Question Video: Swinging a Stone in a Vertical Circle | Nagwa Question Video: Swinging a Stone in a Vertical Circle | Nagwa

# Question Video: Swinging a Stone in a Vertical Circle Physics • First Year of Secondary School

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A stone that has a mass of 1.6 kg is swung in a vertical circle at a constant angular velocity of 6.1 rad/s. The stone is attached to a uniform rope of length 0.33 m, as shown in the diagram. The length of the rope is the same as the radius of the circle throughout the motion of the stone. What is the ratio of the maximum force to the minimum force that the rope can apply to the stone? Give your answer to one decimal place. What is the force that the rope applies to the stone when the rope makes an angle π = 33Β° above the horizontal? Give your answer to the nearest newton.

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### Video Transcript

A stone that has a mass of 1.6 kilograms is swung in a vertical circle at a constant angular velocity of 6.1 radians per second. The stone is attached to a uniform rope of length 0.33 meters, as shown in the diagram. The length of the rope is the same as the radius of the circle throughout the motion of the stone. What is the ratio of the maximum force to the minimum force that the rope can apply to the stone? Give your answer to one decimal place.

In our diagram, we see this stone being swung by a rope around in a vertical circle; that is, the stone only moves in a plane that is entirely vertical. The first part of our question talks about a ratio of forces on the stone. When we think about forces on the stone, at any given moment, there are two.

First, thereβs the tension force β weβll call it capital π β that pulls the stone in towards the center of the circle. The second force always acting on the stone is its weight force; itβs mass times the acceleration due to gravity. While the weight force π times π is always the same regardless of where the stone is along its circular path, the tension force is not like that. Rather, the tension force varies as the stone moves throughout this arc. We could think of the tension force as the force that enforces circular motion on this stone. It always pulls on the stone just hard enough so that the stone does indeed follow a vertical circle.

Part one of our question talks about a maximum and a minimum force that the rope can apply to the stone. That force is the force weβve called capital π, the tension force. Because the tension force always pulls the stone towards the center of its circular arc, we say that it is a centripetal or center-seeking force. In general, the centripetal force πΉ sub c acting on a mass π moving in a circle of radius π is equal to π times π times the angular speed of that object squared.

Notice in our problem statement that weβre told that our stone moves at a constant angular velocity. In other words, all throughout its vertical path, π is the same. In addition, the mass of the stone is constant, and so is the radius π of its circular path. This means that for any position of the stone anywhere along this circular path, the centripetal force πΉ sub c acting on the stone is the same.

But, and here we must be careful, that doesnβt mean that the tension force capital π is always the same. In fact, the tension force varies all throughout the stoneβs circular path. This happens so that the combination of the tension force π and the weight force π times π does yield a constant overall centripetal force.

Since our question asks about the maximum and minimum force that the rope can apply to the stone, we want to think about those places in the stoneβs path where π, the tension force, has a maximum and a minimum value.

Letβs consider the moment where the stone is at its maximum height. As always, the weight force acts directly downward on the stone. And here we see that the weight force is perfectly in line with the tension force. We could say that in this position the weight force on the stone applies a significant center-seeking force. Therefore, at this moment, the tension force π β that is the force of the rope on the stone β does not need to be very great in order for the overall centripetal force to equal the stoneβs mass times its radius of revolution times its angular speed squared.

In fact, since the tension force and the weight force are perfectly aligned at this moment, when the stone is vertically above its rotation center, the tension in the rope is at a minimum. This force that weβve called π sub min is the minimum force that the rope can apply to the stone.

Next, letβs think about the forces involved when the stone is at its lowest position. As always, the tension force in the rope pulls the stone towards the center of the circle. But now the weight force π times π on the stone points in the opposite direction as the tension force. But we recall that the total centripetal force acting on the stone must be constant.

Here, not only must the tension force in the rope keep the stone moving in a circular path, it must also overcome the weight force which is opposed to the tension force. For this reason, the tension in the rope is highest when the stone is at its lowest position. What weβve called π sub max is the maximum force of the rope on the stone.

To start comparing π sub max and π sub min, letβs clear a bit of space on screen. If the centripetal force acting on the stone is πΉ sub c, weβve said that πΉ sub c must be constant all throughout the stoneβs motion. The reason for this is neither the stoneβs mass nor its radius of revolution nor its angular speed vary.

At any given moment, πΉ sub c equals the sum of the tension force in the rope and the weight force on the stone. If we consider downward-directed forces to be forces in the positive direction, then at the stoneβs highest point the centripetal force is equal to π sub min, the minimum tension in the rope, plus π times π. Here, we add these two forces together because they both point in the same direction.

