Circular Motion Examples (College Board AP® Physics 1: Algebra-Based) : Study Guide

Ann Howell

Written by: Ann Howell

Reviewed by: Caroline Carroll

Updated on

Banked surface

  • A banked surface, normally a road or track is a curved surface where the outer edge is raised higher than the inner edge

    • The purpose of this is to make it safer for objects to travel on the curved road at a reasonable speed without skidding

  • The steeper the banked curve the greater the speed an object can travel around the curve without sliding up or down

Banked surface with friction

  • For an object moving on a banked surface with friction:

    • gravitational force F with rightwards arrow on top subscript g acts vertically downwards

    • normal reaction force F with rightwards arrow on top subscript N acts perpendicular to the surface

    • frictional force F with rightwards arrow on top subscript f acts parallel to the surface, either up or down the slope depending on the situation

    • centripetal force F with rightwards arrow on top subscript c is the net force acting horizontally towards the center of the circle

Forces acting on an object on a banked surface

A car on an inclined plane, shows forces: gravitational downward, frictional leftward, normal upward, with angle of incline indicated.
Frictional forces act parallel to and normal forces act perpendicular to the angle of the slope
  • There are two types of friction to consider

    • static friction F with rightwards arrow on top subscript f comma s end subscript prevents the car from skidding

    • kinetic friction F with rightwards arrow on top subscript f comma k end subscript acts if the car starts skidding

  • If an object remains at the same position up the slope of the banked surface while undergoing uniform circular motion then only static friction is present

Kinetic vs static friction on a tire

Comparision of skidding and rolling wheels. Skidding shows kinetic friction (Fk) opposing wheel movement. Rolling shows static friction (Fs) preventing slippage.
Kinetic friction acts when the tire is skidding but static friction acts when the tire is rolling
  • Components of the static friction force and the normal force can contribute to the net centripetal force producing the centripetal acceleration of an object traveling in a circle on a banked surface

    • The net centripetal force acts horizontally towards the center of the circle

  • An object travelling at its ideal speed is travelling at the maximum speed possible before it starts slipping or sliding on the banked surface, and kinetic friction is applied to keep the car in its lane

Net centripetal force

Top view diagram of forces acting on a red rectangle representing a vehicle. Arrows show normal force (n) and friction (fr) from the road and mg (gravity) downward.
Centripetal force is equal to the horizontal normal force and the frictional force components for an object travelling on a banked surface

Faster than ideal speed

  • When the object is moving faster than the ideal speed, static friction acts down the banked surface

    • The static friction prevents the object from skidding upward

  • The net centripetal force is given by the equation

F with rightwards arrow on top subscript c space equals space F with rightwards arrow on top subscript N space sin space theta space plus space F with rightwards arrow on top subscript f comma s end subscript space cos space theta

  • Where:

    • F with rightwards arrow on top subscript c space equals spacenet centripetal force, measured in straight N

    • F with rightwards arrow on top subscript N space equals normal reaction force, measured in straight N

    • theta space equalsangle of incline of banked surface, measured in degree

    • F with rightwards arrow on top subscript f comma s end subscript space equals spacestatic friction, measured in straight N

Components of forces at faster than ideal speed

Forces on an inclined plane, showing gravitational force (mg) downward, normal force (F_N) perpendicular, and frictional force (F_s) opposing motion.
The net centripetal force is the horizontal component of static friction and the normal force

Slower than ideal speed

  • When the object is moving slower than the ideal speed static friction acts up the banked surface in the opposite direction to the centripetal force

    • The static friction prevents the object from skidding downwards

  • The net centripetal force is given by the equation

F with rightwards arrow on top subscript c space equals space F with rightwards arrow on top subscript N space sin space theta space minus space F with rightwards arrow on top subscript f comma s end subscript space cos space theta

Components of forces at slower than ideal speed

Forces on an inclined plane, showing components: Fs and its components Fs*sinθ and Fs*cosθ in green, FN and its components FN*sinθ and FN*cosθ in blue, and mg in red.
The net centripetal force is the horizontal component of static friction and the normal force

Ideally banked surfaces

  • An ideally banked surface has an angle of incline theta so an object can negotiate the curve at a certain speed without the need for friction

