experimental design and data analysis practice circular motion 1

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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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.

Ready to Experiment?

Ask an expert.

Get answers to your questions about how to teach this experiment with our support team.

• Call toll-free: 888-837-6437
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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.

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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Circular Motion - Complete Toolkit

• To describe the direction of the velocity, acceleration, and net force for an object that moves in a circle at a constant speed.
• To use the law of inertia to explain why a person moving in a circle experiences a sensation of being pushed outward and to identify reasons why the outward net force is a fictitious force.
• To mathematically relate the speed, acceleration, radius, mass and net force for an object moving in a circle and to use such relationships to solve physics word problems.
• To construct a free-body diagram for any given situation in which an object is moving in a circle.
• To combine Newton’s second law of motion, free-body diagrams and circular motion equations to determine the value of the acceleration or an individual force for any situation involving an object moving in a circle.

Readings from The Physics Classroom Tutorial

• The Physics Classroom Tutorial, Circular Motion and Satellite Motion Chapter, Lesson 1  
• The Physics Classroom Tutorial, Circular Motion and Satellite Motion Chapter, Lesson 2

Interactive Simulations

Video and Animations

Labs and Investigations

• The Physics Classroom, The Laboratory, Making the Turn Lab Students use simple materials (ball, board, wood block) to discover that an inward force is required for circular motion and without such a force and object continues along a straight-line path.  
• The Physics Classroom, The Laboratory, Loop the Loop Lab Students use a bucket of water and a strong rope to gain a sense as to the relative strength of the tension force on a bucket when swung in a vertical circle; free-body diagrams and calculations are used to explain the observations.  
• The Physics Classroom, The Laboratory, Race Track Lab Students use a computer simulation to explore the direction of forces required to successfully guide a car around an oval track and explain the principles associated with a successful strategy.  

Demonstration Ideas

Minds On Physics Internet Modules

• Circular Motion and Gravitation, Ass’t CG1 -  Speed and Velocity
• Circular Motion and Gravitation, Ass’t CG2 -  Acceleration and Net Force
• Circular Motion and Gravitation, Ass’t CG3 -  Centripetal Force and Inertia
• Circular Motion and Gravitation, Ass’t CG4 -  The Centripetal Force Requirement
• Circular Motion and Gravitation, Ass’t CG5 -  Mathematical Analysis of Circular Motion

Concept Building Exercises

• The Curriculum Corner, Circular Motion, Speed and Velocity
• The Curriculum Corner, Circular Motion, Acceleration and Circular Motion
• The Curriculum Corner, Circular Motion, Circular Motion and Inertia
• The Curriculum Corner, Circular Motion, The Centripetal Force Requirement
• The Curriculum Corner, Circular Motion, Mathematics of Circular Motion

Problem-Solving Exercises

• The Calculator Pad, Circular Motion and Gravitation, Problems #1-15

Science Reasoning Activities

• Science Reasoning Center, Roller Coaster Loops

Real Life Connections

Common Misconceptions

• Velocity, Force and the Inertial Path An object that moves along a curved path experiences a velocity that is directed tangent to the path. At all points along a circle or curved path, the object’s velocity is directed  straight ahead . So this tangent direction or  straight ahead  direction describes the  inertial path  of the object. If all the forces were balanced, then the object would continue along this inertial path. Movement in a circle is a deviation from this straight-ahead direction and thus requires an unbalanced force. Motion in a circle requires a net force directed towards the circle’s center in order to cause the object to deviate from the inertial path. But if this net inward force is removed at any instant, then the object travels along the inertial path tangent to the circle.  
• Acceleration It is a very common belief among students that an object that is moving in a circle at a constant speed is not accelerating. But this is quite untrue. Moving in a circle involves an acceleration … even if it is a constant speed motion. Accelerating objects are changing their velocity. Being a vector, the velocity of an object describes an objects speed and direction. So objects that are changing either their speed or their direction are accelerating. As such, an object that moves in a circle at a constant speed is accelerating due to its change in direction. It is helpful to remind students what the three controls are on a car that allow it to accelerate – brake pedal for slowing down, gas pedal for speeding up, and steering wheel for changing directions.  
• The Notion of an Outward or Centrifugal Force Anyone who has been on a roller coaster car, or a circular ride at an amusement park, or simply taken a sharp left-hand turn as a passenger in a car has experienced what seemed to be a strong push or force that is directed away from the center of the circle. This feelings of being pushed outward is undeniable; yet reality is not always consistent with one’s feelings. The sensation of an outward force is attributed to inertia – the tendency of an object to continue in motion in the same direction that it is heading when not acted upon by an unbalanced force. In these cases – roller coaster, amusement park ride, turning car – you are continuing in a straight line until your inertial path meets up with a car door or wall or some other obstacle. Upon meeting the car door or wall, you begin to push outward upon it. Meanwhile, the door or wall is pushing inward on you – consistent with Newton’s third law. Many have written of these experiences by referring to the notion of a centrifugal or outward force. Those who have generally fall into two categories – those who don’t know what they are writing about and shouldn’t be writing about it and those who do know what they are writing about and must be cautious that they don’t mislead their readers. In the case of the latter, the discussion of a centrifugal force is almost always accompanied by some concluding remark that goes like this: “Such a force is a fictitious force. It does not exist despite the fact that the notion of it may help to explain the sensation of being pushed outward.” If the reader fails to observe the importance of this short phrase, then the reader is misled into believing that motion in a circle is accompanied by a centrifugal or outward force. That’s tragic.

