Video: Mechanical Energy of an Object Undergoing Circular Motion

A ball of mass 1.3 kg at the end of a 1.5-m string swings in a vertical circle. At its lowest point the ball is moving with a speed of 9.5 m/s. What is the speed of the ball at the top of its circular path? What is the magnitude of the tension in the string when the ball is at the top of its circular path? What is the magnitude of the tension in the string when the ball is at the bottom of its circular path?

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Video Transcript

A ball of mass 1.3 kilograms at the end of a 1.5-meter string swings in a vertical circle. At its lowest point, the ball is moving with a speed of 9.5 meters per second. What is the speed of the ball at the top of its circular path? What is the magnitude of the tension in the string when the ball is at the top of its circular path? What is the magnitude of the tension in the string when the ball is at the bottom of its circular path?

In this three-part problem, we can call the speed of the ball at the top of its path 𝑣 sub 𝑑, the tension at that point 𝑇 sub 𝑑, and the tension in the string at the bottom of the ball’s path 𝑇 sub 𝑏. Let’s start by sketching out this vertical circular path. This ball, with a mass of 1.3 kilograms we’ve called π‘š, moves in a circular vertical arc on a string of length π‘Ÿ, which is 1.5 meters. We know that when the ball is at the bottom of the circular arc, its speed is given as 9.5 meters per second. Given all this information, we wanna solve for the speed of the ball when it’s at the top of its arc.

Considering this system as a whole, since no energy is added to it or taken from it, we can use the conservation of energy to solve for the velocity of the ball at the top of its arc. We can say that the initial kinetic plus potential energy of our system is equal to the final kinetic plus potential energy, where our initial point is at the bottom of the circular arc and the final point is at the top. Recalling that kinetic energy is given as one-half an object’s mass times its speed squared and that gravitational potential energy is π‘š times 𝑔 times β„Ž, where the acceleration due to gravity, 𝑔, we’ll treat as exactly 9.8 meters per second squared.

We can write that one-half the ball’s mass times 𝑣 sub 𝑏 squared is equal to one-half its mass times 𝑣 sub 𝑑 squared plus π‘š times 𝑔 times two π‘Ÿ, the radius of the circular arc. We see that the mass π‘š cancels from this expression. And when we rearrange to solve for 𝑣 sub 𝑑, we find it’s equal to the square root of 𝑣 sub 𝑏 squared minus four times 𝑔 times π‘Ÿ. Since we’re given 𝑣 sub 𝑏 and π‘Ÿ and know 𝑔 is a constant, we’re ready to plug in and solve for 𝑣 sub 𝑑. Entering this expression on our calculator, we find that 𝑣 sub 𝑑 is 5.6 meters per second. That’s the speed of the ball at the top of its circular arc.

Next, we wanna solve for the tension of the string when the ball is at point 𝑓, at the top of its arc. If we draw a free body diagram showing the forces acting on the ball at this point, the two forces acting on the ball at this point are the gravitational force, 𝐹 sub 𝑔, and the tension force, 𝑇 sub 𝑑. Recalling Newton’s second law of motion that the net force on an object is equal to its mass times its acceleration and letting motion in the downward direction be positive in our example. We can write that π‘šπ‘”, the gravitational force, plus 𝑇 sub 𝑑, the tension force from the balls at the top of its arc, is equal to the mass of the ball times its acceleration, π‘Ž.

Because the ball is moving in a circle, we know that it accelerates centripetally, that is, towards the center of the circular path, and that the magnitude of this acceleration is equal to 𝑣 squared over π‘Ÿ. So we can write that π‘šπ‘” plus 𝑇 sub 𝑑 is equal to π‘š times 𝑣 sub 𝑑 squared over π‘Ÿ. This expression on the right-hand side is the centripetal force acting on this ball. We want to rearrange and solve for 𝑇 sub 𝑑. When we do, we see it’s equal to π‘š times the quantity 𝑣 sub 𝑑 squared over π‘Ÿ all minus 𝑔. When we plug in for π‘š, 𝑣 sub 𝑑, π‘Ÿ, and 𝑔 and enter this expression on our calculator, we find that, to two significant figures, 𝑇 sub 𝑑 is 15 newtons. That’s the tension in the string when the ball is at the top of its path.

Lastly, we wanna solve for the tension in the string when the ball is at the bottom of its arc. To do this, we can once again draw a free body diagram of the forces on the ball at this point. There are again two forces on the ball: the tension force, 𝑇 sub 𝑏, and the weight force. But now they point in opposite directions. We can write that the gravitational force minus 𝑇 sub 𝑏 is equal to negative π‘šπ‘£π‘ squared over π‘Ÿ, where the centripetal acceleration is negative because it points up. When we rearrange this expression to solve for 𝑇 sub 𝑏, we find it’s equal to π‘š times the quantity 𝑣 sub 𝑏 squared over π‘Ÿ plus 𝑔. When we plug in for these values and enter this expression on our calculator, we find that 𝑇 sub 𝑏 is 91 newtons. That’s the tension in the string when the ball is at the bottom of its arc.

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