Lesson Explainer: Centripetal Force Physics • 9th Grade

In this explainer, we will learn how to analyze the magnitudes, directions, and sources of forces that act on objects moving along circular paths.

We recall that, for an object to change velocity, it must accelerate, and so a nonzero net force must act on it.

A change in velocity does not necessarily involve a change in speed. An object can change the direction of its velocity without changing the magnitude of its velocity. This still corresponds to an acceleration and still requires a nonzero net force.

An object that travels on a curved path at a constant speed must then have a nonzero net force acting on it.

A simple example of a curved path is a circular one. We will show in this explainer that when an object follows a circular path, a force must act on it directed toward the center of the circle that the object moves along. This force is called centripetal force, which means a force that acts toward a center.

Let us consider a satellite that orbits Earth along a circular path, as shown in the following figure.

In this figure, Earth is at the center of the circular path. At the instant shown, a gravitational force, 𝐹, acts on the satellite toward Earth, hence toward the center of the circular path. Gravitational force provides the centripetal force that produces the circular motion of the satellite.

At every point on the path traveled by the satellite, the force acts toward the center of the circle. Two such points are shown in the following figure.

To follow a circular path, the satellite must have a velocity in the direction of the path at each instant. The direction of the circular path at a point on the path is tangent to the radius of the circle at that point. This is shown in the following figure, where velocities π‘£οŠ§ and π‘£οŠ¨ have equal magnitudes and are perpendicular.

We can take a point midway along the circular path between these two points, as shown in the following figure.

Let us consider the direction in which velocity π‘£οŠ§ must change to become velocity π‘£οŠ¨, calling this Δ𝑣. The direction of Δ𝑣 at this point is shown in the following figure.

We see that Δ𝑣 acts toward the center of the circle. The change in velocity is due to the acceleration that is due to the gravitational force. We see that this force acts centripetally.

There is a problem with this model, however, which is of course that the satellite cannot arrive at the point shown if it has velocity π‘£οŠ§. The following figure shows what the actual position of the satellite would be.

The difference between the position of the point on the circular path and that of the point that moves with velocity π‘£οŠ§ is less for a point that travels a smaller distance. This is shown in the following figure.

Part of the perimeter of the circle is shown. Two nearby points, A and B, and a third point C midway between them are shown. The radial lines to each of these points are also shown.

The pink horizontal line from A to C is tangent to the circular path where the radial line intersects A. The length of the arc from A to C is nearly equal to that of the horizontal line.

We can imagine considering pairs of points that are arbitrarily close together. We can also imagine taking an unlimited number of such pairs of points, distributed uniformly around the circle such that the points span the circumference of the circle.

As the number of pairs of points increases, the distance between the points decreases. As the distance between the points of a pair decreases, the straight-line distances between the points of the pair become more nearly equal to the length of the arc between the points.

As the distance between the points becomes negligible, the satellite moves uniformly along a circular path, and the direction of the change in the velocity of the satellite throughout this motion is toward the center of the circle.

We have shown that there is a change of velocity, Δ𝑣, toward the center of the circle. The change of this velocity with time is an acceleration toward the center of the circle, Δ𝑣Δ𝑑.

It is possible to determine how acceleration toward the center of the circle relates to the radius of the circle and the speed of motion along the circumference of the circle.

Consider the following figure.

We see that an object moving in a circle of radius π‘Ÿ has velocity 𝑣A at point A and has velocity 𝑣B at point B. At each point, the object has a speed of 𝑣, where 𝑣=|𝑣|,𝑣=|𝑣|.AB

The angle between A, the center of the circle, and B, changes by Ξ”πœƒ as the object moves from A to B.

The straight-line distance from A to B is Δ𝑑. When A and B are very close to each other, Δ𝑑 is approximately equal to the distance traveled by the object between A and B. We will consider A and B to be sufficiently close together for this to be the case, meaning of course that the figure is not drawn to scale.

The triangle between A, the center of the circle, and the midpoint of A and B (triangle i) is similar to the triangle between A and B, which has one side equal to 𝑣A (triangle ii), as shown in the following figure.

