AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 6: Energy & Momentum of Rotating SystemsRotational Kinetic Energy, Torque & WorkRotational Kinetic Energy

Rotational Kinetic Energy (College Board AP® Physics 1: Algebra-Based)

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Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Rotational kinetic energy equation

  • Rotational kinetic energy is the energy associated with rotational motion

  • It is given by the equation:

K space equals space 1 half I omega squared

  • Where:

    • K = rotational kinetic energy, in straight J

    • I = rotational inertia, in kg times straight m squared

    • omega = angular velocity, in rad divided by straight s

  • Rotational kinetic energy, like other forms of energy, is a scalar quantity

Comparing translational and rotational energy

  • Rotational kinetic energy is not a new type of energy, it is a type of kinetic energy, analogous to translational kinetic energy

  • On a rotating body, the distribution of mass is described by its rotational inertia

  • The rotational inertia of an object about a fixed axis is given by:

I space equals space m r squared

  • Where:

    • m = mass of the object, in kg

    • r = distance to the axis of rotation, in straight m

  • At different points on a rotating body, the translational (linear) velocity varies, but the angular velocity is the same at all points

  • For a single point rotating about a fixed axis, the angular speed is given by:

omega space equals space v over r

  • Where:

    • v = linear velocity, in straight m divided by straight s

  • Substituting these into the equation for rotational kinetic energy:

K space equals space 1 half I omega squared space equals space 1 half open parentheses m r squared close parentheses open parentheses v over r close parentheses squared space equals space 1 half m v squared

  • This shows that the rotational kinetic energy of an object is equivalent to its translational kinetic energy

Total kinetic energy of rigid systems

  • Rigid circular objects, such as wheels, move with both translational and rotational motion

    • For example, the wheels of a car or a bicycle rotate causing it to move forward

  • Newton's laws for rotational motion remain valid as long as:

    • the axis of rotation passes through the object's center of mass

    • the direction of rotation does not change

  • The total kinetic energy K subscript t o t a l end subscript of a rigid system is a combination of:

    • the rotational kinetic energy K subscript r due to its rotation about its center of mass

    • the translational kinetic energy K subscript t due to the linear motion of its center of mass

  • This can be written as:

K subscript t o t a l end subscript space equals space K subscript r space plus space K subscript t

K subscript t o t a l end subscript space equals space 1 half I omega squared space plus space 1 half m v squared

  • A rigid system can have rotational kinetic energy even if its center of mass is at rest

    • This is because individual points within the system can have linear speed and, therefore, kinetic energy

    • As a result, the whole system rotates about the stationary center of mass

  • Examples of this include:

    • flywheels

    • turntables

    • merry-go-rounds

Worked Example

A flywheel of mass M and radius R rotates at a constant angular velocity omega about an axis through its center. The rotational kinetic energy of the flywheel is K.

The rotational inertia of the flywheel is 1 half M R squared.

A second flywheel of mass 1 half M and radius 1 half R is placed on top of the first flywheel. The new angular velocity of the combined flywheels is 2 over 3 omega.

A smaller flywheel is dropped onto a larger flywheel. The flywheel rotates around a central vertical axis in an anticlockwise direction.

What is the new rotational kinetic energy of the combined flywheels?

A      K over 2

B      K over 4

C      K over 8

D      K over 24

The correct answer is A

Answer:

Step 1: Determine the rotational kinetic energy of the first flywheel

K space equals space 1 half I omega squared space equals space 1 half open parentheses 1 half M R squared close parentheses omega squared

K space equals space 1 fourth M R squared omega squared

Step 2: Determine the total rotational inertia of the combined flywheels

I subscript n e w end subscript space equals space I subscript 1 space plus space I subscript 2

I subscript n e w end subscript space equals space 1 half M R squared space plus space 1 half open parentheses 1 half M close parentheses open parentheses 1 half R close parentheses squared

I subscript n e w end subscript space equals space 9 over 16 M R squared

Step 3: Determine the rotational kinetic energy of the combined flywheels

K subscript n e w end subscript space equals space 1 half I subscript n e w end subscript omega subscript n e w end subscript squared space equals space 1 half open parentheses 9 over 16 M R squared close parentheses open parentheses 2 over 3 omega close parentheses squared

K subscript n e w end subscript space equals space 1 half open parentheses 1 fourth M R squared omega close parentheses space equals space 1 half K

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    Author
    Reviewer
    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

    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.