# Physics Tutorial: Centripetal Force

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In this Physics tutorial, you will learn:

• The meaning of centripetal force
• The situations in which centripetal force exists
• What causes the centripetal force?
• Equation of centripetal force
• How does weight change when a vertical centripetal force is present in the system?
• How to calculate the total acceleration of the system in a non-uniform rotational motion

## Introduction

What happens where you are rotating around yourself an object tied to a string? And what happens when the string cuts off? Why?

Why it is dangerous driving at high speeds when you enter a curve?

What keeps the Moon rotating around the Earth? Why the Moon does not collapse to the Earth?

These and many other questions are answered in this tutorial. Please read it carefully as in many textbooks coverage of issues related to this tutorial are either vague or implied.

## The Meaning of Centripetal Force

In the Physics tutorial "Kinematics of the Rotational Motion", we discussed about centripetal acceleration that exists because of the change in direction of velocity vector during rotation. It was stressed that this kind of acceleration exists even when the object is rotating uniformly, i.e. at constant tangential speed. Furthermore, the vector of centripetal acceleration is always directed towards the centre of curvature as shown in the figure below. The equation of centripetal acceleration (in vector form, as discussed in tutorial "Kinematics of the Rotational Motion") is

aC = ∆v/∆t

where ∆v is the change (the difference) in velocity (only in direction, not in magnitude) of the rotating object in two given instants and Δt is the time interval between these two instants.

The scalar form of the equation (as discussed in tutorial "Kinematics of the Rotational Motion") is

aC = v2/r

where r is the radius of curvature. This is the reason why centripetal acceleration is often referred as "radial acceleration, ar".

From Newton's Second Law of Motion we known that where there is an acceleration, there is also a force causing it. Centripetal acceleration makes no exception to this rule as well. Therefore, it is obvious there must be a force causing the centripetal acceleration. This force is known as Centripetal Force, FC and it is the force responsible for keeping an object in rotation. It is in the same direction of centripetal acceleration, i.e. towards the centre of circle. From Newton's Second Law of Motion, we have:

FC = m × aC

where m is the object's mass. In scalar form, the above equation is written as:

FC = m × v2/r

Thus, for example if a 20 kg object rotates at 4 m/s around a circular path of radius 2 m, the centripetal force that keeps the object moving in this circular path is

FC = m × v2/r
= 20 kg × (4 m/s)2/2 m
= 160 N

The scalar form of centripetal force equation is an indicator that this force exists only in rotational motion. Indeed, when the radius increases, magnitude of centripetal force decreases. In this case, any given arc of the circle straightens more and more until it looks like a straight line. This occurs when radius is very long as to be considered as infinity. From mathematics, it is known that when the denominator of a fraction is equal to infinity, the value of fraction is zero. Therefore, centripetal force in linear motion is zero because the denominator of fraction in the formula

FC = m × v2/r

is equal to infinity. This means "linear motion is a special case of rotational motion, i.e. linear motion is a rotational motion in which radius is equal to infinity." When centripetal force stops acting, the object moves linearly in the direction of the tangent line to the circle it had at the last moment when FC was still present in the system. Thus, for example, if you are rotating a stone using a rope and the rope cuts off, the stone will move according the last tangent as shown in the figure. ### What is Centripetal Force?

Centripetal force is not a specific force like those we have dealt with in the past. Centripetal force is like a fictional character whose role can be played by many actors in different cinematographic productions. For example, Earth moves in an almost circular orbit around the Sun due to the gravitational force by which the Sun attracts it. In this case, gravitational force acts as a centripetal force as it keeps the Earth in rotation around the Sun. Another example: When you rotate around yourself a stone tied at the end of a rope, the centripetal force necessary to keep the stone rotating is provided by the tension of the rope. In this case, tension acts like centripetal force, and so on.

#### Example 1

A 2000 kg car enters a curve of radius 60 m as shown in the figure. If the friction coefficient between the car's tires and asphalt is 0.6, calculate the maximum speed the car may have without skidding aside when it is inside the curve. Take g ≈ 10 m/s2.

#### Solution 1

In this case, the frictional force exerted between the car tires and asphalt acts like a centripetal force as it prevents the car from skidding aside (moving in a linear path). Therefore, if we find the value of frictional force, we have found that of centripetal force as well.

We have:

f = μ × N
= μ × m × g
= 0.6 × 2000 × 10
= 12000 N

Therefore, we have FC = 12000 N as well. From the equation of centripetal force, we have

FC = m × v2/r

Substituting the known values, we obtain for the maximum speed allowed:

12000 = 2000 × v2/60
v2 = 360
v = √360
≈19 m/s

## How does Centripetal Force Affect the Weight of Objects?

As discussed in the PHysics tutorial on "Types of Forces I. Gravitational Force and Weight", weight is not always numerically equal to the gravitational force. The only case in which these forces are equal is when an object rests on a flat horizontal surface or moves together with the horizontal basement at constant velocity in any direction. In the abovementioned article, we discussed how the plane slope affects the weight. Thus, steeper the plane, smaller the object's weight.