  • For an object moving on an ideally banked surface without friction

    • gravitational force F with rightwards arrow on top subscript g acts vertically downwards

    • normal reaction force F with rightwards arrow on top subscript N acts perpendicular to the surface

    • centripetal force F with rightwards arrow on top subscript c is the net force acting horizontally towards the center of the circle

  • The net centripetal force is given by the equation

F with rightwards arrow on top subscript c space equals space F with rightwards arrow on top subscript N space sin space theta

Components of the normal force on an ideally banked surface

A car on a sloped road showing forces. Vectors indicate gravity (mg), normal force (FN), and components of the normal force (FN cosθ and FN sinθ), with a labeled angle θ.
A car on an ideally banked surface remains on a banked surface due to the horizontal component of the normal force

Worked Example

What is the minimum angle of an ideally banked road so a car can travel at 40 space straight m divided by straight s and safely negotiate a curve of radius 100 space straight m?

A      0.00 space degree

B      0.03 space degree

C      2.29 space degree

D      58 space degree

The correct answer is D

Answer:

Step 1: Analyze the scenario

  • The car is on an ideally banked road, so there is negligible friction

  • Therefore, the centripetal force acting towards the center of the curve must be equal to the horizontal component of the normal reaction force of the car

Step 2: List the known quantities

  • Tangential speed of car, v space equals space 40 space straight m divided by straight s

  • Radius of curve, r space equals space 100 space straight m

  • Acceleration due to gravity at Earth's surface, g space equals space 10 space straight m divided by straight s squared

Step 3: Determine an expression for the centripetal force as the horizontal component of the normal force

F with rightwards arrow on top subscript c space equals space F with rightwards arrow on top subscript N space sin space theta

F with rightwards arrow on top subscript c space equals space fraction numerator m v squared over denominator r end fraction

F with rightwards arrow on top subscript N space sin space theta space equals space fraction numerator m v squared over denominator r end fraction eq (1)

Step 4: Determine an expression for the net zero vertical normal force

F with rightwards arrow on top subscript N space cos space theta space equals space m g eq (2)

Step 5: Combine equations (1) and (2) to obtain an expression in terms of thetawithout the normal force, stack F subscript N with rightwards arrow on top

fraction numerator eq space open parentheses 1 close parentheses over denominator eq space open parentheses 2 close parentheses end fraction space equals space fraction numerator F with rightwards arrow on top subscript N space sin space theta over denominator F with rightwards arrow on top subscript N space cos space theta end fraction space equals space fraction numerator m v squared over denominator r end fraction times fraction numerator 1 over denominator m g end fraction

tan space theta space equals space fraction numerator v squared over denominator g r end fraction

Step 6: Substitute in the known quantities to obtain the minimum angle of the ideally banked road

theta space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator v squared over denominator g r end fraction close parentheses

theta space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 40 squared over denominator 10 space times space 100 end fraction close parentheses

theta space equals space 58 degree

  • The answer is therefore D

Examiner Tips and Tricks

The value of F with rightwards arrow on top subscript N is not equal to the gravitational force, m g, of the car so it cannot be calculated without the angle of the incline.

Conical pendulum

  • A conical pendulum consists of a mass set in horizontal circular motion suspended on the end of a light inextensible string fixed from a central point above

    • For example, a fairground ride or a ball suspended on a string

Conical pendulum example

Eight people are seated on swings attached to a central pole, which rotates to lift and spin the riders in the air. The swing ride creates a circular motion.
This fairground ride is an example of a conical pendulum where the passengers are suspended from a fixed point and undergo horizontal circular motion
  • It is assumed that:

    • the string is massless

    • the string cannot be stretched

    • air resistance is negligible

  • The behavior of a conical pendulum is similar to that of an object on an ideally banked surface where there is no friction

    • The normal reaction force of the banked surface on the object is equivalent to the tension in the string

  • A component of tension contributes to the net force producing centripetal acceleration experienced by a conical pendulum

    • The tension in the string and the gravitational force on the mass are not equal, so the mass is not in equilibrium, hence there is centripetal acceleration

  • A conical pendulum experiences the following forces:

    • gravitational force F with rightwards arrow on top subscript g acting vertically downwards

