# Classical mechanics: centripetal acceleration

1. Oct 5, 2006

### Parallel

I need some help with centripetal acceleration.

why when an object is whirling in a vertical circle,it must have a centripetal acceleration larger than 'g' in order for it to get to the top of the circle?

what if I give it just enough speed to get to the top(so when it gets to the top,it's speed is zero)will it fall straight down,or will it just "go back"?

it's really hard for me to conceptualize that

2. Oct 5, 2006

### Staff: Mentor

Take a ball whirling on a string as an example. To maintain a vertical circle, the string must be taut at all times. Analyze the forces acting on the ball when at the top of the circle. To just barely have non-zero tension in the string, what's the minimum speed required according to Newton's 2nd law?

3. Oct 5, 2006

### daniel_i_l

For an object to stay in circular motion it must have a force keeping it from flying away. this is the centrifugal force. for example, in the case of a ball on a string the centrifugal force is equal to the tension on the string. so the C force is non 0 even without gravity. with gravity it just has to be bigger.

4. Oct 5, 2006

### Hootenanny

Staff Emeritus
If one considers the situation properly, from an inertial reference frame, there is no need to introduce an imaginary centrifugal force; one should use centripetal force in order to avoid confusion.

5. Oct 5, 2006

### Parallel

So if I look at the forces acting on the object when it's at the top of the circle ,I get that: the string is pulling it down
and the gravitational force is pulling it down

so: T+mg = ma

T = m(a-g)

so the acceleration must be greater than 'g' ,if I want the string to be taut!.
this is the centripetal acceleration, and this is why it must be larger than 'g'.

Am I getting this right?

I'm not so sure how to caclulate the minumum speed required.

6. Oct 5, 2006

### Hootenanny

Staff Emeritus
Your getting close, but the a in your equation above indicates linear acceleration, which is not what you want. The important thing to remember here is that at minimum speed the sum of the forces must be equal to the centripetal force.

7. Oct 5, 2006

### Staff: Mentor

Yes, everything you said is correct.

Use what you know about centripetal acceleration. How does centripetal acceleration relate to speed?

8. Oct 5, 2006

### Parallel

So should I just substitute v^2/r for a?

About the minumum speed,I'm still not getting it.

9. Oct 5, 2006

### Staff: Mentor

Yes, use that expression for centripetal acceleration. Since you know the minimum acceleration, you can solve for the corresponding minimum speed.

10. Oct 5, 2006

### Parallel

I got it.

I want to thank you all for your help,thank you.