Elsewhere on the Web

• HS-PS2.1 :   Analyze data to support the claim that Newton’s second law of motion describes the mathematical relationship among the net force on a macroscopic object, its mass, and its acceleration.
• HS-PS2.A.i :  Newton’s Second Law accurately predicts changes in the motion of macroscopic objects
• Develop and use models (physical, mathematical, and computer models) to predict the behavior of a system and design systems to do specific tasks.
• Use a model to provide mechanistic accounts of phenomena.
• Analyze data using tools, technologies, and/or models (e.g., computational, mathematical) in order to make valid and reliable scientific claims or determine an optimal design solution.
• Analyze data using computational models in order to make valid and reliable scientific claims.
• Use mathematical representations of phenomena to describe explanations.
• Create a computational model or simulation of a phenomenon, designed device, process, or system.
• Design, evaluate, and/or refine a solution to a real-world problem, based on scientific knowledge, student-generated sources of evidence, prioritized criteria, and tradeoff considerations.
• Communicate technical information or ideas (e.g. about phenomena and/or the process of development and the design and performance of a proposed process or system) in multiple formats (including orally, graphically, textually, and mathematically)
• MP.1 – Make sense of problems and persevere in solving them
• MP.2 – Reason abstractly and quantitatively
• MP.4 – Model with mathematics
• A-SSE.1.a :     Interpret parts of an expression, such as terms, factors, and coefficients
• A-SSE.2 :     Use the structure of an expression to identify ways to rewrite it.
• A-CED.2     Create equations in two or more variables to represent relationships between quantities, graph equations on coordinate axes with labels and scales
• A-CED.4     Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations
• F-TF.1     Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle
• F-LE.5     Interpret the parameters in a linear or exponential function in terms of a context
• G-C.4    Construct a tangent line from a point outside a given circle to the circle
• RST.11-12.2    Determine the central ideas or conclusions of a text; summarize complex concepts, processes, or information presented in a text by paraphrasing them in simpler but still accurate terms.
• RST.11-12.3    Follow precisely a complex multistep procedure when carrying out experiments, taking measurements, or performing technical tasks; analyze the specific results based on explanations in the text.
• RST.11-12.9    Synthesize information from a range of sources (e.g., texts, experiments, simulations) into a coherent understanding of a process, phenomenon, or concept, resolving conflicting information when possible.
• RST.11-12.10      By the end of grade 12, read and comprehend science/technical texts in the grades 11-CCR text complexity band independently and proficiently.
• H-3.1.1 – The displacement, or change in position, of an object is a vector quantity that can be calculated by subtracting the initial position from the final position, where initial and final positions can have positive and negative values. Displacement is not always equal to the distance traveled.
• Translate between different representations of the motion of objects: verbal and/or written descriptions, motion diagrams, data tables, graphical representations (position versus time graphs and instantaneous velocity versus time graphs) and mathematical representations.
• Objective M-3.2.2d – A force of constant magnitude acting at right angles to the direction of the object’s motion causes the object to move in a circle at a constant speed. 
• Objective M-3.2.3 – The forces acting on an object can be represented by arrows (vectors) drawn on an isolated picture of the object, called a force diagram. The direction of each arrow shows the direction of the push or pull. Forces are labeled: “(type of interaction) push or pull of (interacting object) on the (object of interest).
• Objective H.3.2.2c – An object moves in a circle when the vector sums of all the forces (net force) is constant in magnitude, always directed at right angles to the direction of motion and always directed toward the same point in space, the center of the circle. The speed of the object does not change; the acceleration causes the continual change in the direction of the change-in-velocity vector.
• Analyze force diagrams to determine if they accurately represent different situations involving multiple contact, gravitational and/or electrical interactions. When appropriate, determine the one-dimensional vector sum of all the forces (net force), and interpret the meaning of the vector sum of all the forces (net force).
• Represent the forces acting on the object of interest by drawing a force diagram showing both the vertical and horizontal forces. When appropriate, use vector addition to determine the relative size and direction of the sum of all the forces (net force), and interpret the meaning of the net force.
• Investigate and explain why an object moving at a constant speed in a circle is accelerating. Justify the explanation by constructing a motion diagram and by using knowledge of acceleration and Newton’s second law.

Teacher Resource Center

Pasco partnerships.

2024 Catalogs & Brochures

Circular motion.

In this lab, students will use force sensors to develop a kinesthetic understanding of circular motion. Students will measure the period of rotation of a mass in uniform circular motion.

Grade Level: High School

Subject: Physics

Student Files

Featured Equipment

PASPORT Force Sensor

This sensor measures both pulling and pushing forces up to ±50 N.

Many lab activities can be conducted with our Wireless , PASPORT , or even ScienceWorkshop sensors and equipment. For assistance with substituting compatible instruments, contact PASCO Technical Support . We're here to help. Copyright © 2018 PASCO

Source Collection: Lab #18

Physics Through Inquiry

More experiments.

• Circuits Experiment Board
• Brewster's Angle

High School

• Charge and Electric Field
• Change in Kinetic Energy
• Determining the Relationship Between Distance and Radiation Count