Video: Comparing Forces and Angular Velocities for an Object Following Different Vertical Circular Paths

A roller coaster travels along a part of a track that forms a vertical loop between two curved tracks, as shown in the diagram. The curved tracks are circular arcs with an arc radius of 24 m. The top half of the loop is a semicircular arc with an arc radius of 14 m. A roller coaster car has a mass of 3,300 kg. At the base of the curved track, the car has a centripetal acceleration of 24 m/s². At the top of the loop, the car has a centripetal acceleration of 16 m/s. What is the ratio of the car’s angular velocity at the top of the loop to the car’s angular velocity at the base of the curved track? Give your answer to one decimal place.

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Video Transcript

A roller coaster travels along a part of a track that forms a vertical loop between two curved tracks, as shown in the diagram. The curved tracks are circular arcs with an arc radius of 24 meters. The top half of the loop is a semicircular arc with an arc radius of 14 meters. A roller coaster car has a mass of 3,300 kilograms. At the base of the curved track, the car has a centripetal acceleration of 24 meters per second squared. At the top of the loop, the car has a centripetal acceleration of 16 meters per second squared. What is the ratio of the car’s angular velocity at the top of the loop to the car’s angular velocity at the base of the curved track? Give your answer to one decimal place.

Right, so in this first part of the question, we’re asked to find the ratio of the angular velocity of a roller coaster car at two different points on the track. This particular segment of track forms a vertical loop between two curved tracks. We’re told that the two curved tracks are circular arcs with an arc radius of 24 meters. So this means that when the car is at the base of the curved track, it’s moving on part of a circle with a radius of 24 meters. Let’s label this radius as lowercase 𝑟 one. The question also tells us that the top half of the loop is a semicircular arc with an arc radius of 14 meters. Let’s label the second radius as lowercase 𝑟 two.

When the car is at the top of the loop, then it’s moving on part of a circle with a radius of lowercase 𝑟 two. So we can see that when the car is in each of these two positions, then it’s undergoing circular motion. In the first case, when it’s at the bottom of the curve track, that circular motion has a radius of 𝑟 one is equal to 24 meters. In the second case, where the car is at the top of the loop, that circular motion has a radius 𝑟 two equals 14 meters. We are also told that at the base of the curved track, the car has a centripetal acceleration of 24 meters per second squared.

Since this is on the circle with a radius of lowercase 𝑟 one, we’ll call this centripetal acceleration 𝑎 one. And we’re then told that, at the top of the loop, the car has a centripetal acceleration of 16 meters per second squared. And since this occurs on a circle with a radius of lowercase 𝑟 two, then we’ll call this centripetal acceleration 𝑎 two.

So just as a recap, the quantities that we’ve labeled with a subscript one are the values at the base of the curved track. And the quantities that we’ve labeled with a subscript two are the values at the top of the loop. Now, this first part of the question is asking us for the ratio of the car’s angular velocity at the top of the loop as compared to the base of the curved track. In each of these two cases, we know the radius of the circular motion and we know the value of the centripetal acceleration.

We can recall that there is a formula that relates the centripetal acceleration, the radius of the circular motion, and the angular velocity. Specifically, the centripetal acceleration, 𝑎 subscript 𝑐, is equal to the square of the angular velocity, 𝜔, multiplied by the radius of the circle, 𝑟. In our case, we’re trying to find the value of the angular velocity 𝜔. So let’s rearrange this formula to make 𝜔 the subject.

If we divide both sides of the equation by 𝑟, then on the right-hand side, the 𝑟’s in the numerator and denominator cancel each other out. Then swapping the left- and right-hand sides of the equation over, we have that 𝜔 squared is equal to 𝑎 subscript 𝑐 divided by 𝑟. Finally, taking the square root of both sides, we have that the angular velocity 𝜔 is equal to the square root of the centripetal acceleration, 𝑎 subscript 𝑐, divided by the radius 𝑟. Now we are ready to substitute in our values for the radius and the centripetal acceleration at each of the two points on the track in order to calculate the angular velocities.

