# Circular Motion: Centripetal Acceleration & Centripetal Force

Show/Hide Sub-topics (Circular Motion | A Level Physics)

Uniform circular motion is a fundamental concept in physics that describes the motion of an object traveling in a circle at a constant speed. This motion is characterized by the object’s angular velocity ($\omega$) and linear velocity ($v$), both of which remain constant throughout the motion. Understanding uniform circular motion is crucial for comprehending various physical phenomena and applications, ranging from planetary orbits to the design of mechanical systems.

## Centripetal Acceleration

One of the key aspects of uniform circular motion is centripetal acceleration, denoted as $a$. Centripetal acceleration is essential for keeping an object in circular motion, and it has several important characteristics:

• Center-seeking: It always points towards the center of the circle along which the object is moving. This inward force is what keeps the object moving in a circular path instead of moving off in a straight line.
• Perpendicular to Velocity: Centripetal acceleration is always perpendicular to the velocity of the object. This perpendicularity ensures that while the direction of the velocity changes, its magnitude remains constant.
• Acts Towards the Centre: The direction of centripetal acceleration is always towards the centre of the circular path. This inward direction is crucial for maintaining circular motion.

### Useful Equations For Centripetal Acceleration

To quantitatively describe centripetal acceleration and the forces involved in uniform circular motion, several equations are useful:

1. $a = \frac{v^2}{r}$: This equation relates centripetal acceleration ($a$) to the square of the linear velocity ($v$) of the object and the radius ($r$) of the circular path. It shows that acceleration increases with velocity and decreases with radius.
2. $a = r\omega^2$: Here, $a$ is also expressed in terms of the radius ($r$) and the square of the angular velocity ($\omega$). This equation underscores the relationship between angular motion parameters and centripetal acceleration. (Hint: You do not need to memorise this – you can “derive” this by substituting the linear velocity equation $v = r \omega$ into equation 1 above.)
3. $a = v\omega$: This formulation links linear velocity ($v$), angular velocity ($\omega$), and centripetal acceleration ($a$), illustrating the interconnectedness of linear and angular motion in circular dynamics. (Hint: You do not need to memorise this – you can “derive” this by substituting the linear velocity equation $v = r \omega$ or more precisely $r = \frac{v}{\omega}$ into equation 1 above.)

### [Optional] Derivation of $a = \frac{v^{2}}{r}$

Consider an object moving in a circle with radius $r$ at a constant speed $v$ – i.e. uniform circular motion.

Step 1: The object covers an arc length $\Delta s$ in a time interval $\Delta t$, so $\Delta s = v\Delta t$.

Step 2: The arc length $\Delta s$ can also be described in terms of the angle $\Delta\theta$ it subtends at the center of the circle: $\Delta s = r\Delta\theta$.

Step 3: Equating the two expressions for $\Delta s$ gives $r\Delta\theta = v\Delta t$, which simplifies to $\Delta\theta = \frac{v}{r}\Delta t$.

Step 4: For very small $\Delta\theta$ (small angle approximation), the change in velocity $\Delta v$ that occurs as the object moves from one point to another on the circle is perpendicular to $v$, and $\Delta\theta$ can also be expressed as $\Delta\theta = \frac{\Delta v}{v}$. (Note: Basically, the arc length formula $s = r \theta$ applied to this case, where arc length is $\Delta v$ and radius is $v$ and angle is $\Delta \theta$.)

Step 5: Combining the expressions for $\Delta\theta$, we get $\frac{\Delta v}{v} = \frac{v}{r}\Delta t$, which simplifies to $\Delta v = \frac{v^2}{r}\Delta t$.

Step 6: The centripetal acceleration $a$ is defined as the change in velocity per unit time, i.e., $a = \frac{\Delta v}{\Delta t}$.

Step 7: Substituting the expression for $\Delta v$ from step 6 into this definition gives $a = \frac{v^2}{r}$, which is the desired equation for centripetal acceleration.

## Centripetal Force

The concept of centripetal force is integral to understanding how circular motion is sustained. This force is the resultant force required to keep an object moving in a circular path at constant speed, and it has several key points:

• Centripetal Force Direction: Always directed towards the center of the circular path, ensuring that the object continues to follow a circular trajectory.
• Not Included in Free Body Diagrams: Centripetal force should not be represented as an independent force in free body diagrams. It is the resultant force acting towards the center.
• No Work Done: Centripetal force does not perform any work on the object in motion since it is always perpendicular to the displacement of the object. This is because work is defined as force acting over a distance in the direction of the force, and in this case, the direction of force and displacement are orthogonal.

