# Loop problem, critical velocity

1. Aug 23, 2009

### Isiudor

loop problem, "critical velocity"

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

First I'd like to ask a general question, if a body enters a loop (like a rollercaster loop)
and PASSES the heighest point, assuming there's no friction, will it neccasserlly complete the loop or can it's orbit around the loop still decay? I was told this is so by my professor.

according to my textbook there's a term called "critical velocity" which it says is the
"minimum speed a body must have WHEN reaching the heighest point in order to complete a full circle", also an equation is given critical speed = square root of g * R. R being the radius of the loop.

In my mind, the latter two statements completely contridict one another, so one of them must be wrong.

bottom line the question is as follows

a block approches a loop of radius = 2 meters, what is the minimal velocity can be, when entering the loop, in order for it to REACH the heighest point of the loop. friction is negliable.

3. The attempt at a solution
my attempt at solution is simple 1/2mv^2 = mgh(enery conservation, since at start of loop height is equel to zero, and at the heighest point the velocity is zero

v^2 = 2 g h
v = 39.2

but according to my text book this is false.

for a reason I do not understand, it the equation must look like this
1/2mv^2 = mgh + squareroot(g * R).

why must it have a speed at the height point in order to cross it? I don't get this "critical speed" thing, I really don't. I'm desperate for help.

2. Aug 23, 2009

### Fightfish

Re: loop problem, "critical velocity"

Seems like you have to go read up on circular motion first, otherwise it may be difficult for you to really understand the issue.
In any case, if the velocity of the block does not reach the 'critical velocity', then the weight of the block will not just provide the centripetal force required to keep it in circular motion; i.e. it exceeds the required centripetal force. This would thus mean that the block would lose contact with the loop (falling off)

3. Aug 23, 2009

### Staff: Mentor

Re: loop problem, "critical velocity"

While energy is certainly conserved, it's not true that the speed at the top will be zero. Realize that you are not tossing the coaster straight up like a ball, but that it's in contact with the track. (If you toss it with that little energy, it will fall off the track long before reaching the top.) To maintain contact--which means have non-zero normal force--it must have some speed so it keeps pressed against the track at all times. To find the minimum speed needed at the top (and understand where the "critical speed" formula comes from) analyze the forces acting at the top and apply Newton's 2nd law. (Hint: What kind of motion does the car experience as it traverses the circular track?)

4. Aug 23, 2009

### Isiudor

Re: loop problem, "critical velocity"

but I WANT the speed to be zero.

my goal is not to make a full circle, just to reach the top, if it falls off the highest point, I've reached my goal. since I want to have the minimal starting speed(at bottom of loop) and reach the highest point in the loop, that's my only goal.

had my goal been to complete a full circle, certainly to find the starting speed then the equation has to be 1/2mv^2 = mg * 2R + sqrt(g * R), but since I just want to REACH the highest point, why isn't the equation just 1/2mv^2 = mg * 2R? in lament terms can't the block reach the top at 0 speed and then fall enough?

anyway, I'd also really like to hear your thoughts on the first two statements of my original post, thanks.

Last edited: Aug 23, 2009
5. Aug 23, 2009

### tiny-tim

Welcome to PF!

Hi Isiudor! Welcome to PF!

You can't reach the top with zero speed …

you'll already have fallen off the loop before then.
Your professor is right, and the book is wrong.

If the body reaches the top, it must have the critical velocity, so yes, it will necessarily complete the loop.

The book should just have said that the critical velocity is "the minimum speed a body can have when reaching the highest point"

6. Aug 23, 2009

### Staff: Mentor

Re: loop problem, "critical velocity"

I don't see any contradiction between the two statements. If it makes it passed the highest point still in contact with the track, will it complete its loop? Yes. When it passes that highest point while still in contact with the track, must it have a minimum speed? Yes. No contradiction. (Did I miss something?)

To restate what I (and tiny-tim) have already stated, you cannot arrange for the car to just make it passed the top with zero speed. It will have fallen off the track long before reaching the top. To maintain contact with the track it must be moving at some minimum speed as it passes the highest point.

7. Aug 24, 2009

### Isiudor

Re: loop problem, "critical velocity"

Yes I get it now, the critical speed is the speed I must have to "stay in orbit" due to the mg + N acting on me, although that at the highest point it's just mg.

and when I "solve" mg = 1/2mv2 I actually get the critical velocity equation.

I guess you are also right regarding the two statements not contradicting each other, but the statement "critical velocity is the speed needed at the highest point in order for the object to complete a full loop" is, although technically correct, is quite misleading. It implies that an object can, for example, reach it with half the speed of critical velocity but simply fall off before completing the loop(which I realize now is impossible due to the laws of circular motion) which is what confused me so much in the first place.

OK so, thanks for your help. cheers.