We’ll begin with the case when the car is at the base of the curved track. At this point, the radius is lowercase 𝑟 one equal to 24 meters and the centripetal acceleration is 𝑎 one equals 24 meters per second squared. If we label the angular velocity at this point as 𝜔 one, then we have that 𝜔 one is equal to the square root of 𝑎 one divided by 𝑟 one, which is the square root of 24 meters per second squared divided by 24 meters. Since 24 divided by 24 gives one and the square root of one is equal to one, then we have that 𝜔 one is equal to one radian per second.

Notice that since the acceleration was measured in its base units of meters per second squared and the radius in its base units of meters, then the angular velocity will come out in its base units radians per second. Now let’s calculate the angular velocity at the top of the loop. In this case, the radius is lowercase 𝑟 two is equal to 14 meters and the centripetal acceleration is 𝑎 two equals 16 meters per second squared. Then, we have that the angular velocity at the top of the curved track, which we’ve labeled 𝜔 two, is equal to the square root of 𝑎 two divided by 𝑟 two. And this is the square root of 16 meters per second squared divided by 14 meters.

When we evaluate this, we find that 𝜔 two is equal to 1.069 and so on with further decimal places in units of radians per second. So we now have values for both 𝜔 one, the angular velocity at the base of the curved track, and for 𝜔 two, the angular velocity at the top of the loop. The question asks us for the ratio of the angular velocity at the top of the loop, so that’s 𝜔 two, to the angular velocity at the base of the curved track, 𝜔 one. To find this ratio, we need to take our value of 𝜔 two and divide it by 𝜔 one. When we do this, there’s a couple of things to notice.

First off, in the denominator, we see that we’re dividing by a numerical value of one, and any number divided by one is just that same number that we started with. So actually, a numerical answer is just going to be this number from the numerator. The second thing to notice is that we’re dividing an angular velocity by an angular velocity. In other words, that’s a quantity with units of radians per second divided by another quantity with units of radians per second. So when we do this, the units cancel each other out, and we’re left with a dimensionless quantity as our answer.

We now have our answer for the ratio of the angular velocity at the top of the loop to the angular velocity at the base of the curved track. But there’s one last step to go because the question asks us to give this answer to one decimal place. When we round our result to one decimal place, we get our answer that the ratio of these angular velocities is given by 1.1.

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

What is the ratio of the normal reaction force 𝑟 one on the car at the base of the curved track to the normal reaction force 𝑟 two on the car at the top of the loop? Give your answer to one decimal place.

Okay, so this part of the question is considering the same two positions on the track. But this time, we’re asked to find the ratio of the normal reaction force at these two places. This normal reaction force is the force exerted on the roller coaster car by the track. As we see from the diagram, the normal reaction force is perpendicular to the direction of the track. And when the car is undergoing circular motion as it is in these two positions, that force is directed towards the center of the circle. We can recall that any object undergoing circular motion experiences a centripetal force, 𝐹 subscript 𝑐, equal to the mass of the object, 𝑚, multiplied by the centripetal acceleration, 𝑎 subscript 𝑐. This centripetal force is the net inward force toward the center of the circle.

In the case of our roller coaster car, moving on this vertical loop, we have two forces contributing to the centripetal force. First, we have the normal reaction force which always points toward the center of the circle. This force has been drawn onto the diagram for us at the two positions we’re asked to consider. At the base of the curved track, it’s labeled 𝑟 one. And at the top of the loop, it’s labeled 𝑟 two. The second force that we need to consider is the force due to gravity. And this always acts vertically downward because gravity acts to pull things down toward the Earth.

Let’s add this force due to gravity onto our diagram at the two points that we are asked to consider. We said that the force due to gravity always acts vertically downward, so we can draw this force as an arrow pointing vertically downward at each position. The magnitude of this gravitational force is equal to 𝑚 multiplied by 𝑔, where 𝑚 is the mass of the object and 𝑔 is the gravitational field strength. We are told in the question that the roller coaster car has a mass of 3,300 kilograms. So in our case, we have that 𝑚 is equal to 3300 kilograms. We can also recall that the gravitational field strength on Earth has a value of 9.8 meters per second squared to two significant figures.

