# Vertical circular motion

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## Main Question or Discussion Point

Let's say I have a stone tied to a string and I'm spinning it vertically. Now, if I provide an initial velocity such that the velocity of the stone = 0 at an angle greater than 0 from the horizontal(visualize it in the "first quadrant") the eq of motion would be: T+mgsin (theta) = mv^2/r. If the velocity equals zero, then tension would end up negative which is not possible. Hence I've come to the conclusion that as soon as The T =0(which would happen before v =0), the stone would exit circular motion and travel somewhat like a projectile. So basically, in the "first quadrant", it's not possible for v to end up 0 while remaining in circular motion since T would = 0 before V would equal 0 and hence the stone would exit circular motion. Is this correct?

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I think you’re correct. If v=0, it’d just fall down.

No. As your equation shows, the string can have zero tangent (at the top) and the stone can still be in circular motion. In this case, sine theta = 1, so that with zero tension, mg = mv^2 / r. If you spin the stone just right so that v^2 = r g, at the top of the swing, the tension will be zero.
When I was teaching we had a lab experiment in freshman physics where we rolled a ball down a ramp and the ball went into a loop the loop. We found if the ball was released at just the right point in the ramp, we could remove a small portion of the track (so the track could not supply the necessary tension i.e. T = 0), without the ball failing to "make the jump" to the remaining track. This shows the tension can be zero, at the top of the track without it falling from the circular path.

The point of my previous post is the lower limit on the velocity is not zero, it is g r, where r is the radius of the circle of the stone on the string.

Yes, of course, but only when theta=pi, so at the top. That is the only position. If sine theta=0,999, v cannot be zero.

But the issue is not v = 0. The stone on the string in circular motion must have v greater than or equal to g times the radius of the path.

correction to my previous note v ^2 greater than g times the radius of the path.

Yes, you’re right :)

I do have to admit in the lab experiment I mentioned the ball travels a parabolic path as soon as it reaches the portion of it's path where the track is missing. The ball recovers the circular path as soon as the ball moves beyond the missing portion of track and the track supplies the necessary normal force. For small lengths of removed track, the parabolic path is sufficiently close to the circular path to "make" the jump, successfully.

The point of my previous post is the lower limit on the velocity is not zero, it is g r, where r is the radius of the circle of the stone on the string.
Yeah but my primary question was:
Can v = 0 in the "first quadrant".
My thought process I described above, where if v = 0, t comes negative. This implies that T =0 some time before v = 0 and as soon as T = 0 it leaves the circular path. This isn't true for theta = 90 because in that case MG provides the centripetal force(temporary)

Yeah but my primary question was:
Can v = 0 in the "first quadrant".
no it cannot
once the bob just crosses horizontal position , its velocity is vertically up so it wants to go up
but that will cause string to break
so it asks string for tension so it can stay in circular motion and string does not break
but mgsin (theta) also increases so if mgsin (theta) is enough to satisfy bob's need of centripetal force
(ie mv^2/r.=mgsin (theta)) or radius of trajectory(ie v^2/gsin(theta)) = length of string
bob will stop asking string for tension , (T=0)
but now the radius of trajectory(ie v^2/gsin(theta)) exceeds length of string (since v decreases and sin(theta) increases)
what this means is that bob exits circular motion , enter projectile motion , string becomes loose
only force on bob is mg down , no force is horizontal dir
bob has non-zero horizontal projection of velocity so net v can never be 0 in 1st quadrant

If I understand it correctly the following holds: if v = 0, the ball falls straight down, and the tension goes to zero (slack), until the distance from hand to bob exceeds the radius of the circle. If v is less than sqrt (2 * g * radius) , the ball does not have the velocity to move in a circle and it falls and follows a non circular path (inward) from the circle and drops. The tension is now zero and the string goes slack. When v = sqrt (2 * g * radius ) exactly, Tension is exactly zero, and the path is circular. When v exceeds sqrt 2 * g * r The path is circular and is kept in "orbit" by a tension that adjusts itself to whatever it needs to keep a bob in circular motion with the high velocity. If the velocity becomes high enough, the striing breaks. Now Tension goes to zero, and the ball follows a ballistic :parabolic: path.

I'm sorry for clarification, In the previous post, I am always considering the tension at the top of the orbit because it is there that the tension is a minimum. below this point for the exactly v = sqrt 2 g r case the Tension is supplied by the string, and T greater than zero. Tension from your equation is m (v^2 / r -g), so you can see Tension is a function of velocity, and the constants for the problem g, and r. I grasp what you are saying in your OP. However the Tension becomes negative when v^2 ls less than 2 g r. If (when) this velocity condition occurs, the ball follows a ballistic path, unconstrained by the tension. I follow your logic in the OP but the lower limit on the velocity for circular motion is not v = 0, it is v less than sqrt ( 2 * g * r) at any point in the circle.

I also have to qualify my earlier answers in that I am assuming, the energy going into the system is conserved. The man, woman, child, or monkey whirling the bob, isn't "pumping" rock to higher and higher velocity. In this case, I'm not sure of my assumption that the Tension would be a minimum at the top of the swing, although I still think it would be. I have to think about it.