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AP · AP C:M · free response

AP Physics C: Mechanics · Question 7

AP Physics C: Mechanics · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
discriminating
Marks
13
Topics
2
Answer
Complete
A uniform rod pivoted at one end, held horizontal before releasepivotuniform rod, mass MLrod when vertical
Fig. 7.1A uniform rod of mass M is held horizontal. Its left-hand end is carried on a hinge at the foot of a short post fixed to a hatched ceiling, and the word "pivot" labels that end. A dimension line below the rod, with a tick at each end, runs from the pivot to the free end and is labelled L. Directly below the pivot a dashed outline of the rod shows the position it will occupy when vertical, and a dashed arc from the free end down to that position, carrying a small arrow, shows the rod swinging down. No forces are marked on the rod.
free response13 marks

A uniform thin rod of mass M and length L is free to rotate in a vertical plane about a frictionless horizontal pivot at one end. The rod is held horizontal and released from rest.

  1. (a)

    Derive Using integration, derive the moment of inertia of the rod about the pivot. Show the mass element you use.

    4 marks
  2. (b)

    Determine Determine the angular acceleration of the rod at the instant it is released, while it is still horizontal.

    3 marks
  3. (c)

    Derive Derive an expression for the angular speed of the rod when it reaches the vertical position.

    3 marks
  4. (d)

    Explain Determine the linear acceleration of the free end of the rod at the instant of release, and explain how a point on the rod can have a downward acceleration greater than g.

    3 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overviewKey answer: 1 point: takes a mass element dm = (M/L)dx at distance x from the pivot 1 point: writes I = ∫x² dm = ∫₀^L x²(M/L)dx 1 point: evaluates the integral to (M/L)(L³/3)
01

(a)

4 marks

Derive Using integration, derive the moment of inertia of the rod about the pivot. Show the mass element you use.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: takes a mass element dm = (M/L)dx at distance x from the pivot

  2. 2

    1 point: writes I = ∫x² dm = ∫₀^L x²(M/L)dx

  3. 3

    1 point: evaluates the integral to (M/L)(L³/3)

  4. 4

    1 point: I = ML²/3

02

(b)

3 marks

Determine Determine the angular acceleration of the rod at the instant it is released, while it is still horizontal.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: the weight acts at the centre of mass, a distance L/2 from the pivot, giving τ = MgL/2

  2. 2

    1 point: α = τ/I = (MgL/2)/(ML²/3)

  3. 3

    1 point: α = 3g/(2L)

03

(c)

3 marks

Derive Derive an expression for the angular speed of the rod when it reaches the vertical position.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: the centre of mass falls a distance L/2, so the energy released is MgL/2

  2. 2

    1 point: sets MgL/2 = ½Iω² with I = ML²/3

  3. 3

    1 point: ω = √(3g/L)

04

(d)

3 marks

Explain Determine the linear acceleration of the free end of the rod at the instant of release, and explain how a point on the rod can have a downward acceleration greater than g.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: a = αL = 3g/2, which is 1.5g

  2. 2

    1 point: the rod is a rigid body, not a free particle — the pivot exerts a force on it, so the net force on the rod is not simply its weight

  3. 3

    1 point: the pivot pushes down on the rod (or, equivalently, the inner part of the rod is accelerating more slowly than g and the rod's rigidity transmits this as a downward force on the outer part), allowing the far end to accelerate faster than free fall

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