# Centripetal acceleration

1. Sep 16, 2010

### Ry122

A car starts from rest and travels along a circular path of radius 200 m. The magnitude of its velocity
increases at a rate of 2 m/s

After 10 s, what is the magnitude of the acceleration of the car (in m/s

my attempt:
v = 10*2 = 20m/s

would the acceleration that the car's undergoing be equal to the centripetal acceleration?
so would it be v^2/r ?

2. Sep 16, 2010

### gabbagabbahey

What is the general expression for velocity in plane polar coordinates? How about acceleration? Can you represent the statment "The magnitude of its velocity increases at a rate of 2 m/s" by an equation?

3. Sep 16, 2010

### Ry122

But you see, no longer is the car engine accelerating the vehicle after 10s and thus the only acceleration that I think might be acting on it at that point is the one keeping the car moving in circular motion ie. the centripetal acceleration.

4. Sep 16, 2010

### gabbagabbahey

Where does it say that?

5. Sep 16, 2010

### Ry122

You're right, it doesn't. So is it going to be the centripetal acceleration combined with the acceleration due to the car's engine?

The acceleration due to the car's engine would be a vector that is tangent to the edge of the circle it's travelling in while the centripetal acceleration vector would be perpendicular to the edge of the circle. Is this correct?

6. Sep 16, 2010

### hiuting

Well according to my knowledge, in rotational kinematics, acceleration is perpendicular to the velocity. Therefore it should be centripedal acceleration.

7. Sep 16, 2010

### diazona

Only centripetal acceleration is perpendicular to the velocity. That is not necessarily the total acceleration.

8. Sep 17, 2010

### gabbagabbahey

It depends on what you mean by "combined" each of these tyoes of acceleration are components of a vector (acceleration)...,.how do you determine the magnitude of a vector?

Yes, you can see this by looking at the general expression for position, velocity and acceleration in polar coordinates:

$$\textbf{r}(t)=r(t)\textbf{e}_r$$

Now, $$\frac{d\textbf{e}_r}{dt}=\dot{\theta}\textbf{e}_{\theta}$$ and $$\frac{d\textbf{e}_\theta}{dt}=-\dot{\theta}\textbf{e}_{r}$$ , so

$$\textbf{v}(t)=\frac{d\textbf{r}}{dt}= \dot{r}\textbf{e}_r+r\dot{\theta}\textbf{e}_{\theta}$$

And

$$\textbf{a}(t) = \frac{d\textbf{v}}{dt}= (\ddot{r}-r\dot{\theta}^2)\textbf{e}_r+(r\ddot{\theta}+2\dot{r}\dot{\theta})\textbf{e}_{\theta}$$

For circular motion, $\dot{r}=\ddot{r}=0$ and you see that the centripetal acceleration $$r\dot{\theta}^2$$ is directed radially inwards, and the tangential acceleration is $$r\dot{\theta}$$...the rate of change of the speed [tex]v=r\dot{\theta}[/itex]