# Can someone explain to me what centripetal acceleration?

1. May 18, 2015

### toesockshoe

1. The problem statement, all variables and given/known data
I also need help on a problem. but FIRST, can someone explain to me how radial acceleration relates to gravity and how gravity can be split up into tangential and radial accelerations and how that can be compared to radial acceleration ? I know the basic stuff..like if a object is moving in a circle with constant speed it has radial acceleration, but what happens when gravity comes into play? Look below for a more specific problem oriented question. I was wondering why g is not part of the centripetal acceleration formula, v^2/r.

A small block of mass m slides along the frictionless loop-the-loop track (of radius r) shown in the diagram. What would the minimum height be such that the block goes all the way around the loop?
Here is a visual representation: http://nebula.deanza.edu/~newton/4A/4AHWSet6.html (its number 2)
2. Relevant equations
$$W= \Delta E$$
KE = 1/2mvf^2=1/2mvi^2
GPE = mgh

3. The attempt at a solution
I chose the system to be the mass and earth. I set the equilibrium point to be the top of the loop.

$$W = \Delta E$$
$$W_{normal} = \Delta GPE + \Delta KE$$
Work done by nomal force is 0.
$$0 = mg(h-2r)+\frac{1}{2}mv_f^2$$
solve for h:
$$h=\frac{v_f^2}{2g}+2r$$

Now Ill do a F=ma problem to find the velocity.
Let my system be the mass at the top of the loop wit the y direction going down.

$$F_g-F_n = ma_c$$
$$F_g -ma_c = F_n$$
$$mg- \frac{mv^2}{r} = F_n$$

HERE IS MY QUESTIONS: Why is the acceleration the same as uniform circular motion acceleration? Shouldnt gravity also be factored into the acceleration. I was wondering why the acceleration can't be:
$$a_c = \frac{mv^2g}{r} [tex] anyway the rest of the problem is straightforward after and I know the answer... but my main question is why is gravity not in the problem. 2. May 18, 2015 ### AlephNumbers The block is not moving in uniform circular motion because the magnitude of the block's speed changes as it moves up the loop the loop; the block's kinetic energy changes as gravity does work on the block. It isn't. Also, you made a sign error in your work with forces. The force of gravity and the normal force are both pointing in the same direction, towards the center of the circle, when the block is at the top of the loop the loop. What would you expect the velocity of the block to be when it is at the top of the loop the loop, if the kinetic energy of the block at the bottom of the loop the loop is just barely sufficient to make the block move all the way through the loop the loop? 3. May 18, 2015 ### toesockshoe so what would the acceleration be? my teacher in class said it is [tex] \frac{v^2}{r}$$

4. May 18, 2015

### AlephNumbers

Do you remember the context of the situation in which your teacher made that statement?

5. May 18, 2015

### toesockshoe

he said that at the top, there is no horizontal force... so its a one d problem. thus he said there is no gravity ONLY at the top, so there will only be centripetal acceleration, which he later went on to say was v^2/r. for any other point on the loop there would be a tangential acceleration, so i guess my question is how would I factor in this tangnetial acceleration if he asked us to find for example the normal force when the mass is 45 degrees from the vertical

6. May 19, 2015

### AlephNumbers

Do you want me to elaborate on what your teacher said?

If you are interested in solving for the tangential acceleration of the mass when it is at a certain angle with the vertical, then you need to break gravity up into tangential and radial components.

7. May 19, 2015

### toesockshoe

alright, i looked at a couple of other threads and here is my new question: why is the tangential acceleration at the top of the loop 0?

8. May 19, 2015

### toesockshoe

alright, so i split up gravity into tangental and radial components. at any point on the sphere, theta above the horizontal, the radial acceleration is Fgsin(theta) and the tangential acceleration is Fgcos(theta) correct?

and because at the top, it is a 90 degree angle with the horizontal, Fgcos(90) is 0 and thus there is no tangential acceleration. however, at the top the radial acceleration is Fgsin(90) which is Fg... so that means Fg has to be the radial acceleration which isnt correct right?