### Useful Equations for Centripetal Force

To calculate the centripetal force, the following equations are essential:

1. $F = \frac{mv^2}{r}$: This equation shows that the centripetal force ($F$) is directly proportional to the mass ($m$) of the object and the square of its velocity ($v$), and inversely proportional to the radius ($r$) of the circle. It highlights how mass, velocity, and radius influence the required force to maintain circular motion. (Hint: You do not need to memorise this – you can “derive” this by remembering Newton’s Second Law $F=ma$ and use the centripetal acceleration equation $a = \frac{v^{2}}{r}$)
2. $F = mr\omega^2$: Similar to the previous equation, but expressed in terms of angular velocity ($\omega$). It provides a direct relationship between centripetal force and angular motion parameters. (Hint: You do not need to memorise this – you can “derive” this by substituting the linear velocity equation $v = r \omega$ into equation 1 above.)

### [Optional] Derivation Of Formula For Car On Banked Icy Road

A car of mass $m$ kg is traveling on a banked curved icy road with radius $r$ m, without slipping upwards or downwards. The velocity of the car is $v$ m s-1. Find the formula that relates the variables $m$, $r$, $v$ and $\theta$, where $\theta$ is the angle of the banked icy road.

Icy road implies that there’s no friction between the tires and the road, which simplifies the forces acting on the car. The forces involved are the gravitational force ($mg$), the normal force ($N$), and the centripetal force required to keep the car moving in a circle ($F_c$). The goal is to find an expression for the banking angle ($\theta$) that ensures the car can travel around the curve without the need for frictional force to prevent slipping.

#### Step 1: Identify the Forces

1. Gravitational Force ($mg$): Acts vertically downward.
2. Normal Force ($N$): Acts perpendicular to the surface of the road.
3. Centripetal Force ($F_c$): Required to keep the car moving in a circular path. In this scenario, it’s provided by the component of the normal force that acts towards the center of the circle.

#### Step 2: Resolve Forces

The normal force ($N$) can be resolved into two components:

• A vertical component ($N \cos(\theta)$) that balances the gravitational force.
• A horizontal component ($N \sin(\theta)$) that provides the centripetal force.

#### Step 3: Force Equilibrium

• Vertical Equilibrium: The vertical component of $N$ balances the weight of the car.
$$N \cos(\theta) = mg$$
• Horizontal (Centripetal) Force: The horizontal component of $N$ provides the necessary centripetal force to keep the car moving in a circular path of radius $r$ at speed $v$.
$$N \sin(\theta) = \frac{mv^2}{r}$$

#### Step 4: Combine Equations

• To find the relationship between $\theta$, $v$, $r$, and $g$, we divide the equation for the horizontal force by the equation for vertical equilibrium.
$$\frac{N \sin(\theta)}{N \cos(\theta)} = \frac{\frac{mv^2}{r}}{mg}$$
• Simplifying, we get the tangent of the angle:
$$\tan(\theta) = \frac{v^2}{rg}$$

This formula shows that the banking angle ($\theta$) necessary for a car to safely navigate a curve on a banked, icy road (where friction is negligible or non-existent) depends only on the velocity of the car ($v$), the radius of the curve ($r$), and the acceleration due to gravity ($g$). This setup ensures that the car can maintain its path around the curve solely through the balance of forces, without relying on friction.

## Worked Examples

### Example 1: Finding Centripetal Acceleration

A toy car moves with a constant speed of 2 meters per second in a circular track of radius 1 meter. What is the centripetal acceleration of the toy car?

• Given: $v = 2$ m/s, $r = 1$ m.
• Formula: $a = \frac{v^2}{r}$.

Calculating $a$:

$$a = \frac{(2)^2}{1} = \frac{4}{1} = 4\, \text{m/s}^2.$$

Conclusion: The centripetal acceleration of the toy car is $4\, \text{m/s}^2$.

### Example 2: Calculating Centripetal Force

A 0.5 kg ball is being swung in a vertical circle with a radius of 2 meters at a constant speed of 5 m/s. What is the centripetal force acting on the ball?