    • tension force F with rightwards arrow on top subscript T acting along the line of the string at an angle theta to the vertical

    • centripetal force F with rightwards arrow on top subscript c is the net force acting horizontally towards the center of the circle

  • The horizontal net centripetal force is given by the equation:

F with rightwards arrow on top subscript c space equals space F with rightwards arrow on top subscript T space sin space theta

  • The vertical component of the tension in the string is equal to the gravitational force acting on the mass

F with rightwards arrow on top subscript g space equals space m g space equals space space F with rightwards arrow on top subscript T space cos space theta

Components of the tension force on a conical pendulum

A pendulum showing forces: tension (F_T) acting at an angle θ, with its components F_T cos θ (vertical) and F_T sin θ (horizontal), and weight (F_g = mg) acting downward.
The tension component in the string can be resolved horizontally to calculate the net centripetal force acting on the mass

Worked Example

A conical pendulum has a mass m suspended from a height h by a light inextensible string with tension F with rightwards arrow on top subscript T at an angle of theta to the vertical. It moves in uniform circular motion of radius, r with tangential velocity, v.

A conical pendulum with a mass m has forces T (tension), mg (gravity), angle θ, radius r, string length l, and height h with circular trajectory.

Verify that the time period of the oscillation of the conical pendulum is given by the equation

T space equals space 2 pi space square root of h over g end root

Answer:

Step 1: Analyze the scenario and determine the required equations

  • The following forces are present on a conical pendulum system

    • The gravitational force acting on the mass m is stack F subscript g with rightwards arrow on top space equals space m g

    • The vertical component of tension in the string is stack F subscript g with rightwards arrow on top space equals space F with rightwards arrow on top subscript T space cos space theta

    • The horizontal component of the tension in the string is stack F subscript c with rightwards arrow on top space equals space F with rightwards arrow on top subscript T space sin space theta

    • The centripetal force is given by the equation stack F subscript c with rightwards arrow on top space equals space fraction numerator m v squared over denominator r end fraction

    • The time period is calculated using T space equals space fraction numerator circumference over denominator tangential space speed end fraction space equals space fraction numerator 2 straight pi r over denominator v end fraction

Step 2: Determine an expression for the centripetal force as the horizontal component of the tension force

F with rightwards arrow on top subscript T space sin space theta space equals space fraction numerator m v squared over denominator r end fraction eq (1)

Step 3: Determine an expression for the net zero vertical tension force

F with rightwards arrow on top subscript T space cos space theta space equals space m g eq (2)

Step 5: Combine equations (1) and (2) to obtain an expression in terms of thetawithout the tension force, stack F subscript T with rightwards arrow on top

fraction numerator eq space open parentheses 1 close parentheses over denominator eq space open parentheses 2 close parentheses end fraction space equals space fraction numerator F with rightwards arrow on top subscript T space sin space theta over denominator F with rightwards arrow on top subscript T space cos space theta end fraction space equals space fraction numerator m v squared over denominator r end fraction times fraction numerator 1 over denominator m g end fraction

tan space theta space equals space fraction numerator v squared over denominator g r end fraction

v space equals space square root of g r space tan space theta end root

Step 6: Determine an expression for tan space theta in terms of h and r

tan space theta space equals space opposite over adjacent space equals space r over h

Step 7: Combine the equations to obtain an expression for v in terms of g, h and r

v space equals space square root of g r space times space open parentheses r over h close parentheses end root

v space equals space square root of fraction numerator g r squared over denominator h end fraction end root

v space equals space r square root of g over h end root

Step 8: Substitute the expression for v into the time period equation

T space equals space fraction numerator 2 straight pi r over denominator v end fraction

T space equals space 2 straight pi r space divided by space open parentheses r square root of g over h end root close parentheses

T space equals space 2 straight pi space square root of h over g end root as required

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Ann Howell

Author: Ann Howell

Expertise: Physics Content Creator

Ann obtained her Maths and Physics degree from the University of Bath before completing her PGCE in Science and Maths teaching. She spent ten years teaching Maths and Physics to wonderful students from all around the world whilst living in China, Ethiopia and Nepal. Now based in beautiful Devon she is thrilled to be creating awesome Physics resources to make Physics more accessible and understandable for all students, no matter their schooling or background.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Physics Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.