Two wires are tied to the 2.0 kg sphere shown in
FIGURE P8.45. The sphere revolves in a horizontal circle at constant speed. a. For what speed is the tension the same in both wires?

Question

Two wires are tied to the 2.0 kg sphere shown in 
FIGURE P8.45. The sphere revolves in a horizontal circle at constant speed. a. For what speed is the tension the same in both wires?

Accepted Answer

Welcome back, everyone. In this problem. A musical toy consists of a wooden ball that rotates in a horizontal circle around the central axis. The mass of the ball is 0.3 kg and is supported by two wires. As shown in the figure, determine the speed at which the tension in both wires is the same if the ball revolves at a constant speed for our answer choices. A says it's 4.42 m per second. B 3.8 m per second, C 6.71 m per second and D 4.52 m per second. Now, if we're, we're really gonna solve this problem, we need to make sure we understand exactly what's going on here. So let's take a quick look at our diagram. OK. Now, in our diagram, you notice here that we have our two wires, OK? So I'm gonna call these wires, L one and L two. OK. Now here notice that for this wire, it makes a 50 degree angle with the horizontal. So that means it's going to make a 40 degree angle with our central axis. OK? As well. If we think about it for L two, it makes a 50 degree angle with our central axis. So that means the horizontal distance from the ball to our central axis. If we call that R, because that's going to be the radius of the ball's path, then it's going to make a 40 degree angle between the ball and our horizontal distance R here. OK. So this is what we know so far and we know that this ball is going in a circular path. So we can imagine it's going our own something like that. OK. Now, if we're going to determine the speed at which the tension in both wires is the same, then once we're talking about tension, it will probably be useful for us to draw a free body diagram to understand what forces are acting on the ball. And that can help us to figure out its speed. So let's go ahead and do that. So here's our ball. OK? And we know that our ball is connected to L one and L two. So let's label the forces from those wires as T one tension in the wire and T two, the tension in L two. OK. And no, we know that the ball's weight is going to be acting downwards as MG. If we label the, the horizontal axis X and or vertical axis Y here. OK. Let me bring that a bit over to the center of the bar, they look a little better. OK? Then no, if we think about both our wires, their tensions are going to have X and Y components. Now, what are those going to be? Well, think about it before. OK. We know that the angle made between L two and the horizontal axis is 40 degrees. OK. So we can label this as 40 degrees here. Now, with that in mind, OK, if we think about it, its horizontal component is adjacent to the 40 degrees. So that's going to be T two cost 40. OK. While its vertical component is opposite to a 40 degree angle, so we can label that as T two sine 40 E. Now when we think about T one, if we go back to our diagram here, T one is the tension for our first wire, L one. OK. And notice that it makes 40 degrees with our central axis. So if we were to draw that vertical axis here for our ball, then the angle between L one and the vertical axis would also be 40 degrees. OK. This angle would also be 40 degrees based on what we know about Z angles. OK? So if that's the case and, and that's alternate angles, I couldn't remember the right name there. So if that's the case, that means this angle is 40 degrees and no for the components of T one that tells us its vertical component, OK? It's a vertical component. Let me put that as a line here and let me get rid of the top. So its vertical component is going to be T one plus 40 degrees and its horizontal component, it is going to be T one S 40 degrees. So these are all of our forces at work here on our ball. And we know that the vertical distance between T one, L one and L two are 2 m. So we can leave that here for T one and T two. Now that we have all of this going on to solve or V, we could resolve our forces in the X and Y directions and then see where it carries us from there. So let's start by resolving our forces in the Y direction. Now, based on Newton's second law of motion here, there is no motion along the Y axis, the acceleration in the Y direction will be zero. So the sum of the forces in the Y direction equals zero in our Y direction going upwards, we have T one cost 40 T two sine 40. OK. Downwards we have the weight of the ball MG. So all of that equals zero. But remember like we said in our problem that we want the speed at which attention in both wires is the same. So in this case, T one equals T two and we can just label those as T and in that case, that means then that T cost for T plus T sign 40 minus MG will equal zero. We can factor ot to be t multiplied by costs 40 plus sine 40 equals mg, add mg to both sides. I know that tells us then that our 10 equals MG divided by cost 40 plus sine 40. Now we know the mass of our ball and we can take the acceleration due to gravity to be 9.8 m per second squared. So this is going to be 0.3 kg, not multiplied by 9.8 m per second squared divided by C 40 plus sine 40 and all of that, OK is equal to 2.1 newtons. So from resolving the forces in the Y direction, we've been able to figure out what the tension is. Next. Let's move on to our voices in the X direction. OK. Now, in the X direction in the X direction, we know that as the wooden ball rotates in a horizontal circle, there's a net net centripetal force acting on it which keeps it moving in horizontal circles. So by Newton's second law that tells us the net force is directly proportional to the acceleration by the formula F equals ma then we can apply that here to say that the sum of the forces in the X direction are going to be equal to the mass of the ball multiplied by the centripetal acceleration. Now, if we go back to our diagram, our forces in the X direction are T two cost 40 T one sine 40. So that means then that T one sin 40 plus T two cost 40 is going to be equal to a mass multiplied by a centripetal acceleration V squared divided by R. Remember we also said the tensions are the same. So it's really tsin 40 plus T cost 40. And if we factor out T OK, then that tells us T multiplied by sine 40 plus co 40 is going to be equal to MV squared divided by R. Now remember we're solving for V. So if we isolate V squared, then V squared is going to be equal to tr multiplied by sine 40 plus costs 40 all divided by our mass. Now, we know almost everything here except the value of R. We don't know what our horizontal distances of the ball from our central axis. However, if we go back to our diagram, if we go back to our diagram notice here that compared to L two, it's really a component of L two, it's really the horizontal component of L two. OK? And since L two or rather, since R is the horizontal component of L two, and from our diagram, we can tell that that's going to be L two K for it. OK? And our triangle is an iso triangle. So that means L two equals 2 m. OK? Then that means we can rewrite our equation for V squared. Then to say V squared equals R which is L two cos 40 multiplied by T multiplied by the sine of 40 plus the cosine of 40 all divided by M. That means then that V itself is going to be equal to the square root of that entire expression. OK. It's the square root of our expression here. I can just copy and paste and now to sulfur V, we can substitute all of what we know here. So by doing that, that means V is going to be equal to the square root of L 2 2 m multiplied by the cosine of 40 degrees, multiplied by 2.1 newtons multiplied by the sine of 40 plus the cosine of 40 all divided by the mass of our wooden ball, which is 0.3 kg. Let's not forget our square root sign. Now, when we put all of this into a calculator to solve then to three significant figures, we should get V to be equal to 3.88 m per second. In other words, the tension, sorry, the ball would have to rotate at a speed of 3.88 m per second. If the tension in both the wires are the same that tells us, then that B is 3.88 m per second. Thanks a lot for watching everyone. I hope this video helped.

Two wires are tied to the 2.0 kg sphere shown in FIGURE P8.45. The sphere revolves in a horizontal circle at constant speed. a. For what speed is the tension the same in both wires?

Verified video answer for a similar problem:

Key Concepts

Centripetal Force

Tension in Wires

Uniform Circular Motion