Now that we’ve got both the normal reaction force and the force due to gravity drawn onto our diagram, let’s see how these relate to the centripetal force 𝐹 subscript 𝑐. We’ve said that this centripetal force is the net force toward the center of the circle. So at the base of the curved track, the centripetal force that we’ve called 𝐹 subscript 𝑐 one is equal to the normal reaction force 𝑟 one, which points toward the center of the circle, minus the gravitational force 𝑚𝑔, which points in the opposite direction. Meanwhile, at the top of the loop, the centripetal force, 𝐹 subscript 𝑐 two, is equal to the normal reaction force, 𝑟 two, plus the gravitational force, 𝑚𝑔, since in this case, both of the two forces act toward the center of the circle.

Now we also know that the centripetal force is equal to the mass of the object multiplied by the centripetal acceleration. Since we know the mass of our roller coaster car and we know the value of the centripetal acceleration at each of the two positions, then it makes sense to use this equation here to replace the centripetal force on the left-hand side of these equations by the mass multiplied by the centripetal acceleration. When we do this, we have two equations, one that applies at the base of the curved track and the other that applies at the top of the loop. And the key point is that now, in each of these equations, we only have one unknown quantity. And that’s the value of the normal reaction force.

So let’s rearrange these equations in order to make the normal reaction force the subject. We’ll begin with this top equation for when the car is at the base of the curved track. First, we add 𝑚 times 𝑔 to each side of the equation, giving us 𝑚 times 𝑎 one plus 𝑚 times 𝑔 is equal to 𝑟 one.

We can tidy this up a bit, swapping the left- and right-hand sides of the equation over and factoring out the mass 𝑚, which is common to both of these two terms. Then we have that the normal reaction force 𝑟 one is equal to 𝑚 multiplied by the sum 𝑎 one plus 𝑔. When we substitute in our values for 𝑚, 𝑎 one, and 𝑔, we get that the normal reaction force 𝑟 one is equal to our mass of 3,300 kilograms multiplied by the sum of our centripetal acceleration, 24 meters per second squared, and the gravitational field strength, 9.8 meters per second squared. Evaluating this expression gives us that 𝑟 one is equal to 111,540 kilogram-meters per second squared, which we can also write as 111,540 newtons.

Now, we’ll look at this lower equation for when the car is at the top of the loop. If we subtract 𝑚 times 𝑔 from both sides of the equation, then we have that 𝑚 times 𝑎 two minus 𝑚 times 𝑔 is equal to 𝑟 two. Factoring out the mass 𝑚 which appears in both these two terms and also swapping the left- and right-hand sides of the equation over, we get this equation here, which says that the normal reaction force 𝑟 two is equal to the mass 𝑚 multiplied by the difference 𝑎 two minus 𝑔. Substituting in our values for 𝑚, 𝑎 two, and 𝑔, we have that 𝑟 two is equal to the mass of 3,300 kilograms multiplied by the difference 16 meters per second squared minus 9.8 meters per second squared. Evaluating this expression gives us that 𝑟 two is equal to 20,460 kilograms-meters per second squared or equivalently 20,460 newtons.

So we now know the value of the normal reaction force, both at the bottom of the curved track and at the top of the loop. The question asks us to find the ratio of this normal reaction force on the car at the base of the curved track, so that’s 𝑟 one, as compared to the top of the loop, so that’s 𝑟 two. This means that we need to find the result of dividing 𝑟 one by 𝑟 two. Substituting these values in gives us that our ratio is equal to 111,540 newtons, so that’s 𝑟 one, divided by 20,460 newtons; that’s 𝑟 two. Evaluating this division gives us that the ratio 𝑟 one over 𝑟 two is equal to 5.4516 and so on with further decimal places.

Notice that since we have a quantity with units of newtons divided by another quantity also with units of newtons, then, as with the first part of the question, when we do the division, the units cancel each other out, and we’re left with a dimensionless quantity. This value here gives us the ratio of the normal reaction force on the car at the base of the curved track to the normal reaction force on the car at the top of the loop, which is what the question was asking us to find. The last remaining step is to round this result to one decimal place. Doing this gives us our final answer of 5.5.

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