### Video Transcript

A 30.0-gram ball at the end of a string is swung in a vertical circle with a radius of 25.0 centimeters. The rotational velocity is 200.0 centimeters per second. Find the tension in the string at the top of the circle. Find the tension in the string at the bottom of the circle. Find the tension in the string at a distance of 12.5 centimeters from the center of the circle when the string is horizontal.

In this statement, weโre told the mass of the ball, 30.0 grams, and weโre told that the ball is swung in a vertical circle with a radius of 25.0 centimeters. We know that the ball is moving at a linear speed of 200.0 centimeters per second.

In part one of the problem, we want to solve for the tension in the string when the ballโs at the top of the circle. Weโll call that ๐ sub ๐ก. In the next part, we want to solve for the tension when the ballโs at the bottom of the circle. Weโll call that ๐ sub ๐. And finally, we want to solve for the tension of the string a distance of 12.5 centimeters from the center of the circle when the string is horizontal. Weโll call that ๐ sub โ.

Letโs begin our solution by drawing a diagram. In this diagram, weโre looking at the rotating string side on. Recall that this is a vertical circle that the ball is moving in. The circle has a radius ๐, and the ball moves around the circle at a linear speed ๐ฃ of 200.0 centimeters per second. The dotted lines represent three snapshots of the ballโs position on the string as it moves around the circle. When the ballโs at the top point, we want to solve for that tension as well as when the ballโs at the bottom point and the wall is horizontal from the origin of the circle.

To get started solving for ๐ sub ๐ก, letโs do three things. First, weโll recall the relationship for centripetal acceleration; next, weโll recall Newtonโs second law of motion; and third, weโll draw a free body diagram of the ball when itโs at the position at the top of the circle. First centripetal acceleration, the centripetal acceleration of an object, ๐ sub ๐, is equal to the objectโs speed squared divided by the radius of the circle in which the object moves. Notice that in our case, weโre given ๐ฃ and ๐ so we can solve for ๐ sub ๐.

Next, letโs remember Newtonโs second law of motion. This law says that the net force acting on an object is equal to the objectโs mass times its acceleration. We can write Newtonโs second law for our scenario as the net force that acts on the ball is equal to the ballโs mass multiplied by its centripetal acceleration. Now letโs draw a free body diagram showing the forces acting on the ball when itโs at the top of the circle. At that point, there are two forces that act on the ball, and they act in the same direction.

There is the force of gravity, ๐น sub ๐, pulling the ball down, and there is the tension in the string, ๐ sub ๐ก, thatโs also at this instant pulling the ball downward. We donโt know whether ๐น sub ๐ is stronger than ๐ sub ๐ก or which one has greater magnitude, but the free body diagram as is helps us to write that the net force acting on the ball at this instant is equal to the sum of the gravitational force on the ball, ๐น sub ๐, plus the tension force.

In this problem, weโll define motion towards the center of the circle as motion in the positive direction. Thatโs why weโve accounted ๐น sub ๐ and ๐ sub ๐ก as positive values. From the second law, we know that the sum of the gravitational and tension forces acting on the ball is equal to the ballโs mass times its centripetal acceleration.

We can now do a bit of substitution for this equation. ๐น sub ๐, the force of gravity on the ball, weโll recall is equal to its mass times the acceleration due to gravity ๐, and ๐ sub ๐, the centripetal acceleration, of the ball we know is equal to ๐ฃ squared divided by ๐.

We want to solve for ๐ sub ๐ก to that end. Letโs subtract ๐ times ๐ from both sides of the equation. This cancels out the term ๐๐ from the left side, leaving us with the equation ๐ sub ๐ก is equal to ๐๐ฃ squared divided by ๐ minus ๐๐.

If we factor in ๐ from the right-hand side, we have a simplified equation for ๐ sub ๐ก. Weโre now ready to plug in for the different variables in this equation; however, ๐, ๐, and ๐ฃ are given in units we donโt typically use. So as we plug them in, letโs convert them to more familiar units of kilograms and meters and meters per second.

A mass of 30.0 grams is equal to 0.030 kilograms; a speed of 200.0 centimeters per second is equal to 0.200 meters per second; and a radius of 25.0 centimeters equals 0.25 meters. ๐, the acceleration due to gravity, weโll treat as exactly 9.8 meters per second squared. When we calculate this value, we find a result for ๐ sub ๐ก of 0.186 newtons. Thatโs the tension in the string when gravity is working in the exact same direction as that tension.

Now that weโve solved for the tension in the string when the ballโs at the top of its arc, letโs solve for that same string tension when the ballโs at the bottom. This time when we draw the free body diagram of the forces acting on the ball, gravity, ๐น sub ๐, and the tension Force, ๐ sub ๐, are acting in opposite directions.

Once again we donโt know the relative magnitude of these two forces, but we do know that per our earlier definition of motion towards the center of the circle being positive and motion away being negative, when we write the force balance equation for the ball in this position, the tension force will consider positive and gravity will subtract from that positive tension force.

Weโll solve for ๐ sub ๐ just the same way that we solved for ๐ sub ๐ก, except that now weโll be subtracting the force of gravity from the centripetal force rather than adding it as before. Once again we factor out an ๐ and plug in the values for ๐, ๐ฃ, ๐, and ๐. With these values substituted in for our variables, when we compute ๐ sub ๐, we find a value of 0.774 newtons.

So the string tension is much greater when the ballโs at the bottom of its arc compared to when itโs at the top. This is because the ballโs inward or center-seeking acceleration is constant throughout this arc. Yet at the bottom, this inward acceleration has to overcome gravity, whereas at the top gravity aids that center-seeking acceleration. Weโre finally ready to solve for the tension in the string at a point 12.5 centimeters away from the circle center when the string is horizontal.

We are essentially simulating a shortened string of 12.5 centimeters, now in radius, with the same tangential velocity, 200.0 centimeters per second. Letโs draw a free body diagram of the forces acting on the ball when the string is horizontal. As always, gravity acts down, and now the tension of the string is acting horizontally to the left towards the center of the circle. These two force vectors are at right angles to one another. That means theyโre separable, independent.

When we write a horizontal force balance equation to solve for ๐ sub โ, we wonโt even include ๐น sub ๐ because the forces are perpendicular to one another and do not affect one another. In the horizontal direction, the net force acting on the ball equals the tension force ๐ sub โ, which is equal to the mass of the ball times its centripetal acceleration ๐ฃ squared over ๐, where ๐ is our revised and updated value of 12.5 centimeters while ๐ฃ continues to be 200.0 centimeters per second.

We can solve for ๐ sub โ now simply by plugging in for ๐, ๐ฃ, and ๐. When we do that, again being careful to use units of kilograms, meters, and meters per second, we calculate a value for ๐ sub โ of 0.960 newtons. This value is greater than ๐ sub ๐ and ๐ sub ๐ก, but remember this value is computed for a radius of half the value we used before. That is what accounts for the greater tension in the string.