Another factor that affects the weight is the vertical acceleration an object may have besides the gravitational acceleration. This extra acceleration may be linear as in the case when a lift starts moving upwards and due to the inertia (tendency to keep the previous state of motion) a person inside the lift weighs more than usual, or centripetal such as when a car moves along a vertical curve. In any case the equation of weight is

W = Fg - Fextra (vertical)

where Fg = m × g is the gravitational force exerted by the object and Fextra (vertical) is the other vertical force acting on the object. Thus, in the example of lift that starts moving upwards, this extra force is the upward force exerted by the lift's engine. However, we are not interested here for such scenarios but for those in which a centripetal force is involved, i.e. in scenarios similar to the abovementioned example in which a car is moving along a vertical curve. In this case, centripetal force is the extra vertical force discussed above and consequently, centripetal acceleration is the extra vertical acceleration involved.

Thus, when an object is moving at the lowest point of a circular path, the centripetal force is directed upwards (in the opposite direction to the gravitational force) and when it is moving at the highest point, the centripetal force is directed downwards (in the direction of gravitational force) as shown in the figure. Therefore, the equation of weight becomes

W = Fg - FC

which means that when the object is at the highest point, it weighs less than usual as the Fg and FC are subtracted while when it is at the lowest point, it weighs more than usual as Fg and FC are added (two consecutive minuses become plus).

#### Example 2

A 10 t truck passes on a circular bridge as shown in the figure. Calculate the maximum velocity the truck can have at the lowest point of the bridge if its radius of curvature is 40 m. The bridge can hold up to 12 t of load. For convenience, take g ≈ 10 m/s2.

#### Solution 2

We have 10 t = 10 000 kg and 12 t = 12 000 kg.

Gravitational force of the truck is

Fg = m × g
= 10 000 kg × 10 m/s2
= 100 000 N

The maximum weight the bridge can hold is

Wmax = mmax × g
= 12 000 kg × 10 m/s2
= 120 000 N

From the equation of weight, we have

Wmax = Fg - FC

Centripetal force is in the opposite direction of gravity, thus, it must be negative. The scalar form of the above equation therefore is:

Wmax = Fg - (-FC)
= Fg + FC

Therefore, the maximum centripetal force allowed to avoid harms in the bridge is

FC = Wmax - Fg
= 120 000 N - 100 000 N
= 20 000 N

Hence, we obtain for the maximum velocity allowed in the lowest point of the bridge:

FC = m × v2/r
v2 = FC × r/m
v = √FC × r/m
= √20 000 × 40/10 000
= √80
≈ 8.94 m/s

Thus, when an object is at the lowest point of a circular trajectory we have W > Fg and when it is at the highest point of a circular trajectory, then W < Fg.

## Total Acceleration during a non-Uniform Rotational Motion

As discussed in this tutorial and in tutorial "Kinematics of Rotational Motion", when an object is rotating uniformly around a fixed point, the only acceleration present in the system is the centripetal acceleration, aC. However, when the object's speed increases or decreases during rotation, there also exists another acceleration due to speeding up or slowing down - the standard acceleration we have discussed in Section 4: Dynamics. Since the direction of this acceleration complies with the direction of velocity (it is a kind of instantaneous acceleration whose direction is collinear with the vector of instantaneous velocity), we call it "tangential acceleration, at" and it is perpendicular to the centripetal acceleration as shown in the figure below. The positions in the above figure are taken a bit distant from each other for visual purpose. This bring a higher degree of inaccuracy in our approach as we don't want the velocity vectors have different direction. It is obvious that the smaller the time interval, the closer the velocity vectors are. As a result, their direction will fit more and more. Here, we have taken a kind of intermediate direction for the tangential acceleration vector but it is obvious that when the time interval is very small, the direction of all three vectors: initial velocity vector, final velocity vector and tangential acceleration vector will be the same. This direction will be perpendicular to that of centripetal acceleration vector.

We have explained in the tutorial "The Meaning of Acceleration. Constant and Non-Constant Acceleration. Gravitational Acceleration" that acceleration caused by the change in the magnitude of velocity (here the tangential acceleration at) is calculated by the equation

at = v - v0/∆t

where v and v0 are the magnitudes of final and initial velocities respectively and Δt is the time interval this process occurs.