• Given: $m = 0.5$ kg, $v = 5$ m/s, $r = 2$ m.
• Formula: $F = \frac{mv^2}{r}$.

Calculating $F$:

$$F = \frac{0.5 \times (5)^2}{2} = \frac{0.5 \times 25}{2} = \frac{12.5}{2} = 6.25\, \text{N}.$$

Conclusion: The centripetal force acting on the ball is $6.25$ N.

### Example 3: Determining Speed from Centripetal Acceleration

Question: A cyclist moves around a circular path with a radius of 50 meters. If the cyclist experiences a centripetal acceleration of $2\, \text{m/s}^2$, what is the speed of the cyclist?

• Given: $a = 2 \text{m/s}^2$, $r = 50$ m.
• Formula to rearrange: $a = \frac{v^2}{r}$ to solve for $v$.

Solving for $v$:

$$v = \sqrt{a \times r} = \sqrt{2 \times 50} = \sqrt{100} = 10\, \text{m/s}$$

Conclusion: The speed of the cyclist is 10 m/s.

### Example 4: Comparing Centripetal Forces

Question: Two cars, Car A and Car B, are racing in circles with the same speed. Car A has a mass of 1200 kg and moves in a circle of radius 30 meters. Car B has a mass of 1500 kg and moves in a circle of radius 40 meters. Which car experiences a greater centripetal force, and by how much?

• Given: $v$ is constant, $m_A = 1200$ kg, $r_A = 30$ m, $m_B = 1500$ kg, $r_B = 40$ m.
• Formula: $F = \frac{mv^2}{r}$ (since $v$ is constant, focus on the ratio of $m/r$).

Calculating for both:

• $F_A = \frac{1200}{30} = 40$ (unit of force per unit of velocity squared).
• $F_B = \frac{1500}{40} = 37.5$ (unit of force per unit of velocity squared).

Conclusion: Car A experiences a greater centripetal force. The difference in centripetal force (per unit of velocity squared) between Car A and Car B is 2.5 units of force per unit of velocity squared.

### Example 5: Angular Velocity and Centripetal Acceleration

A Ferris wheel with a radius of 10 meters rotates at a rate of 0.5 revolutions per minute (rpm). Calculate the centripetal acceleration of a passenger at the edge of the Ferris wheel.

• Given: $r = 10$ m, rotational speed = $0.5$ rpm.
• Convert rpm to radians per second: $0.5$ rpm = $\frac{0.5 \times 2\pi}{60}$ rad/s.

Calculating angular velocity ($\omega$):

$$\omega = \frac{0.5 \times 2\pi}{60} = \frac{\pi}{60}\, \text{rad/s}$$

• Formula for centripetal acceleration: $a = r\omega^2$.

Calculating $a$:

$$a = 10 \times \left(\frac{\pi}{60}\right)^2 = 10 \times \frac{\pi^2}{3600} \approx 0.027\, \text{m/s}^2$$

Conclusion: The centripetal acceleration of a passenger at the edge of the Ferris wheel is approximately $0.027\, \text{m/s}^2$.

### Example 6: Circular Motion With Elastic Cord

A mass of 0.050 kg is attached to one end of a piece of elastic cord of unstretched length 0.50 m. The force constant of the cord is 40 N m-1. The mass is rotated steadily on a smooth table in a horizontal circle of radius 0.70 m. What is the approximated speed of the mass?

1. 11 m s-1
2. 15 m s-1
3. 20 m s-1
4. 24 m s-1

To find the approximated speed of the mass, we first need to determine the force exerted by the elastic cord when it is stretched. The force exerted by an elastic cord can be calculated using Hooke’s Law, which is given by:

$$F = k \Delta x$$

where:

• $F$ is the force exerted by the cord,
• $k$ is the force constant of the cord, and
• $\Delta x$ is the extension of the cord from its unstretched length.

Given that the unstretched length of the cord is 0.50 m and the radius of the circle (which is also the length of the cord when stretched) is 0.70 m, the extension of the cord $\Delta x$ is $0.70 \, \text{m} – 0.50 \, \text{m} = 0.20 \, \text{m}$.

The force constant $k$ is given as $40 \, \text{N m}^{-1}$.

