string circular motion experiment

6.2 Uniform Circular Motion

Section learning objectives.

By the end of this section, you will be able to do the following:

• Describe centripetal acceleration and relate it to linear acceleration
• Describe centripetal force and relate it to linear force
• Solve problems involving centripetal acceleration and centripetal force

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.
• (D) calculate the effect of forces on objects, including the law of inertia, the relationship between force and acceleration, and the nature of force pairs between objects.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Circular and Rotational Motion, as well as the following standards:

Section Key Terms

Centripetal Acceleration

[BL] [OL] Review uniform circular motion. Ask students to give examples of circular motion. Review linear acceleration.

In the previous section, we defined circular motion . The simplest case of circular motion is uniform circular motion , where an object travels a circular path at a constant speed . Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity , either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force . The sharper the curve and the greater your speed, the more noticeable this effect becomes.

[BL] [OL] [AL] Demonstrate circular motion by tying a weight to a string and twirling it around. Ask students what would happen if you suddenly cut the string? In which direction would the object travel? Why? What does this say about the direction of acceleration? Ask students to give examples of when they have come across centripetal acceleration.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δ s Δ s becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration a c because centripetal means center seeking .

Consider Figure 6.7 . The figure shows an object moving in a circular path at constant speed and the direction of the instantaneous velocity of two points along the path. Acceleration is in the direction of the change in velocity and points toward the center of rotation. This is strictly true only as Δ s Δ s tends to zero.

Now that we know that the direction of centripetal acceleration is toward the center of rotation, let’s discuss the magnitude of centripetal acceleration. For an object traveling at speed v in a circular path with radius r , the magnitude of centripetal acceleration is

Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that a c is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than at 50 km/h.

We can also express a c in terms of the magnitude of angular velocity . Substituting v = r ω v = r ω into the equation above, we get a c = ( r ω ) 2 r = r ω 2 a c = ( r ω ) 2 r = r ω 2 . Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is

Tips For Success

The equation expressed in the form a c = rω 2 is useful for solving problems where you know the angular velocity rather than the tangential velocity.

Virtual Physics

Ladybug motion in 2d.

In this simulation, you experiment with the position, velocity, and acceleration of a ladybug in circular and elliptical motion. Switch the type of motion from linear to circular and observe the velocity and acceleration vectors. Next, try elliptical motion and notice how the velocity and acceleration vectors differ from those in circular motion.

Grasp Check

In uniform circular motion, what is the angle between the acceleration and the velocity? What type of acceleration does a body experience in the uniform circular motion?

• The angle between acceleration and velocity is 0°, and the body experiences linear acceleration.
• The angle between acceleration and velocity is 0°, and the body experiences centripetal acceleration.
• The angle between acceleration and velocity is 90°, and the body experiences linear acceleration.
• The angle between acceleration and velocity is 90°, and the body experiences centripetal acceleration.

Centripetal Force

[BL] [OL] [AL] Using the same demonstration as before, ask students to predict the relationships between the quantities of angular velocity, centripetal acceleration, mass, centripetal force. Invite students to experiment by using various lengths of string and different weights.

Because an object in uniform circular motion undergoes acceleration (by changing the direction of motion but not the speed), we know from Newton’s second law of motion that there must be a net external force acting on the object. Since the magnitude of the acceleration is constant, so is the magnitude of the net force, and since the acceleration points toward the center of the rotation, so does the net force.

Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop.

The component of any net force that causes circular motion is called a centripetal force . When the net force is equal to the centripetal force, and its magnitude is constant, uniform circular motion results. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton’s second law of motion, a net force causes the acceleration of mass according to F net = m a . For uniform circular motion, the acceleration is centripetal acceleration: a = a c . Therefore, the magnitude of centripetal force, F c , is F c = m a c F c = m a c .

By using the two different forms of the equation for the magnitude of centripetal acceleration, a c = v 2 / r a c = v 2 / r and a c = r ω 2 a c = r ω 2 , we get two expressions involving the magnitude of the centripetal force F c F c . The first expression is in terms of tangential speed, the second is in terms of angular speed: F c = m v 2 r F c = m v 2 r and F c = m r ω 2 F c = m r ω 2 .

Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton’s second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r , you get

From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

Watch Physics

Centripetal force and acceleration intuition.

This video explains why centripetal force, when it is equal to the net force and has constant magnitude, creates centripetal acceleration and uniform circular motion.

Misconception Alert

Some students might be confused between centripetal force and centrifugal force. Centrifugal force is not a real force but the result of an accelerating reference frame, such as a turning car or the spinning Earth. Centrifugal force refers to a fictional center fleeing force.

• The yoyo will fly inward in the direction of the centripetal force.
• The yoyo will fly outward in the direction of the centripetal force.
• The yoyo will fly to the left in the direction of the tangential velocity.
• The yoyo will fly to the right in the direction of the tangential velocity.

Solving Centripetal Acceleration and Centripetal Force Problems

To get a feel for the typical magnitudes of centripetal acceleration, we’ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we’ll calculate the force required to make a car round a curve.

Estimating Centripetal Acceleration

In this activity, you will measure the swing of a golf club or tennis racket to estimate the centripetal acceleration of the end of the club or racket. You may choose to do this in slow motion. Recall that the equation for centripetal acceleration is a c = v 2 r a c = v 2 r or a c = r ω 2 a c = r ω 2 .

• One tennis racket or golf club
• One ruler or tape measure
• Work with a partner. Stand a safe distance away from your partner as he or she swings the golf club or tennis racket.
• Describe the motion of the swing—is this uniform circular motion? Why or why not?
• Try to get the swing as close to uniform circular motion as possible. What adjustments did your partner need to make?
• Measure the radius of curvature. What did you physically measure?
• By using the timer, find either the linear or angular velocity, depending on which equation you decide to use.
• What is the approximate centripetal acceleration based on these measurements? How accurate do you think they are? Why? How might you and your partner make these measurements more accurate?

The swing of the golf club or racket can be made very close to uniform circular motion. For this, the person would have to move it at a constant speed, without bending their arm. The length of the arm plus the length of the club or racket is the radius of curvature. Accuracy of measurements of angular velocity and angular acceleration will depend on resolution of the timer used and human observational error. The swing of the golf club or racket can be made very close to uniform circular motion. For this, the person would have to move it at a constant speed, without bending their arm. The length of the arm plus the length of the club or racket is the radius of curvature. Accuracy of measurements of angular velocity and angular acceleration will depend on resolution of the timer used and human observational error.

Was it more useful to use the equation a c = v 2 r a c = v 2 r or a c = r ω 2 a c = r ω 2 in this activity? Why?

• It should be simpler to use a c = r ω 2 a c = r ω 2 because measuring angular velocity through observation would be easier.
• It should be simpler to use a c = v 2 r a c = v 2 r because measuring tangential velocity through observation would be easier.
• It should be simpler to use a c = r ω 2 a c = r ω 2 because measuring angular velocity through observation would be difficult.
• It should be simpler to use a c = v 2 r a c = v 2 r because measuring tangential velocity through observation would be difficult.

Worked Example

Comparing centripetal acceleration of a car rounding a curve with acceleration due to gravity.

A car follows a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h). What is the magnitude of the car’s centripetal acceleration? Compare the centripetal acceleration for this fairly gentle curve taken at highway speed with acceleration due to gravity ( g ).

Because linear rather than angular speed is given, it is most convenient to use the expression a c = v 2 r a c = v 2 r to find the magnitude of the centripetal acceleration.

Entering the given values of v = 25.0 m/s and r = 500 m into the expression for a c gives

To compare this with the acceleration due to gravity ( g = 9.80 m/s 2 ), we take the ratio a c / g = ( 1.25  m/s 2 ) / ( 9.80 m/s 2 ) = 0.128 a c / g = ( 1.25  m/s 2 ) / ( 9.80 m/s 2 ) = 0.128 . Therefore, a c = 0.128 g a c = 0.128 g , which means that the centripetal acceleration is about one tenth the acceleration due to gravity.

Frictional Force on Car Tires Rounding a Curve

• Calculate the centripetal force exerted on a 900 kg car that rounds a 600-m-radius curve on horizontal ground at 25.0 m/s.
• Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
• If the car would slip if it were to be traveling any faster, what is the coefficient of static friction between the tires and the road? Could we conclude anything about the coefficient of static friction if we did not know whether the car could round the curve any faster without slipping?

