Question Video: Determining the Variation in the Component of a Centripetal Force Due to Tension for an Object Following a Vertical Circular Path Physics • 9th Grade

A stone that has a mass of 2.4 kg is swung in a vertical circle at a constant angular velocity of 7.2 rad/s. The stone is attached to a uniform rope of length 0.25 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 difference between the magnitude of the force that the rope applies to the stone at point 𝐴, where the rope points vertically upward, and the magnitude of the force that the rope applies to the stone at point 𝐡, where the rope points horizontally? Give your answer to the nearest newton.

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Video Transcript

A stone that has a mass of 2.4 kilograms is swung in a vertical circle at a constant angular velocity of 7.2 radians per second. The stone is attached to a uniform rope of length 0.25 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 difference between the magnitude of the force that the rope applies to the stone at point 𝐴, where the rope points vertically upward, and the magnitude of the force that the rope applies to the stone at point 𝐡, where the rope points horizontally? Give your answer to the nearest newton.

Okay, so in this first part of the question, we’re asked to find the difference between the magnitude of the force that the rope applies to the stone at point 𝐴, so that’s this point here, as compared to point 𝐡, that’s this point here. We know that at point 𝐴, the rope points vertically upward, while at point 𝐡, the rope points horizontally. The question tells us that the mass of the stone is 2.4 kilograms. If we label this mass as π‘š, then we have that π‘š is equal to 2.4 kilograms. We’re also told that the rope which the stone is attached to has a length of 0.25 meters and that this length is the same as the radius of the circle throughout the motion of the stone. Therefore, this length gives us the radius of the circular motion of the stone. We’ll label this radius π‘Ÿ so that we have π‘Ÿ is equal to 0.25 meters.

The final piece of information that we are given is that the stone is swung in a vertical circle at a constant angular velocity of 7.2 radians per second. We’ll label this angular velocity as πœ” so that we have πœ” is equal to 7.2 radians per second. Since our stone is undergoing circular motion, we know that it will experience a centripetal force. We can recall that the centripetal force experienced by an object in circular motion, which here we’ve labeled 𝐹 subscript 𝐢, is equal to the mass of that object, π‘š, multiplied by the radius of the circle, π‘Ÿ, multiplied by the square of the angular velocity, πœ”.

Now, in our case, the radius of the circle is constant at 0.25 meters. The angular velocity is constant at 7.2 radians per second. And of course the mass of the stone is also constant at 2.4 kilograms. Since all three of the quantities on the right-hand side of this equation are constant, then this means that the value of 𝐹 subscript 𝐢, the centripetal force experienced by the stone, must also remain constant. In other words, this centripetal force must have the same magnitude at all points around the circle. In particular, this means that it must have the same value at point 𝐴 as it does at point 𝐡.

Now, this centripetal force on the stone arises from two different sources. We have the tension force acting on the stone provided by the rope, and we also have the force due to gravity. The force from the rope will always act towards the center of the circle since the rope extends from the center of the circle out to the radius of the trajectory of the stone. Meanwhile, we know that gravity acts to pull things down toward the earth. So, the force due to gravity will always act vertically downward, no matter where on the circle the stone is. So, let’s add these two forces onto our diagram at the points marked 𝐴 and 𝐡.

First, let’s consider point 𝐴. We have the tension force from the rope, which we know acts towards the center of the circle, which in this case is vertically downward. Let’s label this force from the rope at point 𝐴 as 𝐹 subscript 𝐴. Then, we also have our force due to gravity, which we know always acts vertically downward. This force has a magnitude of π‘š multiplied by 𝑔, where π‘š, which is 2.4 kilograms, is the mass of the stone and 𝑔 is the gravitational field strength. The value of 𝑔 on Earth to two significant figures is equal to 9.8 meters per second squared.

Now, let’s consider the forces acting on the stone at point 𝐡. First off, we have the force from the rope, which we know acts towards the center of the circle. Let’s label this force from the rope at point 𝐡 as 𝐹 subscript 𝐡. Then, we also have the force due to gravity, which we know always acts vertically downward. The magnitude of this force is equal to π‘š multiplied by 𝑔, just the same as it was at point 𝐴.