If we then turn to consider the stone at its lowest elevation, we can say that that same centripetal force is equal to π times π β thatβs positive because itβs directed downward β minus π sub max. What weβre finding then is that π sub min plus ππ is equal to ππ minus π sub max. What we basically have then is two equations, one for π sub min and one for π sub max.

In our equation for π sub min, if we subtract π times π from both sides, then on the right-hand side, π times π minus π times π cancels out. And we find that negative π times π plus π times π times π squared equals π sub min. Factoring out an π and rearranging this slightly, we find that π sub min is equal to π times the quantity π times π squared minus π.

Now, letβs find a similar expression for π sub max. If we add π sub max to both sides of the second equation, then π sub max cancels on the right. And next, we subtract π times π times π squared from both sides, canceling these two terms on the left. Finally, we can factor out an π on the right-hand side. And we find that π sub max equals π times the quantity ππ squared plus π. Notice then that the only difference between π sub min and π sub max in these equations is the sign in front of the acceleration due to gravity π.

Anyway, we want to find the ratio of these forces. That is, we want to solve for π sub max divided by π sub min. Notice that when we write out this fraction, the mass π of our stone is common to numerator and denominator, and so it cancels out. Our result will be the same regardless of the mass of our stone. Now, the radius π of our stoneβs path is 0.33 meters. π, the magnitude of our stoneβs angular velocity, is given a 6.1 radians per second. And π, the acceleration due to gravity, is 9.8 meters per second squared.

With all these values substituted in, weβre now ready to calculate our ratio. The answer, to one decimal place, is 8.9. Note that there are no units in this result because itβs a ratio of two values that have the same units. The ratio then of the maximum force to the minimum force that the rope can apply to the stone is 8.9.

Letβs look now at part two of our question.

What is the force that the rope applies to the stone when the rope makes an angle π equals 33 degrees above the horizontal? Give your answer to the nearest newton.

We can now think of the stone being in this position so that we want to solve for the tension π along the rope at this angle π of 33 degrees. To help us do this, letβs consider this up-close sketch. The tension in the rope acts towards the center of the stoneβs rotation, thatβs here, and the weight force as always acts straight downward. Weβve said that for the stone to follow a circular arc, it must experience a constant center-seeking force. At all points in the stoneβs motion, this force is equal to some combination of the tension in the rope and the weight force π times π. At this particular position of our stone, we want to see how these two forces interact.

Notice that the weight force on the stone can be broken up into components that are parallel and perpendicular to the rope. This parallel component, because it points in the same direction as the tension in the rope, contributes to the overall centripetal force on the stone. If we call this component of the weight force πΉ sub π€, we can write that this component added to the tension in the rope is equal to the total centripetal force on the stone. The reason we add these two forces together is because we see they point in the same direction, towards the center of the stoneβs rotation.

In this part of our question, we want to solve for what weβve called π in this equation, the tension in the rope. Rearranging this equation a bit, we see that π equals π times π times π squared minus πΉ sub π€.

From what weβre told in our problem statement, we know the mass of the stone, the radius of its circular arc, and the angular speed with which it moves. What we donβt yet know is πΉ sub π€. If we consider the pink triangle of which πΉ sub π€ is one side, we note that that right triangle is similar to this right triangle weβve sketched in in orange. Therefore, this angle here in the orange triangle, 33 degrees, is the same as this angle here in the pink one.

Now, if we have a right triangle,and one of the other interior angles is called π, then the side of the triangle opposite π β weβve called it π β is equal to the hypotenuse of the right triangle β times the sin on the angle π. In our pink triangle, πΉ sub π€ is the side opposite this known angle of 33 degrees. For this right triangle, the hypotenuse is π times π, the mass of the stone times the acceleration due to gravity.

We can write then that πΉ sub π€, the component of the weight force that acts in the same direction as the tension force, equals the mass of the stone times the acceleration due to gravity times the sin of 33 degrees. We can substitute this result for πΉ sub π€ into our equation for the tension π. If we do that and then also factor out the mass of our stone, common to both terms on the right-hand side, we find that the tension in the rope in this position is equal to the mass of the stone multiplied by the quantity ππ squared minus π times the sin of 33 degrees.

Clearing a bit of space so we can substitute in for these variables, we know that the mass of our stone is 1.6 kilograms. The radius of the stone circle is 0.33 meters. And the angular speed of the stone is 6.1 radians per second. And π is 9.8 meters per second squared.

Calculating this entire expression, rounded to the nearest newton, we get a result of 11 newtons. This is the force that the rope applies to the stone when the rope makes an angle of 33 degrees above the horizontal.

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