Circular Motion in a Vertical Loop (College Board AP® Physics 1: Algebra-Based)

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Ann Howell

Written by: Ann Howell

Reviewed by: Caroline Carroll

Circular motion in a vertical loop

  • An object such as a ball on a string is an example of circular motion in a vertical loop

  • The forces acting on the ball are:

  • As the ball moves around the circle:

    • the direction of the tension will change continuously

    • the magnitude of the tension will change continuously, reaching:

      • a maximum value at the bottom

      • a minimum value at the top

  • The direction of the gravitational force on the ball never changes

    • Therefore, the resultant centripetal force on the ball changes depending on its position as it rotates around the circle

Forces in a vertical circle

An object in circular motion has forces and velocities at the top and bottom of its path. The gravitational force on the object (Fg=mg), tension (T), radius (r), and angle (θ) are labeled.
Gravitational force is constant for an object rotating in a vertical circle but tension changes depending on the position of the object within the circle

Minimum speed

  • At the top of a vertical, circular loop, an object requires a minimum speed to maintain circular motion

  • The tension in the string is zero at the top of the circular loop

  • At the top of the loop at this minimum speed, the gravitational force is the only force that causes the centripetal acceleration

    • So, a space equals space g

Forces at the top of a vertical circle

A pendulum in circular motion with forces labeled. The tension (T) acts along the string of length (r), and gravitational force (mg) acts downward.
At the top of a vertical circle both tension and gravitational force act downwards towards the center of the circle

Derived equation

  • The minimum speed required to maintain circular motion at the top of the circular loop is given by:

v space equals space square root of g r end root

  • Where:

    • v space equals minimum speed at the top of the vertical circular loop, measured in straight m divided by straight s

    • g space equals spaceacceleration due to gravity at Earth's surface, measured in straight m divided by straight s squared

    • r space equals spaceradius of circle, measured in straight m

Derivation:

Step 1: Identify the fundamental principles

F with rightwards arrow on top subscript n e t end subscript space equals space m a with rightwards arrow on top

  • Where:

    • F with rightwards arrow on top subscript n e t end subscript = net centripetal force exerted on the object

    • m = mass of the object

    • a with rightwards arrow on top = centripetal acceleration of the object

  • The net force in this case is the gravitational force or the weight force

  • The acceleration in this case is the acceleration due to gravity at the Earth's surface

  • Therefore:

Weight space equals space F with rightwards arrow on top subscript g space equals space m g

  • Where:

    • F with rightwards arrow on top subscript g = gravitational force exerted on the object

    • m = mass of the object

    • g = acceleration due to gravity

  • The magnitude of the centripetal acceleration of an object moving in a circular path is given by:

a subscript c space equals space v squared over r

  • Where:

    • a subscript c = magnitude of the centripetal acceleration

    • v = tangential or linear speed

    • r = radius of circular path

Step 2: Apply the specific conditions

  • When an object is moving in a circular loop, the net centripetal force is calculated using the tangential speed and the radius of the circular loop

    • Substitute the centripetal acceleration into Newton's second law

F with rightwards arrow on top subscript n e t end subscript space equals space m open parentheses v squared over r close parentheses

  • At the top of a vertical circular loop, the gravitational force is the only force that causes the centripetal acceleration

    • The tension in the string is zero

F with rightwards arrow on top subscript g space equals space space F with rightwards arrow on top subscript n e t end subscript space

m g space equals space m open parentheses v squared over r close parentheses

Step 3: Rearrange to obtain an equation for the minimum speed at the top of a vertical circular loop

up diagonal strike m g space equals space up diagonal strike m open parentheses v squared over r close parentheses

g space equals space open parentheses v squared over r close parentheses

g r space equals space v squared

v space equals space square root of g r end root

Worked Example

A bucket of mass 8.0 kg is filled with water and is attached to a string of length 0.5 m.

What is the minimum speed the bucket must have at the top of the circle so no water spills out?

A box hanging 0.5 meters from the top of a dashed circular path representing its motion. Copyright notice at the bottom.

A      2.24 space straight m divided by straight s

B      3.16 space straight m divided by straight s

C      6.32 space straight m divided by straight s

D      50 space straight m divided by straight s

The correct answer is A

Answer:

Step 1: Draw the forces on the bucket at the top

non-uniform-circular-motion-we-ans
  • Although tension is in the rope, at the very top, the tension is 0

Step 2: Recall the equation for minimum speed at the top of the vertical loop

v space equals space square root of g r end root

Step 4: Substitute in values to calculate

v space equals space square root of 10 space cross times space 0.5 end root space equals space 2.24 space straight m space straight s to the power of negative 1 end exponent

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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

Author: 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.