Looking at point 𝐴, we see that both of the two forces act toward the center of the circle. This means that both forces contribute to the centripetal force 𝐹 subscript 𝐢. So, at point 𝐴, we can say that the centripetal force 𝐹 subscript 𝐢 is equal to the force from the rope 𝐹 subscript 𝐴 plus the force due to gravity π‘š multiplied by 𝑔. Meanwhile, if we look at point 𝐡, then we see that only the force from the rope acts toward the center of the circle. The force due to gravity is completely perpendicular to this direction. This means that this force due to gravity π‘š multiplied by 𝑔 does not contribute to the centripetal force at point 𝐡. So, at point 𝐡, we simply have that the centripetal force 𝐹 subscript 𝐢 is equal to the force provided by the rope 𝐹 subscript 𝐡.

Recall that the question was asking us to find the difference between the magnitude of the force applied by the rope at point 𝐴 and point 𝐡. That’s the difference between the value of 𝐹 subscript 𝐴 and 𝐹 subscript 𝐡. So, let’s calculate these two values and find their difference. Beginning with 𝐹 subscript 𝐴, we’ll take this equation and rearrange it to make 𝐹 subscript 𝐴 the subject. We can do this by subtracting π‘š multiplied by 𝑔 from both sides of the equation. Then, we have that 𝐹 subscript 𝐢 minus π‘š times 𝑔 is equal to 𝐹 subscript 𝐴. We can then use this equation for the centripetal force to replace 𝐹 subscript 𝐢 by π‘š times π‘Ÿ times πœ” squared. Then we have that π‘š times π‘Ÿ times πœ” squared minus π‘š times 𝑔 is equal to 𝐹 subscript 𝐴.

Since we know the values of the quantities π‘š, π‘Ÿ, πœ”, and 𝑔, then we can substitute them in to the left-hand side of this equation in order to calculate 𝐹 subscript 𝐴. When we do these substitutions, our first term on the left-hand side becomes 2.4 kilograms, that’s our mass π‘š, multiplied by 0.25 meters, our radius π‘Ÿ, multiplied by the square of 7.2 radians per second, that’s our angular velocity πœ”. The second term, which we subtract from the first, becomes 2.4 kilograms, our mass π‘š, multiplied by 9.8 meters per second squared, our value of the gravitational field strength 𝑔. Then, the first term evaluates to 31.104 and the second term evaluates to 23.52.

Now, we’ve gone ahead and given these terms units of newtons. Since this expression is equal to our force 𝐹 subscript 𝐴, then we know that the terms must have units of force. And since all of the quantities in the line above are expressed in terms of their base units, then we know that the force must also come out in its base units, which is units of newtons. It’s worth pointing out that this first value on the left-hand side is the value of π‘š times π‘Ÿ times πœ” squared, which is just the value of the centripetal force 𝐹 subscript 𝐢. We said already that 𝐹 subscript 𝐢 has the same magnitude at all points around the circle. And now we can put a number on that magnitude. Specifically, it’s equal to 31.104 newtons. Finally, when we evaluate this subtraction, we find that 𝐹 subscript 𝐴, the value of the force provided on the stone by the rope, is equal to 7.584 newtons.

Now, let’s work out 𝐹 subscript 𝐡. In fact, this step is really easy since we know that 𝐹 subscript 𝐡 is equal to 𝐹 subscript 𝐢. And we’ve already worked out that 𝐹 subscript 𝐢, the centripetal force, is equal to 31.104 newtons. So, we therefore know that 𝐹 subscript 𝐡 is equal to 31.104 newtons.

The final step is to find the difference between these two values 𝐹 subscript 𝐴 and 𝐹 subscript 𝐡. If we subtract 𝐹 𝐴 from 𝐹 𝐡, then that gives us 31.104 newtons minus 7.584 newtons, which is equal to 23.52 newtons. And this value is the difference between the magnitude of the force that the rope applies to the stone at point 𝐴 as compared to point 𝐡. But the question actually asked us to give this answer to the nearest newton. So, rounding our result to the nearest newton, we get an answer of 24 newtons.

Now, it turns out that there was actually a quicker way that we could have got to this answer. Let’s go back to the point where we’d just worked out the equations for the forces acting on the stone at point 𝐴 and point 𝐡. At point 𝐴, we knew that the centripetal force 𝐹 subscript 𝐢 was equal to 𝐹 subscript 𝐴 plus π‘š multiplied by 𝑔, while at point 𝐡, we knew that the centripetal force was equal to 𝐹 subscript 𝐡. But we’d already said that we knew that the magnitude of the centripetal force 𝐹 subscript 𝐢 would be the same at all points around the circle.

So, looking at these two equations, we can see that both of them must have the same value on the left-hand side, since 𝐹 subscript 𝐢 has the same value at point 𝐴 as it does at point 𝐡. If the two equations have the same left-hand side, then the right-hand sides of these two equations must also be equal. Setting these two right-hand sides equal to each other gives us an equation that says 𝐹 subscript 𝐡 is equal to 𝐹 subscript 𝐴 plus π‘š times 𝑔.