Strategy and Solution for (a)

We know that F c = m v 2 r F c = m v 2 r . Therefore,

Strategy and Solution for (b)

The image above shows the forces acting on the car while rounding the curve. In this diagram, the car is traveling into the page as shown and is turning to the left. Friction acts toward the left, accelerating the car toward the center of the curve. Because friction is the only horizontal force acting on the car, it provides all of the centripetal force in this case. Therefore, the force of friction is the centripetal force in this situation and points toward the center of the curve.

Strategy and Solution for (c)

If the car is about to slip, the static friction is at its maximum value and f = μ s N = μ s m g f = μ s N = μ s m g . Solving for μ s μ s , we get μ s = 938 900 × 9.8 = 0.11 μ s = 938 900 × 9.8 = 0.11 . Regardless of whether we know the maximum allowable speed for rounding the curve, we can conclude this is a minimum value for the coefficient.

Since we found the force of friction in part (b), we could also solve for the coefficient of friction, since f = μ s N = μ s m g f = μ s N = μ s m g . The static friction is only equal to μ s N μ s N when it is at the maximum possible value. If the car could go faster, the friction at the given speed would still be the same as we calculated, but the coefficient of static friction would be larger.

Practice Problems

Calculate the centripetal acceleration of an object following a path with a radius of a curvature of 0.2 m and at an angular velocity of 5 rad/s.

Check Your Understanding

• Uniform circular motion is when an object accelerates on a circular path at a constantly increasing velocity.
• Uniform circular motion is when an object travels on a circular path at a variable acceleration.
• Uniform circular motion is when an object travels on a circular path at a constant speed.
• Uniform circular motion is when an object travels on a circular path at a variable speed.

Which of the following is centripetal acceleration?

• The acceleration of an object moving in a circular path and directed radially toward the center of the circular orbit
• The acceleration of an object moving in a circular path and directed tangentially along the circular path
• The acceleration of an object moving in a linear path and directed in the direction of motion of the object
• The acceleration of an object moving in a linear path and directed in the direction opposite to the motion of the object
• Yes, the object is accelerating, so a net force must be acting on it.
• Yes, because there is no acceleration.
• No, because there is acceleration.
• No, because there is no acceleration.

Identify two examples of forces that can cause centripetal acceleration.

• The force of Earth’s gravity on the moon and the normal force
• The force of Earth’s gravity on the moon and the tension in the rope on an orbiting tetherball
• The normal force and the force of friction acting on a moving car
• The normal force and the tension in the rope on a tetherball

Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the formative assessment will help identify which objective is causing the problem and direct students to the relevant content.

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Investigating Circular Motion ( OCR A Level Physics )

Revision note.

Investigating Circular Motion

Equipment & method.

• Tie a bung of mass m,  to a piece of string, which sits horizontally
• Thread it though a glass tube and a paper clip, which sits vertically
• At the other end of the string a heavier mass,  M is suspended vertically
• This provides the centripetal force,  F = Mg when the tension in the string is constant
• The time taken for several rotations is recorded and repeated to remove any random errors
• The masses in the experimental set up are changed before the experiment is repeated again

Explanation

• When the force it provides is equal to the centripetal force , Mg 
• This is the centripetal force required to make the bung travel in a circular path
• The  weight , and hence the centripetal force , required for different masses, radii and speeds can be investigated
• The tension in the string
• The weight of the bung downwards
• If the centripetal force required is greater than its weight then the suspended mass moves upwards
• The paperclip will move accordingly to make this movement clearer
• As the bung moves around the circle, the direction of the tension will change continuously
• This is because the direction of the weight of the bung never changes, so the resultant force will vary depending on the position of the bung in the circle
• At the bottom of the circle, the tension must overcome the weight, this can be written as:

• As a result, the acceleration, and hence, the speed of the bung will be slower at the top
• At the top of the circle, the tension and weight act in the same direction, this can be written as:

• As a result, the acceleration, and hence, the speed of the bung will be faster at the bottom

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Uniform Circular Motion Lab

This experiment investigates how the force necessary to maintain a body in uniform circular motion (i.e. in a circle of constant radius and with a constant speed) is related to the speed of the object experiencing UCM. You should note that there are other possibilities for the independent variable. We are keeping all of the possible independent variables fixed save the applied force. 

