# Question Video: Determining the Centripetal Force for a Car Travelling around a Circular 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 𝑟 = 22 m to the center of its circular path. The angle of the track above the horizontal 𝜃 = 28°. What is the magnitude of the force that acts on the car along 𝑟? Answer to the nearest newton.

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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 𝑟 equals 22 meters to the center of its circular path. The angle of the track above the horizontal 𝜃 equals 28 degrees. What is the magnitude of the force that acts on the car along 𝑟? Answer to the nearest newton.

This question asks what the magnitude of the force acting along the radius 𝑟 of a circular path that a car moves along is. This force on the car is due to the normal reaction force on the car due to the contact between the car and an inclined surface.

Let’s begin by recording the information we’re given. The mass of the car, which we’ll call 𝑚, is 660 kilograms. The car travels on a circular track with radius 𝑟 equals 22 meters. And the track is angled at 𝜃 equals 28 degrees above the horizontal.

Clearing some space on screen, let’s recall that when an object moves in a circle, it is acted on by a force toward the center of that circle known as centripetal force. Represented by 𝐹 sub 𝑐, centripetal force equals an object’s mass 𝑚 times its linear speed 𝑣 squared divided by the radius of the circular path the object follows 𝑟. In our example, the force on the car towards the center of the track is centripetal. So we’ll label it with a subscript 𝑐.

Let’s consider viewing the car as though it is moving directly toward us. Two forces act on the car: the weight of the car downward and the normal reaction force that we’ll call 𝑅. The slope is smooth, so friction forces up or down the slope of the track can be ignored. Notice that there is no force we’ve called 𝐹 sub 𝑐 acting on the car. Nonetheless, there is a center-seeking force on the car.

From our perspective, the center of the track is somewhere off to the right. And a component of the normal reaction force points this way. Since it always points toward the track’s center, it is a centripetal force. Centripetal forces are always caused by real forces acting on an object. In this case, we see that force is the normal reaction force. This is the physical reason for 𝐹 sub 𝑐.

Rather than solving for 𝐹 sub 𝑐 using the equation above, we’re going to do it using our free body diagram. First, let’s express 𝐹 sub 𝑐 in terms of the normal reaction force 𝑅. To do this, we can set up coordinate axes like this so that 𝐹 sub 𝑐 is entirely in the positive 𝑥-direction. If we divide 𝑅 into its 𝑥- and 𝑦-components, we see that a right triangle is formed by these vectors. This right triangle is actually similar to the right triangle formed by the track, the ground, and the vertical line connecting them. Therefore, because the track is inclined at an angle 𝜃 from the horizontal, the topmost angle in the triangle of force vectors is also 𝜃.

We can say then that 𝐹 sub 𝑐 is equal to the force vector 𝑅 times the sin of 𝜃. We know 𝜃; it’s 28 degrees. But we don’t yet know 𝑅. To solve for it, we’ll compare the normal reaction force with the weight force 𝑊. We’ll be helped in doing this by resetting our coordinate axes so that now positive 𝑥 points down the track and positive 𝑦 points in the direction of 𝑅, perpendicularly away from the track.

If we separate 𝑊 into its 𝑥- and 𝑦-components using these new axes, we find that, first, we once again have a right triangle similar to the one formed by the track, where here 𝜃 is the angle between 𝑊 and 𝑊 sub 𝑦. And second, the component 𝑊 sub 𝑦 is equal in magnitude to the normal reaction force 𝑅.

Using our right triangle of weight force vectors, we see that 𝑊 sub 𝑦 equals 𝑊 times the cos of 𝜃. Combining these equations, we see that 𝑅 equals 𝑊 cos 𝜃. So we can substitute the right side of this equation in for 𝑅 in our earlier expression. In general, the weight force 𝑊 of an object equals its mass 𝑚 times the acceleration due to gravity 𝑔. We can make that substitution and now have an expression for 𝐹 sub 𝑐 in terms of known quantities. 𝑚 is 660 kilograms, 𝑔 is 9.8 meters per second squared, and 𝜃 is 28 degrees.

Calculating this expression gives a result, to the nearest newton, of 2681 newtons. This is the magnitude of the centripetal force on the car, the force which acts in the direction of 𝑅.