We can now calculate the force exerted by the cord:

$$F = 40 \, \text{N m}^{-1} \times 0.20 \, \text{m}$$

This force provides the centripetal force necessary to keep the mass moving in a circle, which can be calculated using the formula for centripetal force:

$$F_{\text{centripetal}} = \frac{mv^2}{r}$$

where:

• $m$ is the mass of the object,
• $v$ is the speed of the mass, and
• $r$ is the radius of the circle.

Given that $m = 0.050 \, \text{kg}$ and $r = 0.70 \, \text{m}$, we can solve for $v$ using the calculated force $F$ as the centripetal force. Let’s calculate it.

The approximated speed of the mass is approximately $10.58 \, \text{m s}^{-1}$. Among the given options, this value is closest to $11 \, \text{m s}^{-1}$, so the correct answer is:

1. $11 \text{ m s}^{-1}$.

### Example 7: Stone In Circular Motion

A 0.500 kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the path at a constant frequency of 2.20 rev s-1. At the moment the stone is overhead, the stone is released. What is the magnitude and direction of the velocity of the stone when it leaves the circular path?

To find the magnitude and direction of the velocity of the stone when it is released from the circular path overhead, we first need to calculate the speed of the stone at that moment. The stone is moving in a circular path with a constant frequency, and it is released when it is overhead, so its velocity will be tangential to the circle at the point of release.

The speed ($v$) of the stone can be calculated using the formula that relates the speed to the angular velocity – $v = r \omega$. Then angular velocity is given by $\omega = 2 \pi f$, hence $v = 2 \pi r f$.

Given:

• The length of the string (which is also the radius of the circular path) is $r = 75.0 \, \text{cm} = 0.75 \, \text{m}$.
• The frequency of revolution is $f = 2.20 \, \text{rev s}^{-1}$.

Substituting the given quantities into the above equation, we find that the magnitude of the velocity of the stone when it leaves the circular path is approximately $10.37 \, \text{m s}^{-1}$. The direction of this velocity will be tangential to the circular path at the point of release, which, since the stone is released when it is overhead, means the velocity direction is horizontally outward from the circle at that point.

### Example 8: Child Swinging On A Rope

A 30.0 kg child swings on a rope with a length of 4.00 m that is hanging from a tree. At the bottom of the swing the child is moving at a speed of 4.00 m s-1. What is the tension in the rope?

To find the tension in the rope when the child is at the bottom of the swing, we need to consider both the gravitational force acting on the child and the centripetal force required to keep the child moving in a circular path.

At the bottom of the swing, the tension in the rope ($T$) is the sum of the force due to gravity acting on the child ($mg$, where $m$ is the mass of the child and $g$ is the acceleration due to gravity) and the centripetal force ($F_{\text{centripetal}}$) required to keep the child moving in a circular path. The centripetal force can be calculated using the formula:

$$F_{\text{centripetal}} = \frac{mv^2}{r}$$

where:

• $m$ is the mass of the child,
• $v$ is the speed of the child, and
• $r$ is the length of the rope.

Thus, the total tension in the rope at the bottom of the swing is:

$$T = mg + \frac{mv^2}{r}$$

Given:

• $m = 30.0 \, \text{kg}$,
• $g = 9.81 \, \text{m/s}^2$,
• $v = 4.00 \, \text{m/s}$,
• $r = 4.00 \, \text{m}$.

Let’s calculate the tension in the rope:

$$T = (30)(9.81) + \frac{30 \times 4^{2}}{4} = 414.3 \text{ N}$$

The tension in the rope when the child is at the bottom of the swing is approximately $414.3 \, \text{N}$.

### Example 9: Car On A Banked Road With No Friction

A 2000 kg car is traveling on a banked curved icy road with radius 40 m. The velocity of the car is 25.0 m s-1. What will be the angle of the banked road if the car is to travel on the icy road without sliding?

We’ll use the formula:

$$\tan(\theta) = \frac{v^2}{r \cdot g}$$

where:

• $v = 25.0$ m/s is the velocity of the car,
• $r = 40$ m is the radius of the curve,
• $g = 9.8 \text{ m/s}^2$ is the acceleration due to gravity,
• $\theta$ is the angle of the banked road we need to find.

