e tanθ = tan(x/a)

θ = x/a

dθ/dt = (1/a)dx/dt

dx/dt=2a

Is it correct now ?

10. May 27, 2015

### SammyS

Staff Emeritus
Looks good!

11. May 27, 2015

### D H

Staff Emeritus
What's that two doing there? (That's a devil's advocate kind of question.)

I find that it's best to work symbolically until the very, very end. If needed (sometimes it's not needed), I'll plug in the numerical values, but only at the very, very end.

12. May 27, 2015

### Tanya Sharma

I have worked symbolically only . In fact what I did in post#7 is also correct , since 'a' is a constant .

Last edited: May 27, 2015
13. May 27, 2015

### Tanya Sharma

Is the tangential acceleration equal to $4m/s^2$ ?

Is net acceleration equal to $2\sqrt{6}m/s^2$ ?

14. May 27, 2015

### SammyS

Staff Emeritus
I don't mind that 2 so much, since early in the problem we're told the tangent line rotates with constant angular speed of 2 rad/s . Otherwise, we need to carry around some ω0, which I suppose isn't so bad. Plugging in a = 1/2 so early in the game could be a bad idea.

15. May 27, 2015

### SammyS

Staff Emeritus
I did not get that.

Can you show some intermediate steps?

16. May 27, 2015

### ehild

The curve is given in Cartesian coordinates, so the acceleration can be expressed by $\ddot x(t), \ddot y(t)$. We also know that dx/dt =2a, constant.

17. May 27, 2015

### Tanya Sharma

Is $4m/s^2$ the net acceleration ?

18. May 27, 2015

### ehild

I think yes. Why aren't you sleep yet?

19. May 27, 2015

### Tanya Sharma

Hooray!!! Now I can sleep peacefully .

Thanks ehild , Sammy , D H .

20. May 27, 2015

### D H

Staff Emeritus