So now we can see straightaway that the difference between the force provided by the rope at point 𝐡 as compared to point 𝐴 is equal to this value π‘š times 𝑔. We can make this more explicit by rearranging the equation, subtracting 𝐹 subscript 𝐴 from both sides. Then we have that 𝐹 subscript 𝐡 minus 𝐹 subscript 𝐴 is equal to π‘š times 𝑔. Putting in the values that π‘š is equal to 2.4 kilograms and 𝑔 is equal to 9.8 meters per second squared, we calculate that 𝐹 subscript 𝐡 minus 𝐹 subscript 𝐴 is equal to 23.52 newtons. And again, rounding this to the nearest newton gives us our same answer of 24 newtons.

This second approach was a whole lot quicker. We didn’t even need to know the value of 𝐹 subscript 𝐢. We simply used the fact that it was a constant in order to find the difference between 𝐹 subscript 𝐴 and 𝐹 subscript 𝐡.

Now, let’s have a look at the second part of the question.

What is the difference between the magnitude of the force that the rope applies to the stone at point 𝐴, where the rope points vertically upward, and the magnitude of the force that the rope applies to the stone at point 𝐢, where the rope points vertically downward? Give your answer to the nearest newton.

So, in this second part of the question, we’re being asked to find the difference between the magnitude of the force that the rope applies to the stone at point 𝐴 as compared to point 𝐢. Now, we already know that at point 𝐴, the rope is pointing vertically upward. And both of our forces, so that’s the force due to the rope 𝐹 subscript 𝐴 and the force due to gravity π‘š times 𝑔, are acting toward the center of the circle. Now, at point 𝐢, the rope is vertically downward.

We know that the force from the rope always acts along the direction of the rope toward the center of the circle. So, we can draw this force at point 𝐢 acting vertically upward toward the center of the circle. Now, we won’t use the same naming convention as we did for this force at points 𝐴 and 𝐡 since 𝐹 subscript 𝐢 is already taken for the centripetal force. Instead, we’ll call this force from the rope at point 𝐢 𝐹 subscript 𝑅. Then, the other force that we need to consider is the force on the stone due to gravity. We know that this force always acts vertically downward and that it has a magnitude of π‘š times 𝑔, where π‘š is the mass of the stone and 𝑔 is the gravitational field strength.

We already have an equation for the forces acting on the stone at point 𝐴. So, let’s write down an equivalent equation at point 𝐢. Recalling that the centripetal force 𝐹 subscript 𝐢 is the resultant force toward the center of the circle, then at point 𝐢, we have that the centripetal force 𝐹 subscript 𝐢 is equal to the force from the rope 𝐹 subscript 𝑅 minus the force due to gravity π‘š times 𝑔. This is because the force from the rope acts inward toward the center of the circle, while the force due to gravity acts in the opposite direction outward away from the center.

Now that we have these two equations at point 𝐴 and point 𝐢, we can use the fact that 𝐹 subscript 𝐢 has the same magnitude at all points around the circle. So, these two equations have the same left-hand side. This means that the right-hand sides of these two equations must also be equal to each other. So we can write that 𝐹 subscript 𝐴 plus π‘š times 𝑔 is equal to 𝐹 subscript 𝑅 minus π‘š times 𝑔. Subtracting π‘š times 𝑔 from both sides of this equation gives us that 𝐹 subscript 𝐴 plus two times π‘š times 𝑔 is equal to 𝐹 subscript 𝑅. Then, subtracting 𝐹 subscript 𝐴 from both sides gives us that two times π‘š times 𝑔 is equal to 𝐹 subscript 𝑅 minus 𝐹 subscript 𝐴. In other words, the difference between the force provided to the stone by the rope at point 𝐢 and point 𝐴 is equal to two times π‘š times 𝑔.

Substituting in our numbers for the mass of the stone and the value of the gravitational field strength, we have that this difference is equal to two multiplied by our mass of 2.4 kilograms multiplied by our value of 𝑔 of 9.8 meters per second squared. Evaluating this gives a result of 47.04 newtons. And this number is the difference between the magnitude of the force from the rope at point 𝐴 as compared to point 𝐢. But as with the first part of the question, we’re asked to give this answer to the nearest newton. Rounding our value of 47.04 newtons to the nearest newton gives us our answer of 47 newtons.

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