Rotational Kinetic Energy

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

$E_{k} = \frac{1}{2} m v^{2}$

$E_{k} = \frac{p^{2}}{2 m}$

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

$E_{k} = \frac{1}{2} I ω^{2}$

$E_{k} = \frac{L^{2}}{2 I}$

Where:
- $E_{k}$ = rotational kinetic energy (J)
- $I$ = moment of inertia (kg m²)
- $ω$ = 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 $(v \propto r)$
- 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 $v = ω r$
- The top point has a velocity of $2 v$ or $2 ω r$

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

$E_{p} = m g ∆ h$

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

$E_{K t o t a l} = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

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 $E_{p}$ and $E_{K t o t a l}$
- Using the equation for the moment of inertia of the object
- Using the relationship between linear and angular velocity $v = ω r$

For example, for a ball (a solid sphere) of mass m and radius r, its moment of inertia is

$I = \frac{2}{5} m r^{2}$

Equating the equations for $E_{p}$ and $E_{K t o t a l}$ and simplifying gives

$m g ∆ h = \frac{1}{2} m {(ω r)}^{2} + \frac{1}{2} (\frac{2}{5} m r^{2}) ω^{2}$

$m g ∆ h = \frac{1}{2} m ω^{2} r^{2} + \frac{1}{5} m ω^{2} r^{2}$

$m g ∆ h = \frac{7}{10} m ω^{2} r^{2}$

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.

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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 $I = \frac{1}{2} M R^{2}$

The gum is a solid sphere with a moment of inertia $I = \frac{2}{5} m r^{2}$

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

$B e f o r e E_{T} = \frac{1}{2} I_{T} {ω_{1}}^{2}$

Where

$I_{T} = \frac{1}{2} M R^{2}$

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

$E_{G} = \frac{1}{2} I_{G} {ω_{2}}^{2}$

Where

$I_{G} = \frac{2}{5} m r^{2}$

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

$A f t e r E_{T} = \frac{1}{2} (\frac{1}{2} M R^{2}) {ω_{2}}^{2}$

The total rotational kinetic energy after is

$E_{k} = \frac{1}{2} (\frac{1}{2} M R^{2} + \frac{2}{5} m r^{2}) {ω_{2}}^{2}$

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

$∆ E_{k} = b e f o r e E_{k} - a f t e r E_{k}$

$∆ E_{k} = \frac{1}{4} M R^{2} {ω_{1}}^{2} - \frac{1}{2} (\frac{1}{2} M R^{2} + \frac{2}{5} m r^{2}) {ω_{2}}^{2}$

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 × 10⁻³kg
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

$E_{k} = \frac{1}{4} (550 \times 10^{- 3}) \times {(120 \times 10^{- 3})}^{2} \times 3 . 28^{2} - \frac{1}{2} (\frac{1}{2} (550 \times 10^{- 3}) \times {(120 \times 10^{- 3})}^{2} + \frac{2}{5} (0.009 \times {(117 \times 10^{- 3})}^{2})) \times 3 . 11^{2}$

$E_{k} = 0.0213 - 0.01938$

$E_{k} = 1.92 \times 10^{- 3} J$

Examiner 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.

Rotational Kinetic Energy (AQA A Level Physics)

Revision Note