A 500 g ball moves in a vertical circle on a 102-cm-long string. ...

Question

A 500 g ball moves in a vertical circle on a 102-cm-long string. If the speed at the top is 4.0 m/s, then the speed at the bottom will be 7.5 m/s. (You'll learn how to show this in Chapter 10.) (b) What is the tension in the string when the ball is at the top?

Accepted Answer

Hi, everyone in this practice problem, we are being asked to find the tension on the cable at the highest vertical position. We will have an object with a mass of 0.125 kg attached to a cable revolving in a vertical circular motion with a radius of 0.5 m. The magnitude of the velocity is 4.5 m per second at the highest vertical position and 7.8 m per second at the lowest vertical position. We're being asked to find a cable extension at the highest vertical position and the options given are a 3.0, Newton B 11, Newton C 12, Newton and D 22 Newton. So the object will actually undergo a vertical circular motion and the speed of the object as well as the position of the factors will be con constantly changing. The object will actually be subjected to two different forces which are going to be represented by this diagram of our system right here. So we'll have the circle here which is representing the vertical circular motion trajectory. And I am going to represent our object with the sphere located at the highest vertical position because that is the tension that we are interested to look at. So the object right here will be subjected to two different forces. The first one is going to be the weight which is represented by this red arrow right here, which is going to just be M multiplied by G and going to be pointing vertically downward. And this weight is actually going to be the same at any position in this vertical circular motion trajectory. So next, we have the tension of the court which I'm going to represent with this uh blue arrow right here. And because the object is at the highest vertical position, and I'm going to represent the court with this black uh line right here. And because the object is at the highest vertical position, the tension will actually be pointing downward just like. So, however, the tension direction is going to be constantly changing throughout each and every single position along this vertical circular motion trajectory. Um We are asked to find the tension at the highest vertical position. So this is going to be the only one that we are interested to look at. Next. What we wanna do is to actually apply Newton's second law in order for us to actually find what the tension uh is. So according to Newton's second law, the total forces or the sigma of the forces acting upon our system in the radial direction will equals to the mass of the system, which is, in this case is the mass of the object multiplied by the A uh acceleration in the graph in the radial direction. We want to recall that A R will equals to the linear velocity divided by the radius of the vertical circular motion trajectory. And we wanna substitute that into our Newton second law. So Sigma F R will become negative M G because the gravitational force is pointing downwards. And we wanna plus that with the tension force which is also negative T. So I'm just gonna directly uh write negative T here. So negative M G minus T will equals to negative M multiplied by A R. We wanna include the negative value or negative sign there as well. And then we want to substitute that with the V squared divided by R. So negative M G minus T equals negative M multiplied by V squared divided by R. We wanna just rearrange this so that we get the uh equation for T because that is what we're interested to look at. So this will then be M multiplied by V squared divided by R minus M multiplied by G. And we can pull both MS out. So T will then be M multiplied by parentheses, open parenthesis V squared divided by R minus G Just like. So, and that will essentially be the equation that we are going to use to actually find or solve for T. So now we can substitute all the values that we know the M is going to be 0.125 kg. Given in the problem statement multiply that by open parenthesis V squared, the V is going to be the velocity at the highest vertical position which is 4.5 m per second squared divided by the radius of the f of vertical circular motion trajectory, which is 0.5 m minus G which is going to be 9.81 m per second squared, close parentheses just like. So, So that will essentially finally, after calculating give us the tension value of 3.0 Newton. So 3.0, Newton is going to be the cable tension at the highest vertical position of the object. And that will actually correspond to option a in our answer choices. So answer a is going to be the answer to this particular practice problem. So that'll be all for this one. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topic and that will be it. Thank you.

A 500 g ball moves in a vertical circle on a 102-cm-long string. If the speed at the top is 4.0 m/s, then the speed at the bottom will be 7.5 m/s. (You'll learn how to show this in Chapter 10.) (b) What is the tension in the string when the ball is at the top?

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