Question Video: Analyzing Object Motion around a Circular Inclined Track Physics • 9th Grade

A car with a mass of 660 kg drives at constant speed along a smooth circular track, as shown in the diagram. The car follows a path along the center of the track, maintaining a constant distance π‘Ÿ = 22 m to the center of its circular path. The angle of the track above the horizontal πœƒβ‚ = 28Β°. What is the angle πœƒβ‚‚ from the vertical at which 𝑅, the reaction force on the car from the surface of the track, acts? What is the magnitude of the force that acts on the car along π‘Ÿ? How much time does the car take to return to a point along its path?

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Video Transcript

A car with a mass of 660 kilograms drives at constant speed along a smooth circular track, as shown in the diagram. The car follows a path along the center of the track, maintaining a constant distance π‘Ÿ equals 22 meters to the center of its circular path. The angle of the track above the horizontal, πœƒ one, equals 28 degrees. What is the angle πœƒ two from the vertical at which 𝑅, the reaction force on the car from the surface of the track, acts?

In our diagram, we see this angle πœƒ one, which is a measure of the angle between the inclination of our track and a horizontal plane. When a car drives around the track, it experiences a reaction force called 𝑅. In the first part of our question, we want to solve for the angle between this reaction force and a vertical line. That angle is called πœƒ two. To get started, let’s clear some space on screen, and let’s imagine that we’re looking at a cross section of this inclined plane with a car on it. From this perspective, the car will be coming out of the screen towards us. The normal reaction force 𝑅 that acts on the car is indeed normal to or perpendicular to this inclined surface. And the angle between the reaction force and the vertical line is the one we want to solve for, πœƒ two.

If we consider the motion of our car, we know that the car is neither moving in this direction off of the track nor in this direction into the track. That means there must be an equal and opposite force to the reaction force 𝑅, and indeed there is. It’s a component of what we can call the weight force equal to the mass of the car times the acceleration due to gravity. If the overall weight force is represented by this pink arrow, then its perpendicular components are represented by these two orange ones. Since these two orange arrows are perpendicular, we know that they’re separated from one another by 90 degrees. Therefore, this triangle of the weight force and its components is a right triangle, and indeed so is the larger blue triangle we’ve drawn here. In fact, these two right triangles are similar to one another.

In that similarity, this right angle here is copied here in the smaller triangle. Then, this interior angle in the larger triangle, whatever this angle is, is copied here in the smaller triangle, and the angle πœƒ one in our larger triangle is copied right here. Knowing that that angle in our smaller triangle is πœƒ one, let’s see how that compares to πœƒ two. If we extend the line along which the pink arrow lies and we also draw a dashed line in line with the larger component of the weight force of the car, we see that the angle between those two line extensions, the two pink dashed lines, is πœƒ two. We’ve seen in our smaller triangle though that this angle is πœƒ one. We can write then that πœƒ one equals πœƒ two. And let’s recall from our problem statement that πœƒ one is given as 28 degrees. This answers our question then as to the value of πœƒ two. πœƒ two is equal to 28 degrees.

Let’s look now at part two of our question.

What is the magnitude of the force that acts on the car along π‘Ÿ?

We can find lowercase π‘Ÿ on our larger diagram. It’s the distance between the car when it’s on the middle of this curved track and the center of that track. On our sketch then, the direction of lowercase π‘Ÿ points like this. Now, let’s assume that the only two forces acting on our car at this moment are the reaction force, capital 𝑅, and the weight force. We’ll call this the mass of the car times the acceleration due to gravity. We can see that the direction of lowercase π‘Ÿ and the weight force π‘š times 𝑔 are separated by 90 degrees. This means there is no component, no part of the weight force that acts along the direction of lowercase π‘Ÿ. So the only force that can have some component along this direction is the reaction force capital 𝑅.

Let’s now consider an up-close sketch of this reaction force as it relates to the direction of lowercase π‘Ÿ. We can divide the reaction force into vertical and horizontal components. Here’s the vertical component of that force, and here’s the horizontal component. Notice that the horizontal component of capital 𝑅 is indeed in the direction of lowercase π‘Ÿ. Whatever this magnitude is then is the answer to this part of our question. We know that the angle between our reaction force capital 𝑅 and a vertical line is 28 degrees. Since this vertical dashed line is parallel with this vertical component of the force capital 𝑅, the angle here between that vertical component of the reaction force and the reaction force itself is also 28 degrees. This angle of 28 degrees we’ve just discovered is part of a right triangle.

We can now remember that given a right triangle, where we know one of the other interior angles, we’ll say it’s called πœƒ, the length of the triangle side opposite πœƒ, here it’s called π‘œ, is equal to the hypotenuse of that triangle, called β„Ž, times the sin of πœƒ. We can make use of this result over here in our right triangle. If we call this length of the side of our right triangle 𝐹 sub π‘Ÿ, for the magnitude of the force that acts along the direction of lowercase π‘Ÿ, by the result we recalled over here, we can say that 𝐹 sub π‘Ÿ is equal to 𝑅, the magnitude of the reaction force times the sin of 28 degrees. This is a step towards our answer, but we don’t yet know what the reaction force capital 𝑅 is.