9. May 19, 2015

### AlephNumbers

The sin and cos would be switched. When the block gets to the upper half of the loop the loop, I would flip the circle so that the origin is centered on the top of the loop the loop, and then switch the signs on the terms.

The centripetal force at the top of the loop the loop would be the force of gravity plus the normal force. The specific problem you stated above uses this.

Now, this is important: If the kinetic energy of the block when it is at the bottom of the loop the loop is just barely sufficient to keep the block in the loop the loop when it goes through the top of the loop the loop, what is the velocity of the block when it is at the top of the loop the loop?

10. May 19, 2015

### toesockshoe

it is NOT 0... i know that for sure. i do know that the normal force has to equal the gravitational force at the top of the loop. so the normal force would be 0. i found veloicty using F=ma... how can you do it using energy thinking?

11. May 19, 2015

### AlephNumbers

Your solution to the problem using energy says otherwise. If all of the kinetic energy the block has at the bottom of the loop is converted into gravitational potential energy by the time the block is at the top of the loop, then the kinetic energy the block has at the top of the loop is zero. Since the mass of the block is presumably invariant, this means that the speed of the block when it is at the top of the loop is indeed zero.

The correct equation for the centripetal force of the block when it is at the top of the loop is mv2/r = Fn + mg. Since the speed of the block when it is at the top of the loop is zero, we can say that Fn = -mg. Does the block make contact with the top of the loop? What conclusions can you draw concerning the normal force on the block?

12. May 19, 2015

### haruspex

Hmmm.. no wonder you're confused.
In an inertial frame, the gravitational force acts equally at all points in the loop.
The normal force varies, but can never act radially outwards.
The centripetal force is the resultant of those two forces. It always acts radially inwards. It varies because the speed varies.
Sounds like your teacher is thinking of the frame of reference of the block.
You don't mean that, I hope.
If it's at the minimum speed to retain contact, yes, the normal force will be zero. But that is the reason it will be zero, no other. Were it going faster the normal force would never be zero.

13. May 19, 2015

### toesockshoe

but if the speed was 0 at the top, then the mass would fall off the loop.

14. May 19, 2015

### haruspex

It does? Can you point out exactly where?
As toesockshoe writes, the velocity at the top cannot be zero (or it wouldn't reach the top).

15. May 19, 2015

### AlephNumbers

Yes, I am wrong. It is just has the minimum speed necessary to retain contact with the loop.

16. May 19, 2015

### vela

Staff Emeritus
Well, for one thing, the units don't work out. The units on the righthand side would be kg m2/s4, which aren't the units of acceleration.

A question for you to consider is, why do you think gravity should appear in the acceleration? The acceleration of an object is, by definition, $\vec{a} = \frac{d^2 \vec{r}}{dt^2}$, where $\vec{r}$ is the position of the object. It's completely determined by the path the object takes. In particular, if an object is following a circular path of radius $R$ and has an instantaneous speed $v$, it will experience a radial acceleration of magnitude $v^2/R$. This is true regardless of whether it's speeding up, slowing down, or moving at constant speed.

The only way gravity enters into the analysis is that the block's weight contributes to the net force acting on the block.

17. May 19, 2015

### toesockshoe

also:
I said the following earlier:
" and because at the top, it is a 90 degree angle with the horizontal, Fgcos(90) is 0 and thus there is no tangential acceleration. however, at the top the radial acceleration is Fgsin(90) which is Fg... so that means Fg has to be the radial acceleration which isnt correct right?" but someone pointed out that the sins and cosines are mixed up.

i do understand that i made a mistake and acceleration should actually be Fgcos(theta/m and Fgsin(theta)/m ... but did i really switch the sines and cosines. i drew a picture and i think its right.

18. May 19, 2015

### haruspex

AlephNumbers defined theta as the angle to the vertical (which is more usual), and might not have noticed that you are defining it as angle to the horizontal.

19. May 19, 2015

### toesockshoe

OHH. thanks. that helps a lot!