We can see that the sine of the angle πœƒ in triangle i is given by ο€»ο‡π‘Ÿ.ο‹²οŒ½οŠ¨

We can also see that the angle in triangle ii that is opposite to the shortest side is given by 𝑣.ο‹²ο“οŠ¨

Hence, Ξ”π‘‘π‘Ÿ=Δ𝑣𝑣.

This equation can be rearranged as follows: Δ𝑣=Ξ”π‘‘β‹…π‘£π‘Ÿ.

Suppose that the time interval in which the object moves from A to B is Δ𝑑.

The centripetal acceleration of the object, π‘Ž, is therefore given by π‘Ž=Δ𝑣Δ𝑑.

The value of Δ𝑣 can be substituted into this equation: π‘Ž=π‘£β‹…Ξ”π‘‘π‘Ž=π‘£Ξ”π‘‘β‹…Ξ”π‘‘π‘Ÿπ‘Ž=𝑣Δ𝑑Δ𝑑⋅1π‘Ÿ.ο‹²οŒ½οŽ

Recall that we have said Δ𝑑 can be approximated as being equal to the distance traveled from A to B. Therefore, we can say that 𝑣=Δ𝑑Δ𝑑.

Substituting this equation, we obtain π‘Ž=𝑣⋅𝑣⋅1π‘Ÿπ‘Ž=π‘£π‘Ÿ.

This is an equation for π‘Ž that only depends on the radius of the circle and the speed of motion along the circumference of the circle.

This acceleration can also be defined using πœƒ.

The angular speed, πœ”, of the object is given by πœ”=Ξ”πœƒΞ”π‘‘.

As πœƒ is very small, we can approximate Ξ”πœƒ as Ξ”πœƒ=Ξ”π‘‘Ξ”π‘Ÿ.

We then find that πœ”=ο€»ο‡Ξ”π‘‘πœ”=Δ𝑑Δ𝑑⋅1π‘Ÿπœ”=𝑣⋅1π‘Ÿπœ”=π‘£π‘Ÿπ‘£=π‘Ÿπœ”.ο‹²οŒ½ο‹²οŽ

This value of 𝑣 can be substituted into π‘Ž=π‘£π‘ŸοŠ¨ to obtain π‘Ž=π‘Ÿβ‹…πœ”π‘Ÿπ‘Ž=π‘Ÿπœ”.

The centripetal force on an object in uniform circular motion is the product of the object’s centripetal acceleration and mass.

Equation: The Centripetal Force on an Object in Uniform Circular Motion

The centripetal force, 𝐹, on an object of mass π‘š that moves uniformly with speed 𝑣 along the circumference of a circle of radius π‘Ÿ is given by 𝐹=π‘šπ‘£π‘Ÿ.

Centripetal forces are not necessarily gravitational.

For example, a centripetal force can be provided by a tension. The following figure shows a stone attached to the end of a string being made to move along a circular path.

The tension in the string acts on the stone to accelerate it toward the center of the circle.

In the case of a car moving in a circle, the centripetal force is due to the friction between the car’s wheels and the road surface.

Let us now look at some examples involving centripetal force.

Example 1: Determining a Centripetal Force

What is the magnitude of the centripetal force that must act on an object of mass 1.0 kg to make it move along a circular path of diameter 1.0 m, completing a circle every 1.0 s? Give your answer to the nearest newton.

Answer

The centripetal force, 𝐹, on an object of mass π‘š that moves uniformly with speed 𝑣 along the circumference of a circle of radius π‘Ÿ is given by 𝐹=π‘šπ‘£π‘Ÿ.

The question states the values of π‘š and π‘Ÿ but not that of 𝑣.

The value of 𝑣 is the speed of the object. This can be found by finding the distance the object moves to completely travel the circumference of the circle once, divided by the time taken to do this.

The circumference, 𝑙, of the circle is given by 𝑙=πœ‹π‘‘, where 𝑑 is the diameter of the circle.

