AQA A Level Physics

Revision Notes

A LevelPhysicsAQARevision Notes11. Engineering Physics11.1 Rotational Dynamics11.1.9 Rotational Kinetic Energy

11.1.9 Rotational Kinetic Energy

Rotational Kinetic Energy

  • A body moving with linear velocity has an associated linear kinetic energy given by

E subscript k space equals space 1 half m v squared

E subscript k space equals space fraction numerator p squared over denominator 2 m end fraction

  • Similarly, a rotating body with angular velocity has an associated rotational kinetic energy given by

E subscript k space equals space 1 half I omega squared

E subscript k space equals space fraction numerator L squared over denominator 2 I end fraction

  • Where:
    • E subscript k = rotational kinetic energy (J)
    • I = moment of inertia (kg m2)
    • omega = angular velocity (rad s−1)
    • L = angular momentum (kg m2 s−1)

Rolling without slipping

  • Circular objects, such as wheels, are made to move with both linear and rotational motion
    • For example, the wheels of a car, or bicycle rotate causing it to move forward
  • Rolling motion without slipping is a combination of rotating and sliding (translational) motion
  • When a disc rotates:
    • Each point on the disc has a different linear velocity depending on its distance from the centre open parentheses v space proportional to space r close parentheses
    • The linear velocity is the same at all points on the circumference 
  • When a disc slips, or slides:
    • There is not enough friction present to allow the object to roll
    • Each point on the object has the same linear velocity
    • The angular velocity is zero
  • So, when a disc rolls without slipping:
    • There is enough friction present to initiate rotational motion allowing the object to roll
    • The point in contact with the surface has a velocity of zero
    • The centre of mass has a velocity of v space equals space omega r
    • The top point has a velocity of 2 v or 2 omega r

E16CHGH6_1-4-9-rotational-kinetic-energy-rolling-without-slipping

Rolling motion is a combination of rotational and translational motion. The resultant velocity at the bottom is zero and the resultant velocity at the top is 2v

Rolling down a slope

  • Another common scenario involving rotational and translational motion is an object (usually a ball or a disc) rolling down a slope
  • At the top of the slope, a stationary object will have gravitational potential energy equal to

E subscript p space equals space m g increment h

  • As the object rolls down the slope, the gravitational potential energy will be transferred to both translational (linear) and rotational kinetic energy
  • At the bottom of the slope, the total kinetic energy of the object will be equal to

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

1-4-9-rotational-kinetic-energy-rolling-down-a-slope-ib-2025-physics

The GPE store of the ball is transferred to the translational and rotational kinetic energy store as it rolls down the slope

  • The linear or angular velocity can then be determined by
    • Equating E subscript p and E subscript K space t o t a l end subscript
    • Using the equation for the moment of inertia of the object
    • Using the relationship between linear and angular velocity v space equals space omega r
  • For example, for a ball (a solid sphere) of mass m and radius r, its moment of inertia is

I space equals space 2 over 5 m r squared

  • Equating the equations for E subscript p and E subscript K space t o t a l end subscript and simplifying gives

m g increment h space equals space 1 half m open parentheses omega r close parentheses squared space plus space 1 half open parentheses 2 over 5 m r squared close parentheses omega squared

m g increment h space equals space 1 half m omega squared r squared space plus space 1 fifth m omega squared r squared

m g increment h space equals space 7 over 10 m omega squared r squared

Worked example

A student conducts an experiment to determine the moment of inertia of a turntable. The diagram shows the turntable with a small lump of gum held above it. An optical sensor connected to a data recorder measures the angular speed of the turntable.

11-1-9-rotational-kinetic-energy-we

The turntable has a mass of 550 g and a radius of 120 mm. It is made to rotate before it then rotates freely. The lump of gum is dropped from a small height above the turntable and sticks to it. The results from the experiment are as follows:

mass of gum = 9.0 g

radius at which gum sticks to the turntable = 117 mm

Angular speed of turntable immediately before gum is dropped = 3.28 rad s–1

Angular speed of turntable immediately after gum is dropped = 3.11 rad s–1

The turntable is modelled as a solid disc with a moment of inertia I space equals space 1 half M R squared

The gum is a solid sphere with a moment of inertia I space equals space 2 over 5 m r squared

Calculate the magnitude of the change in rotational kinetic energy of the turntable from the instant before the gum is dropped until immediately after it sticks to the turntable. 

Answer:

Step 1: State the rotational kinetic energy equation of the system before the gum is dropped

  • The gum is not moving before it is dropped, so only the turntable has rotational kinetic energy

B e f o r e space E subscript T space equals space 1 half I subscript T omega subscript 1 squared

  • Where

I subscript T space equals space 1 half M R squared

Step 2: State the rotational kinetic energy equation of the system after the gum is stuck

  • The gum also has inertia when on the turntable

E subscript G space equals space 1 half I subscript G omega subscript 2 squared

  • Where

I subscript G space equals space 2 over 5 m r squared

  • The rotational kinetic energy of the turntable after the gum has stuck is

A f t e r space E subscript T space equals space 1 half open parentheses 1 half M R squared close parentheses omega subscript 2 squared

  • The total rotational kinetic energy after is

E subscript k space equals space 1 half open parentheses 1 half M R squared space plus thin space 2 over 5 m r squared close parentheses omega subscript 2 squared

Step 3: State the equation that will give the difference in the rotational kinetic energies

increment E subscript k space equals space b e f o r e space E subscript k space minus space a f t e r space E subscript k

increment E subscript k space equals space 1 fourth M R squared omega subscript 1 squared space minus space 1 half open parentheses 1 half M R squared plus thin space 2 over 5 m r squared close parentheses omega subscript 2 to the power of 2 space end exponent

Step 4: Substitute in the values

  • ω1 = 3.28 rad s–1 (before gum dropped)
  • ω2 = 3.11 rad s–1 (after gum dropped)
  • mass of gum, m = 9.0 g = 9 × 10−3 kg
  • radius at which gum sticks to the turntable, r = 117 mm = 117 × 10−3 m
  • Mass of turntable, M = 550 g = 550 × 10−3 kg
  • Radius of turntable, R = 120 mm = 120 × 10−3 m

E subscript k space equals space 1 fourth open parentheses 550 space cross times space 10 to the power of negative 3 end exponent close parentheses space cross times space open parentheses 120 space cross times space 10 to the power of negative 3 end exponent close parentheses squared cross times space 3.28 squared space minus space 1 half open parentheses 1 half open parentheses 550 space cross times space 10 to the power of negative 3 end exponent close parentheses space cross times space open parentheses 120 space cross times space 10 to the power of negative 3 end exponent close parentheses squared space plus space 2 over 5 open parentheses 0.009 space cross times space open parentheses 117 space cross times space 10 to the power of negative 3 end exponent close parentheses squared close parentheses close parentheses space cross times space 3.11 squared

E subscript k space equals space 0.0213 space minus space 0.01938 

E subscript k space equals space 1.92 space cross times space 10 to the power of negative 3 end exponent space straight J

Exam Tip

Questions with a long calculation like this are best to split sections when putting them into your calculator. This makes it easier to check and type, ensuring the brackets are in the correct place. It is best to type it in exactly as you have it written down so it is input correctly.

The question asks for the magnitude of the change in rotational kinetic energy. If you found the KE after − KE after then the answer is negative. No worries because the change is the same, no matter which way you find the difference, 

Looking at the angular velocity before and after, when it is smaller, tells us that the rotational kinetic energy will also be less. 

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