IB Physics HL · guided topic map
Rotational mechanics for IB Physics HL
Rotational mechanics for IB Physics HL, organized into 1 syllabus topic and 5 mapped concept guides.
- Syllabus topics
- 1
- Mapped concept guides
- 5
- Educational level
- IB Diploma Physics Higher Level
Syllabus to lesson
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A.4Rigid body mechanics
Space, time and motion
5 guides+
Rigid body mechanics
Space, time and motion
- 01Torque and rotational dynamicsHL only
- 02Moment of inertiaHL only
- 03Rotational work, energy, and powerHL only
- 04Angular momentumHL only
- 05Angular-momentum conservationHL only
Diagrams
Rotational mechanics as IB Physics HL draws it
The figures from the IB Physics HL practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.
01Figure 5Rigid body mechanics · Galilean and special relativityIB
Figure comment
Figure 5Side view. A straight hatched slope runs down from the upper left to a horizontal surface at the lower right. A cylinder is drawn end-on resting on the slope near the top, labelled solid cylinder, 2.0 kg, released from rest, with a line from its centre to the rim labelled radius 0.15 m. A dashed horizontal line runs to the right from the level of the cylinder's centre, and a dimension line marks 1.2 m between that level and the horizontal surface at the bottom of the slope.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. The 1.2 m is measured from the dashed line through the axis, so it is the drop of the centre of mass and not the length of the slope; 0.15 m is a radius, not a diameter.
aState State the relationship between the translational speed of the cylinder's centre and its angular speed while it rolls without slipping, using the dimension marked on the drawing.
Check answer 2 marks
- v = ωR
- with R = 0.15 m from the figure, v = 0.15ω
bDetermine The cylinder reaches the bottom of the slope with a translational speed of 3.96 m s⁻¹. Determine its angular speed there.
Check answer 3 marks
- rolling without slipping, so ω = v / R
- ω = 3.96 / 0.15
- ω = 26 rad s⁻¹
cDetermine The slope is inclined at 25° to the horizontal. Determine the acceleration of the cylinder's centre down the slope, and the time it takes to reach the bottom from rest.
Check answer 4 marks
- for a solid cylinder a = g sin θ / (1 + I/MR²) = g sin θ / 1.5 = (2/3)g sin θ
- a = (2/3) × 9.81 × sin 25° = 2.76 m s⁻²
- the distance travelled is along the slope, 1.2 / sin 25° = 2.84 m, not 1.2 m
- t = √(2 × 2.84 / 2.76) = 1.43 s, which checks against v = at = 2.76 × 1.43 = 3.96 m s⁻¹
dShow (that) Friction is the only force exerting a torque about the axis. Show that the coefficient of static friction must be at least (tan θ)/3 for the cylinder to roll without slipping, and evaluate this for the 25° slope.
Check answer 4 marks
- torque about the axis: fR = Iα = ½MR²(a/R), so f = ½Ma
- substituting a = (2/3)g sin θ gives f = (1/3)Mg sin θ
- the normal force is N = Mg cos θ, so μ_min = f/N = (tan θ)/3
- for θ = 25°, μ_min = 0.466 / 3 = 0.16
Transfer challenge
A block of mass 1.5 kg hangs from a light string wound round the rim of a solid cylindrical pulley of mass 2.0 kg and radius 0.15 m, free to turn about a fixed horizontal axis. Determine the acceleration of the block and the tension in the string.
Check answer 4 marks
- pulley: TR = Iα = ½MR²(a/R), so T = ½Ma = 1.0a
- block: mg − T = ma, so 1.5 × 9.81 − 1.0a = 1.5a
- a = 14.7 / 2.5 = 5.9 m s⁻², and T = 1.0 × 5.9 = 5.9 N
- the acceleration is below g because part of the released potential energy goes into spinning the pulley, exactly as it went into spinning the cylinder on the slope