The diameter of the circle is 1.0 m, so we see that 𝑙=1.0Γ—πœ‹π‘™β‰ˆπœ‹.mm

The time taken for the object to move this distance is 1.0 s. We see then that the speed, 𝑣, of the object is 𝑣=πœ‹1.0/𝑣=πœ‹/.msms

We can now substitute this value of 𝑣 into 𝐹=π‘šπ‘£π‘Ÿ.

Hence, we obtain 𝐹=1.0Γ—(πœ‹/)0.5𝐹=2(πœ‹).kgmsmN

To the nearest newton, 𝐹=20N.

Example 2: Comparing Centripetal Accelerations at Different Points

A uniform rope is rotated horizontally around one of its ends, as shown in the diagram. The end of the rope opposite to the fixed end returns to its position every 0.65 s. The free end of the rope moves at constant speed from a point A to a point B.

  1. What is the ratio of the magnitude of the centripetal acceleration of the point A to the magnitude of the centripetal acceleration at point B?
  2. What is the ratio of the magnitude of the centripetal acceleration of the point A to the magnitude of the centripetal acceleration at point D?

Answer

Part 1

The magnitude of the centripetal acceleration, π‘Ž, of a point on the rope is given by π‘Ž=π‘£π‘Ÿ, where 𝑣 is the speed of the point and π‘Ÿ is the straight-line distance from the center of the circle to the point.

The question states that the free end of the rope moves at constant speed. This means that the value of 𝑣 is equal for points A and B.

The free end of the rope maintains a constant distance of 0.22 m from the center of the circle. This value does not change between points A and B.

We see then that neither of the factors that the centripetal acceleration depends on change between points A and B. The value of the centripetal acceleration therefore does not change between the points.

The ratio of the magnitude of the centripetal acceleration at point A to that at point B must therefore be 1.

Part 2

The magnitude of the centripetal acceleration, π‘Ž, of a point on the rope is given by π‘Ž=π‘£π‘Ÿ, where 𝑣 is the speed of the point and π‘Ÿ is the straight-line distance from the center of the circle to the point.

Comparing the values of 𝑣 and π‘Ÿ at points A and D, we see that they both change.

The change in π‘Ÿ is stated. The ratio of the value of π‘Ÿ at A to that at D is given by 0.220.16=1.375π‘Ÿπ‘Ÿ=1.375, where π‘ŸοŠ§ is the value of π‘Ÿ at A and π‘ŸοŠ¨ is the value of π‘Ÿ at D.

The change in 𝑣, however, is not stated.

The value of 𝑣 for points A and D can be determined as the time taken for the free end of the rope to return to its initial position is stated. The time taken is 0.65 s.

The distance traveled, π‘™οŠ§, by the free end of the rope in 0.65 s is given by 𝑙=πœ‹π‘‘, where π‘‘οŠ§ is the diameter of the circular path followed, 0.22 m.

The distance traveled, π‘™οŠ¨, by the free end of the rope in 0.65 s is given by 𝑙=πœ‹π‘‘, where π‘‘οŠ¨ is the diameter of the circular path followed, 0.16 m.

We see then that 𝑙=0.22πœ‹,𝑣=0.22πœ‹0.65.ms

We also see that 𝑙=0.16πœ‹,𝑣=0.16πœ‹0.65.ms

We can now find the ratio between the values of 𝑣 for points A and D: =1.375𝑣𝑣=1.375𝑣=1.375𝑣.οŠ¦οŽ–οŠ¨οŠ¨οŽ„οŠ¦οŽ–οŠ¬οŠ«οŠ¦οŽ–οŠ§οŠ¬οŽ„οŠ¦οŽ–οŠ¬οŠ«οŠ§οŠ¨οŠ§οŠ¨msms

We see then that π‘Ž=(1.375𝑣)1.375π‘Ÿπ‘Ž=1.375×𝑣1.375π‘Ÿπ‘Ž=1.375π‘£π‘Ÿπ‘Ž=1.375π‘Ž.