Let’s calculate the angle $\theta$,

$$\tan(\theta) = \frac{25^{2}}{40 \times 9.8}$$

The angle of the banked road, for the car to travel on the icy road without sliding, should be approximately $57.90$ degrees. This angle ensures that the centripetal force needed for the car to navigate the curve safely at a speed of 25.0 m/s is provided by the component of the gravitational force acting perpendicular to the surface of the banked road.

### Example 10: Apparent Weight

Assuming the Earth to be a uniform sphere rotating about an axis through the poles, the apparent weight of a body at the Equator compared with it’s apparent weight at the poles would be

1. greater, because the angular velocity of the Earth is greater at the Equator than at the pole
2. greater, because the weight at the Equator is given by the sum of the gravitational attraction of the Earth and the centripetal force due to the circular motion of the body
3. smaller, because the gravitational attraction of the Earth must provide both the weight and the centripetal force due to the circular motion of the body
4. smaller, because the gravitational attraction at the pole is greater than that at the Equator

To determine how the apparent weight of a body at the Equator compares with its weight at the poles, considering the Earth as a uniform sphere rotating about its axis, we must understand the effects of Earth’s rotation and gravity on the body’s weight.

#### Step 1: Understanding Weight and Gravitational Force

The weight of an object on the surface of the Earth is the gravitational force exerted on it by the Earth. This force is given by Newton’s law of universal gravitation:

$$F_{\text{gravity}} = \frac{G M m}{r^2}$$

where:

• $F_{\text{gravity}}$ is the gravitational force (weight),
• $G$ is the gravitational constant,
• $M$ is the mass of the Earth,
• $m$ is the mass of the object, and
• $r$ is the distance from the center of the Earth to the object.

#### Step 2: Effect of Earth’s Rotation

The rotation of the Earth causes objects at the Equator to experience a centrifugal force outward due to the rotational motion. This force effectively reduces the net gravitational force (apparent weight) acting on the object. The centrifugal force is given by:

$$F_{\text{centrifugal}} = m \omega^2 r$$

where:

• $m$ is the mass of the object,
• $\omega$ is the angular velocity of Earth’s rotation, and
• $r$ is the radius of the Earth’s rotation at the location of the object (maximum at the Equator).

#### Step 3: Apparent Weight Calculation

At the poles, the centrifugal force is zero because the radius of rotation ((r)) is zero; the object does not move in a circle but merely rotates in place. Thus, the apparent weight is purely the gravitational force.

At the Equator, the apparent weight is the gravitational force minus the centrifugal force due to Earth’s rotation:

$$F_{\text{apparent}} = F_{\text{gravity}} – F_{\text{centrifugal}}$$

Since $F_{\text{centrifugal}} > 0$ at the Equator, $F_{\text{apparent}}$ at the Equator is less than $F_{\text{gravity}}$.

#### Step 4: Evaluating the Statements

1. Greater, because the angular velocity of the Earth is greater at the Equator than at the pole: This statement is incorrect. The angular velocity ($\omega$) of the Earth is constant; what changes is the effect of this rotation, which reduces the apparent weight at the Equator, not increases it.
2. Greater, because the weight at the Equator is given by the sum of the gravitational attraction of the Earth and the centripetal force due to the circular motion of the body: This statement is incorrect. The centrifugal force (not centripetal) due to the Earth’s rotation reduces the apparent weight at the Equator, it does not add to it.
3. Smaller, because the gravitational attraction of the Earth must provide both the weight and the centripetal force due to the circular motion of the body: This statement is correct.
4. Smaller, because the gravitational attraction at the pole is greater than that at the Equator: This statement is true but incorrect. The gravitational attraction itself is slightly stronger at the poles due to the oblate shape of the Earth (it’s not a perfect sphere but slightly flattened at the poles). However, the main reason for the difference in apparent weight is the centrifugal force at the Equator, as discussed.

### Conclusion

The correct explanation is option 3.

##### Mini Physics

As the Administrator of Mini Physics, I possess a BSc. (Hons) in Physics. I am committed to ensuring the accuracy and quality of the content on this site. If you encounter any inaccuracies or have suggestions for enhancements, I encourage you to contact us. Your support and feedback are invaluable to us. If you appreciate the resources available on this site, kindly consider recommending Mini Physics to your friends. Together, we can foster a community passionate about Physics and continuous learning.

This site uses Akismet to reduce spam. Learn how your comment data is processed.