To determine that, we need to go back to our original sketch of the scenario. We said that the reaction force capital 𝑅 is equal and opposite to this component of the weight force. That component we see is also part of a right triangle, where the hypotenuse of that triangle is equal to the mass of this car times the acceleration due to gravity. To find the magnitude of our orange arrow that represents the larger of the two components of the weight force, we can note that the length of that arrow is equivalent to the side length π‘Ž over here on our right triangle. Unlike the side length π‘œ, the side length π‘Ž is given by the relation β„Ž times the cos of πœƒ. Here, we can note that π‘Ž is the side of the triangle adjacent to the angle πœƒ, while π‘œ is the side opposite πœƒ.

Now, how can we use this result to solve for the length of our larger component of the weight force? In this right triangle, the hypotenuse is not β„Ž, but rather it’s π‘š times 𝑔. So we have π‘š times 𝑔. And then, that’s multiplied by what we’ve called over here the cos of the angle πœƒ. In our diagram, that angle is πœƒ one, which we know is 28 degrees. So π‘š times 𝑔 times the cos of 28 degrees is equal to the length of the larger component of the weight force. And as we’ve said, that’s equal in magnitude to the reaction force capital 𝑅. We can now take this expression for capital 𝑅 and substitute it in for capital 𝑅 in this equation.

We now see that the magnitude of the force that acts on the car in the direction of lowercase π‘Ÿ is π‘š, the car’s mass, times 𝑔, the acceleration due to gravity, times the cos of 28 degrees times the sin of 28 degrees. In our problem statement, we’re told that the mass of the car, we’ve called it π‘š, is 660 kilograms. We can then recall that 𝑔, the acceleration due to gravity, is 9.8 meters per second squared. We have then that 𝐹 sub π‘Ÿ equals 660 kilograms times 9.8 meters per second squared times the cos of 28 degrees times the sin of 28 degrees.

Computing this value, we get an answer of about 2681 newtons. We note though that in our problem statement, we’re given values, such as the mass of the car and the radial distance lowercase π‘Ÿ, up to two significant figures. Therefore, we’ll want to round our final answer to two significant figures. When we do that, 2681 rounds to 2700 newtons. This is the magnitude of the force that acts on the car along the direction of lowercase π‘Ÿ.

Let’s look now at the last part of our question.

How much time does the car take to return to a point along its path?

Here, we’re thinking of the car completing one revolution on this circle. We want to know the time it takes to do that, and we’ll call that time capital 𝑇. One way to solve for capital 𝑇 is to think about the centripetal force, the center-seeking force that acts on the car. Generally, centripetal force is represented by 𝐹 sub 𝑐. And it’s equal to the mass of an object moving in a circular path multiplied by the radius of that path times the angular speed of the object squared. In our case, the centripetal force that acts on our car is equal to the force on the car in the direction of lowercase π‘Ÿ that we solved for in part two. Recall that we called that force 𝐹 sub lowercase π‘Ÿ. And we found it was equal to 2700 newtons.

So then, the force on the car in the direction of lowercase π‘Ÿ is the centripetal force on the car. And our equation tells us that 𝐹 sub 𝑐 equals the car’s mass times the radius of the circular path it follows times its angular speed squared. In general, angular speed πœ” equals a change in an object’s angular position, Ξ”πœƒ, divided by the time it took to change that position, Δ𝑑. In this part of our question, we’re thinking about a change in angular position of one complete revolution around this path. In units of radians, one revolution is two πœ‹ radians, and the total time it takes to do this is capital 𝑇. Recall that this is the value we want to solve for.

To help us do that, let’s replace πœ” in this equation with two πœ‹ radians divided by capital 𝑇. We can now simplify the way this expression looks a bit by removing 𝐹 sub 𝑐, the centripetal force. We can write simply that 𝐹 sub π‘Ÿ, the force whose magnitude we calculated, equals π‘š times π‘Ÿ times πœ” squared. If we clear some space and then multiply both sides of this equation by capital 𝑇 squared, then capital 𝑇 cancels out from the right-hand side. As a next step, we divide both sides of the equation by 𝐹 sub lowercase π‘Ÿ. That makes this factor cancel out on the left.

And then, as a last step to get an equation where capital 𝑇 is the subject, we take the square root of both sides. The square root of capital 𝑇 squared equals simply capital 𝑇. Then, on the right-hand side of our expression, we have the square root of π‘š times π‘Ÿ over 𝐹 sub lowercase π‘Ÿ all multiplied by two πœ‹ radians. Just like with capital 𝑇, when we squared and then took the square root of this factor, we ended up with simply the result in the parentheses, two πœ‹ radians.

In this expression for capital 𝑇, we know the mass of the car π‘š. We know the force 𝐹 sub lowercase π‘Ÿ. And we’re also told the radial distance lowercase π‘Ÿ. In our problem statement, that’s given as 22 meters. We now substitute all these values into our expression. Rounding our answer to two significant figures, we get a result of 15 seconds. This is how long it takes the car to return to a point along its path. This is also known as the car’s period.

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