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Uniform Circular Motion Example

by Jared Rovny
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    00:01 We can best analyze this with an example, so supposed you're swinging an apple around your head on a one-meter string, as one does.

    00:08 And we could ask a few questions about this for example, what is the slowest speed that the apple can be swinging and still maintain the circular path? Or we could also ask how much tension would be in the string or this rope that I'm using to swing an object right at the bottom of the path if we know the mass of our object, in this case, say a one-kilogram object? So first give this a try using what we've just introduced the new force for something to go into a circular motion and to uniform circular motion and see if you can solve these two questions by analyzing the scenario especially with pictures and using a new equation.

    00:42 If you've done this, be sure to get something that looks sort of like this analysis.

    00:47 Let's first take that first question, what is the slowest speed that's apple could be going or swinging and still maintain its circular path.

    00:55 In this picture here, you can see that we have our radius of one-meter circle and we also have our apple with mass, m with the force mg pulling it downwards towards the center of our circle.

    01:06 As this object is moving around in a circle, what we have is two forces on it mg acting downwards and then also the tension in the rope pulling it downward as well.

    01:15 So we can call this F-sub-T for the tension force pulling downwards on our object.

    01:20 So these two forces pulling downwards on our object, when it's at the top of the circle, when it's at the bottom of the circle, as you swinging it, we have the gravitational force still acting downwards and this is always what it does, another tension force is acting upwards, so we have gravity acting in opposite direction of tension at the bottom of the path and at the same direction of the tension at the top of the path.

    01:42 What we're interested in, is what is the slowest speed that this apple can be moving.

    01:47 Let's first write our acceleration equation that the acceleration must be equal to v squared divided by the radius of our path, further we could write the force equation m times a equals mv squared over r, looking at this equation, we can see that if we want the smallest velocity for our object we need the force on the left hand side to also be minimized.

    02:12 What is the minimum force that can keep this object in a circular path, we're looking at this object right at the top of this path we have mg and the force of tension acting downwards on the object as opposed to the bottom or we have mg and I'll call it force of tension to keep our notation consistent, pulling upwards on the object, looking at just the scenario, we can see that our object has a force downwards from tension, a force downwards from gravity, so if I want to minimize this force which has two components tension and gravity, then what I would like to do is make the tension, let's call this one the tension force as small as possible so that this object will still go in a circle.

    02:52 What I can do in that case, so let's just say that the tension is zero that would be the smallest it could possibly give, at the top of the path I can't give it any less than zero with the string.

    03:01 And this would mean that the only force acting on the object downward at the top is mg this is the only force acting on our object, in this case, an apple when it's right at the top of that path.

    03:11 And since we know this to be the case we can now solve for the slowest speed using Newton's second law, we have F equals ma, the F, the force that we've mentioned is at the top of this path is mg, so we have mg acting towards the center of our circle is equal to the mass time the acceleration and since we know this object must be going in a circle we have the acceleration v squared over r.

    03:38 A few important things here, one is that you can see that I've chosen a particular sign convention, so for problems with things that are going in a circular motion what I'll be doing here is calling the center of our circle the positive direction, so if you're moving towards the center this is the positive direction.

    03:55 So in this case we have mg equals mv squared over r, you can be divide both size by the mass and then solve for the velocity and see that the velocity is equal to the square root of g times the radius of our circle.

    04:06 So this velocity, the velocity where we minimized, the total force pulling the object towards the center by completely getting rid of the force that we had control over.

    04:14 The velocity of the result in motion would be the square root of g times r which we can be approximate as about 10 for g times r which is 1 meter times 1.

    04:27 So this is the square root of 10 which is approximately 3.1 meters per second.

    04:33 So in this kind of problem we had to appeal to some possibly difficult logic to follow if we've seen this for the first time, because we're doing again a sort of minimization rule, so this object got right to the top of this path and basically alway says we want to minimize the forces acting on our object.

    04:51 It's more difficult, the different parts of the path to consider this but if you really think about you're swinging something on the rope, right at the top of the path, where you give the least tension pulling downwards and then you feel you have to pull it against the bottom and then the least tension again at the top.

    05:03 And the reason is when that object is at the top just what we've seen in this example, gravity is helping you pull that object towards the center of your circle, whereas at the bottom you have to fight gravity to keep the force towards the center of the circle.

    05:15 And so we pick the top of this path because we knew right then we can let go of our object to give it the minimum speed that needed to stay in uniform circular motion.

    05:24 What we can do now is go to the second part of this problem and ask if this is the case, if it's at the slowest velocity how much tension would be in the string when the object, this apple, is right at the bottom of the path.

    05:35 Well once again we've seen that at the bottom of the path we have gravity downwards and we have the tension upwards, so we have a different direction for a force in this time.

    05:44 So writing one more time Newton's second law, we have F equals ma using the sign convention one more time or I'm calling towards the center of the circle the positive direction meaning away from the center would be a negative direction then we can see that we have f tension minus mg is equal to mv squared over r.

    06:08 And then again we put the mv squared over r on the right hand of our equation because we're trying to tell the mathematics, we're trying to tell the equation that all the forces put together at the end of the day, had better be equal to mv squared over r, because we know our object is going in a circle.

    06:22 This is something we can simplify very quickly, because we saw from the prior part that mv squared over r is also equal to mg, we saw this by analyzing the top of the path.

    06:35 And that right at the top of the path the only force acting downward on the object was the gravitational force and we had stopped pulling on the rope.

    06:42 We had to stop applying any tension to the string, and so the only force acting was mg, and we had this equal to mv squared and over r.

    06:49 So now we can say that the force of tension minus mg is equal to from our analysis before mg.

    06:56 So in the force of tension that we have to apply at the bottom of the path adding mg in both sides is twice the mass times the gravitational acceleration.

    07:05 And in this particular example, we had a 2 kilogram object, or sorry a 1 kilogram object for this apple, so what we have is this is equal to 2 times the mass which is 1 kilogram and then again we're going to use the approximation for g as being 10 meters per second squared.

    07:20 This is equal to 2 times 10, for the numbers 20 and then our units work out as we help kilograms meters per second squared giving as 20 Newtons, the units of force.

    07:31 So this would be the force of the tension right when the object is at the bottom of the path.

    07:37 So you can see that when you're swinging something around in a circle, what we got mathematically actually follows our intuition that when something at the bottom of the path, the gravitational forces at you have to not only pick gravity by applying your own mg, we have to apply one more unit of mg in order to keep it pulled towards the center of the circle, right at the top of the path, you don't have to apply any tension at all.

    07:58 Gravity will apply the mg itself and keep your object going into a circle.


    About the Lecture

    The lecture Uniform Circular Motion Example by Jared Rovny is from the course Force.


    Included Quiz Questions

    1. Tension at the top = 34N, Tension at the bottom = 74N.
    2. Tension at the top = 3.4N, Tension at the bottom = 74N.
    3. Tension at the top = 34N, Tension at the bottom = 7.4N.
    4. Tension at the top = 3N, Tension at the bottom = 7N.
    5. Tension at the top = 0.34N, Tension at the bottom = 0.74N.
    1. 4.1 x 103 N.
    2. 41x 103 N.
    3. 4.1x 104 N.
    4. 41N.
    5. 4.1 x 102.
    1. 9.2 x 103 m/s2.
    2. 9.2 x 102 m/s2.
    3. 9.2 m/s2.
    4. 92 m/s2.
    5. 920 m/s2.

    Author of lecture Uniform Circular Motion Example

     Jared Rovny

    Jared Rovny


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