On the other hand, we have the centripetal (radial) acceleration whose formula is

aC = v2/r

where the velocity v in fact is a kind of average velocity if the time interval is not very small; it is not meant to be the magnitude of final velocity as in the case of tangential acceleration. Therefore, we will write the above formula as

aC = < v >2/r

where usually we have

< v > = v0 + v/2

As a result, since aC and at are perpendicular, we can use the Pythagorean Theorem to calculate the magnitude of the total acceleration a in a non-uniform rotational motion:

a = √a2t + a2C

The direction of total acceleration is obtained by applying the parallelogram rule for the addition of vectors as shown in the figure: ### Example 3

An object rotating around a circular path of radius equal to 4 m increases its angular velocity from 0.2 rad/s to 1.4 rad/s in 12 s as shown in the figure. Calculate the magnitude of total acceleration experienced by the object at the end of this process.

### Solution 3

First, let's work out the initial and final (tangential) velocities of the object. Thus, from the linear-to-rotational quantities relationship we have

v = ω × r

Therefore, for the magnitude of the initial velocity v0, we obtain

v0 = ω0 × r
= 0.2rad/s × 4 m
= 0.8 m/s

and for that of final velocity v, we obtain

v = ω × r
= 1.4rad/s × 4 m
= 5.6 m/s

Thus, for the tangential acceleration at we obtain

at = v - v0/t
= 5.6 m/s - 0.8 m/s/12 s
= 0.4 m/s2

As for the centripetal acceleration, since we are interested what happens at the end of process, we consider only the value of final velocity in the formula. Thus, we have

aC = v2/r
= 5.62/4
= 7.84 m/s2

Hence, the magnitude of total acceleration at the end of process is

a = √a2t + a2C
= √0.42 + 7.842
= 7.85 m/s2

## Summary

From Newton's Second Law of Motion we known that where there is an acceleration, there is also a force causing it. Centripetal acceleration makes no exception to this rule as well. Therefore, it is obvious there must be a force causing the centripetal acceleration. This force is known as Centripetal Force, FC and it is the force responsible for keeping an object in rotation. It is in the same direction of centripetal acceleration, i.e. towards the centre of circle.

As centripetal acceleration point towards the centre of curvature in the direction of circle's radius is often referred as "radial acceleration, ar".

The equation of Centripetal Force based on the Newton's Second Law of Motion, is:

FC = m × aC

where m is the object's mass. In scalar form, the above equation is written as:

FC = m × v2/r

When centripetal force stops acting, the object moves linearly in the direction of the tangent line to the circle it had at the last moment when FC was still present in the system.

Centripetal force is not a specific force in itself; many forces such as frictional, gravitational, tension force etc. act as centripetal forces in certain conditions.

Weight is not always numerically equal to the gravitational force. As an example in this regards we can mention situations when an object is moving according a vertical rotational path. In this case, the equation of weight is

W = Fg - Fextra (vertical)

and since this extra vertical force is nothing else but the centripetal force, we can write

W = Fg - FC

When an object is at the lowest point of a circular trajectory we have W > Fg and when it is at the highest point of a circular trajectory, then W < Fg.

When an object is rotating uniformly around a fixed point, the only acceleration present in the system is the centripetal acceleration, aC. However, when the object's speed increases or decreases during rotation, there also exists another acceleration due to speeding up or slowing down. Since the direction of this acceleration complies with the direction of velocity (it is a kind of instantaneous acceleration whose direction is collinear with the vector of instantaneous velocity), we call it "tangential acceleration, at" and it is perpendicular to the centripetal acceleration aC.

The acceleration caused by the change in the magnitude of velocity (here the tangential acceleration at) is calculated by the equation

at = v - v0/∆t

where v and v0 are the magnitudes of final and initial velocities respectively and Δt is the time interval this process occurs.

On the other hand, we have the centripetal (radial) acceleration whose formula is

aC = v2/r

As a result, since aC and at are perpendicular, we can use the Pythagorean Theorem to calculate the magnitude of the total acceleration a in a non-uniform rotational motion:

a = √a2t + a2C

The direction of total acceleration is obtained by applying the parallelogram rule for the addition of vectors.

## Centripetal Force Revision Questions

1) A small sphere is moving laterally at 0.4 m/s inside a cylindrical glass as shown in the figure below. What is the maximum diameter the glass must have in order to keep the sphere rotating through the lateral face without touching the lower base of glass? Take g ≈ 10 m/s2. 1. 1.6 cm
2. 3.2 cm
3. 6 cm
4. 60 cm

Correct Answer: B

2) What is the minimum velocity a vertical roller coaster must have so that a 40 kg child not exerting any pushing force on the seat when she is at the highest position during rotation? The radius of curvature of the roller coaster is 80 m.

1. 2 m/s
2. 8.9 m/s
3. 28.3 m/s
4. 80 m/s

Correct Answer: C

3) The rotor of an engine increases its angular velocity from 2 rad/s to 5 rad/s in 0.1 s. What is the total acceleration of peripheral points of the rotor at the end of this process if its diameter is equal to 20 cm? 1. 3.9 m/s2
2. 0.39 m/s2
3. 3 m/s2
4. 2.5 m/s2

Correct Answer: A

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