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Option C: The Physics of Sports

Rotational energy and energy conservation in sport

  • 3 things to know
  • A2 Unit 4 Section B

Overview

Recall questions on Rotational energy and energy conservation in sport.

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Answer each one before you reveal it.

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Key term

Explain why a pole vaulter does not clear a bar at exactly the height their run-up speed would predict.

Key term

Show that a rolling solid sphere carries two sevenths of its kinetic energy as rotation.

Key term

What fraction of the kinetic energy of a rolling thin spherical shell is rotational, given I = ⅔mr²?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Write the expression for the total kinetic energy of a ball rolling without slipping, and state the relationship between its linear and angular speed.

    Model answer

    The total kinetic energy is ½mv² + ½Iω², where m is the mass in kg, v the speed of the centre in m s⁻¹, I the moment of inertia in kg m² and ω the angular velocity in rad s⁻¹. Rolling without slipping means v = rω, where r is the radius in m.

  2. 2

    Show that a rolling solid sphere carries two sevenths of its kinetic energy as rotation.

    Model answer

    Substituting I = ⅖mr² and ω = v/r into ½Iω² gives ⅕mv², so the total is ½mv² + ⅕mv² = 0.7mv². The rotational share is 0.2 divided by 0.7, giving two sevenths or 28.6%.

  3. 3

    What fraction of the kinetic energy of a rolling thin spherical shell is rotational, given I = ⅔mr²?

    Model answer

    The rotational term becomes ⅓mv², so the total is ½mv² + ⅓mv² = ⅚mv². The rotational fraction is ⅓ divided by ⅚, giving 0.40 or 40%.

  4. 4

    A solid ball rolls from rest down a slope of vertical height 1.2 m without slipping. Calculate its speed at the bottom, taking g as 9.81 m s⁻².

    Model answer

    Energy conservation gives mgh = 0.7mv², so v = √(9.81 × 1.2 ÷ 0.7) = √16.8 = 4.10 m s⁻¹. Substituting back, 0.7 × 4.10² = 11.8 J per kilogram. That is equal to gh, since 9.81 × 1.2 is also 11.8.

  5. 5

    A hollow ball behaving as a thin spherical shell rolls from rest down a slope of height 1.2 m. Calculate its speed at the bottom, taking g as 9.81 m s⁻².

    Model answer

    Energy conservation gives mgh = ⅚mv², so v = √(9.81 × 1.2 × 6 ÷ 5) = √14.1 = 3.76 m s⁻¹. Substituting back, ⅚ × 3.76² = 11.8 J per kilogram. That is equal to gh for this slope.

  6. 6

    A solid ball and a hollow ball of the same mass and radius are released together at the top of a slope. Which reaches the bottom first, and why?

    Model answer

    The solid ball wins, because it has the smaller moment of inertia and so gives up a smaller share of the released potential energy to spinning. More of that energy is left as translational kinetic energy, so its centre moves faster all the way down.

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