Your apparatus, shown in figure 1, consists of a glass tube, some string, a rubber stopper, and an alligator clip. The force that keeps the rubber stopper traveling in a circle at constant speed is supplied by the string. The source of the force is gravity pulling down on a mass attached to the lower end of the string. If we ignore the friction of the string on the glass tube, the force downward on the mass due to gravity is the same as the force inward on the stopper. Hence the force is M hanging g and the velocity can be calculated by measuring the time for thirty complete circles, calculating the distance traveled (30 times the circumference), and then the average speed.

• Measure the mass of your stopper.  (You won't need it until the end, but measuring it now makes sure you don't forget to do it.)
• Thread the string through the tube.
• Attach the stopper to one end and 150 g of mass to the lower end.
• Stretch the apparatus out on your lab table and position the stopper so that the distance between the middle of the stopper and the top of the tube is 50 cm.
• Attach the alligator clip about 2 cm below the bottom of the tube.
• Holding the tube in one hand, start swinging the stopper slowly in a circle above your head. As you swing faster, the clip will rise.  This happens because the suspended mass can't supply enough force to keep the stopper from moving out. Thus, the radius increases and the force needed to maintain UCM will then decrease (due to the larger radius the needed acceleration is smaller). After a little practice, you should be able to maintain the alligator clip about 2 cm below the bottom of the tube.
• Have a partner start a stopwatch and time the amount of time needed for thirty complete revolutions. You should concentrate on trying to keep the alligator clip in a steady position 2 cm below the bottom of the tube.
• Once the time has been measured, record your data in a table (similar to the example below) and proceed on to other masses.
• You should acquire additional data for hanging masses of: 20, 30, 50, 100, 200, 250, and 280 grams.

For each case, we know the radius of the circle is equal to the value set earlier, and that the applied force is the weight of the hanging mass. Calculate the force applied, and the resulting velocity for each case and complete a table like the one shown in your notebook.

Once your table is complete,  plot a graph of velocity vs. applied force. What is the shape of the graph? If it is straight, draw in the best fit straight line and determine the equation for the best fit line. If the graph is not straight, use its shape to figure out how to re-plot the graph in such a manner as to obtain a straight (or straighter) line. Draw in the best fit line for this graph and determine the equation which describes the line.

Use Newton's second law ( with the net force being the weight of the suspended mass) and your knowledge of circular motion to predict what the slope of your line should be. ( i.e. solve for v in terms of F net .   Calculate this theoretical slope.   Compare this theoretical slope with the slope that you got from your best fit line.   How well do they agree? Make a verbal comparison and also calculate the percentage difference between the experimental and theoretical slopes. Which slope is larger? What are some possible experimental reasons that the two might not agree? For one of the possible sources of errors (i.e experimental uncertainty not mistakes), think about whether it would tend to increase or to decrease the measured slope? Explain your reasoning.

Barney Taylor Physics Dept. Miami University - Hamilton 1601 University Blvd., Hamilton, OH 45011 (513) 785-3016 [email protected]

last modified Aug, 2013

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Shop Experiment Circular Motion Experiments​

Circular motion.

Experiment #8 from Physics Explorations and Projects

Introduction

The goal of this activity is for students to determine the relationship between the (angular or linear) velocity, radius, and mass on the centripetal force or acceleration necessary to keep an object moving in a circular path.

In the Preliminary Observations, students will observe an object that is swung on a string in a circular path. Students then explore force (or acceleration) and circular motion using a method of their choosing, though instructors may provide a limited number of choices.

During the subsequent inquiry process, students may use a variety of tools ranging from a mass on a string to the Centripetal Force Apparatus. This activity can also be done virtually using Pivot Interactives.

Students should finish the activity having evaluated data graphically and developed an expression relating a center-pointed (centripetal) net force to the properties of circular motion.

• Design and perform an investigation.
• Draw a conclusion from evidence.
• Understand that the centripetal force acting on (or the centripetal acceleration of) an object moving in a circular pattern is governed by one of two relationships

Sensors and Equipment

This experiment features the following sensors and equipment. Additional equipment may be required.