Hence, we see that the ratio of the magnitude of the centripetal acceleration at A to that at D is 1.375.

This result can be more easily obtained using the formula π‘Ž=π‘Ÿπœ”, where πœ” is the angular speed of the rope, which is the same for all points along the rope, assuming that the rope does not curve.

The value of πœ” can be found by taking the time for the string to return to its original position: πœ”=2πœ‹0.65πœ”=2πœ‹0.65/.radsrads

We see then that, for π‘Ÿ of 0.16 m, π‘Ž=0.16Γ—ο€Ό2πœ‹0.65/mrads and that, for π‘Ÿ of 0.22 m, π‘Ž=0.22Γ—ο€Ό2πœ‹0.65/.mrads

From this, we can see that the ratio of the values of π‘Ž is the same as the ratio of the values of π‘Ÿ, 1∢1.375.

Example 3: Comparing the Angular and Linear Speeds of an Object in Uniform Circular Motion

Which of the lines on the graph correctly shows how the linear speed of an object varies with the radius of the circular path followed by the object? Assume that the angular velocity of the object is constant.

Answer

The greater the radius, π‘Ÿ, of a circular path, the greater the length of that path. The circumference, 𝑙, of the circular path is given by 𝑙=2πœ‹π‘Ÿ.

We see then that a change in π‘Ÿ produces a directly proportional change in 𝑙.

The speed, 𝑣, of an object along a circular path is given by 𝑣=Δ𝑑Δ𝑑, where Δ𝑑 is the distance traveled in a time interval Δ𝑑.

The angular velocity of the object is stated to be constant.

This means that whatever the radius of the circular path that the object moves along, the object will take the same time to return to the same point on the path.

We can then compare two speeds, π‘£οŠ§ and π‘£οŠ¨, that correspond to different values of Δ𝑑, Ξ”π‘‘οŠ§ and Ξ”π‘‘οŠ¨, but the same value of Δ𝑑: 𝑣=Δ𝑑Δ𝑑,𝑣=Δ𝑑Δ𝑑.

Suppose that Δ𝑑=2Δ𝑑.

We see then that 𝑣=2Δ𝑑Δ𝑑𝑣=2𝑣.

We can define the lengths of two circular paths, π‘™οŠ§ and π‘™οŠ¨, as 𝑙=2πœ‹π‘Ÿ,𝑙=2πœ‹π‘Ÿ.

Suppose that π‘Ÿ=2π‘Ÿ.

Hence, we see that 𝑙=2𝑙.

If we suppose that π‘™οŠ§ and π‘™οŠ¨ are the distances traveled in the time interval Δ𝑑, we see that, for objects with equal angular velocities, the speed of the object that travels a distance π‘™οŠ§ is twice that of the object that travels a distance π‘™οŠ¨.

We can take two points on the graph from the question and compare the changes in π‘Ÿ and 𝑣 for these points, as shown in the following figure.

Here, we can see that the gray line intersects both of these points and that none of the other lines do so. Therefore, the gray line is the correct option.

Let us now summarize what we have learned in this explainer.

Key Points

  • A centripetal force is a force that acts toward the center of a path that an object follows.
  • An object at a point on a curved path that the object moves along has an instantaneous velocity tangent to the direction of the path at that point.
  • The centripetal acceleration, π‘Ž, of an object in uniform motion at speed 𝑣 along a circular path of radius π‘Ÿ is given by π‘Ž=π‘£π‘Ÿ.
  • The centripetal force, 𝐹, on an object of mass π‘š in uniform motion at speed 𝑣 along a circular path of radius π‘Ÿ is given by π‘Ž=π‘šπ‘£π‘Ÿ.
  • The centripetal acceleration, π‘Ž, of an object in uniform motion along a circular path of radius π‘Ÿ is given by π‘Ž=π‘Ÿπœ”, where πœ” is the angular speed of the object.
  • Centripetal forces can be due to gravity, tension, friction, or any other force that acts toward the center of a path followed by an object.

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