Correlations

Teaching to an educational standard? This experiment supports the standards below.

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This experiment is #8 of Physics Explorations and Projects . The experiment in the book includes student instructions as well as instructor information for set up, helpful hints, and sample graphs and data.

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Using the Interactive

Circular Motion: Mass on a String

Hsc physics syllabus.

analyse the forces acting on an object executing uniform circular motion in a variety of situations, for example:

Circular Motion Involving Tension Force

This video discusses the analysis of uniform circular motion scenarios involving a mass on a string or rope. 

Understanding Circular Motion with a String

When an object is tethered to a string or rope and set into circular motion, like a ball swung in a circle, it's the tension in the string that provides the necessary centripetal force to keep the object moving in that circle.

Role of Tension in Circular Motion

Tension is a force exerted by a string or rope when it is stretched. In circular motion with a string:

Tension Force : Always pulls inward, towards the centre of the circular path. It contributes to the centripetal force, ensuring the object remains in its circular trajectory.

Centripetal Force : This isn’t a new or separate force. Instead, in the context of our spinning object, the tension in the string provides the centripetal force.

The relationship between tension and centripetal force depends on the angle the string or rope makes with either the vertical or horizontal axis.

When String is Horizontal

Consider a ball or mass attached to a string, undergoing uniform circular motion in a horizontal path as shown. 

It is important to understand that this situation is only possible if the weight force of the mass is relatively negligible compared to the tension force. This is because the weight force is not balanced by any other forces, and would result in a net force acting downward on the mass, causing the string to create an angle with the horizontal axis.

In this case, the centripetal force is provided by the tension force:

$$F_{\text{centripetal}} = T$$

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

This equation suggests that the magnitude of tension force depends on:

• Mass of the object : A heavier object requires more centripetal force, hence more tension, to keep it in a circular path.
• Radius of the circle : A larger circle or greater distance from the centre reduces the tension required, while a smaller circle increases it.
• Speed of the object : The faster the object moves, the greater the necessary centripetal force, and thus the greater the tension in the string.

In order to create such a circular path, the magnitude of tension needs to be substantially greater than the object's weight force. This can be achieved by increasing the object's speed, decreasing the length of the string (thus the radius), or using a smaller mass.

When String is at an Angle

Consider a ball or mass attached to a string, undergoing uniform circular motion in a horizontal path as shown. The string makes an angle  θ with the vertical axis.

When the string or rope is at an angle, the direction of the tension force is also at an angle. This means we must resolve the tension force vector into its horizontal and vertical components.

$$T_x = T\sin{\theta}$$

$$T_y = T\cos{\theta}$$

In this case, the only force vector directed towards the centre of the circular path is the horizontal component of tension force. Therefore, the centripetal force is provided by the horizontal component of tension:

$$F_{\text{centripetal}} = T\sin{\theta}$$

$$\frac{mv^2}{r} = T\sin{\theta}$$

The vertical component of tension must balance the downward weight force of the mass if undergoes a horizontal circular motion.

$$mg = T\cos{\theta}$$

By dividing the two equations, we get:

$$\frac{v^2}{rg} = \tan{\theta}$$

Rearranging this equation, we can derive an expression for velocity:

$$v = \sqrt{rg\tan{\theta}}$$

Therefore, the velocity of the mass on a string in circular motion depends on

• radius of circular path
• acceleration due to gravity
• angle the string makes with the vertical axis; when angle increases, velocity increases

Vertical Circular Motion

Mass on a string or rope can be swung to create circular motion in a vertical orientation. In this instance, uniform circular motion can be achieved by ensuring that the centripetal force (net force) acting on an object remains constant.

It is important to understand that the direction and magnitude of different forces acting on the mass in a vertical circular motion change depending on its position.

For example, while the magnitude and direction of weight force remain unchanged, the direction and magnitude of tension change throughout the motion. At the top of the circular path, tension force acts downward. At the bottom of the circular path, tension force acts upward. 

When an object attached to a rope is swung vertically, the tension in the rope changes throughout the motion. More tension force is needed at the bottom of the uniform circular motion compared to at the top because gravitational force acts in the opposite direction to centripetal force.  

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