IGCSE Physics · guided topic map

Rotational mechanics for Cambridge IGCSE Physics

Rotational mechanics for IGCSE Physics, organized into 2 syllabus topics and 2 mapped concept guides.

Syllabus topics: 2
Mapped concept guides: 2
Educational level: Cambridge IGCSE Core and Extended

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1.5.2

Turning effect of forces

Motion, forces and energy

1 guide

01Turning effect of forcesMapped lesson

1.5.3

Centre of gravity

Motion, forces and energy

1 guide

01Centre of gravityMapped lesson

Diagrams

Rotational mechanics as IGCSE Physics draws it

The figures from the IGCSE Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 4.1Turning effect of forcesIGCSE

Figure comment

Fig. 4.1A uniform beam rests on a pivot standing on the ground, the pivot placed at the centre of the beam. A 4.0 N weight hangs from the beam at a point 0.80 m to the left of the pivot, and an 8.0 N weight hangs to the right of the pivot at a distance marked d. Both distances are measured along the beam from the pivot. The drawing is not to scale.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

1.5.2 Turning effect of forces Open IGCSE question · Motion, forces and energy Q4

Guided questions 5 parts

Reading cue. Both distances are measured from the pivot, not from the ends of the beam, and the pivot is at the centre of a uniform beam, so the beam's own weight pulls straight down through it.

aIdentify Identify which of the two weights in Fig. 4.1 hangs the nearer to the pivot when the beam balances, and state the reason.
recall2 marks
Check answer 2 marks
1. the 8.0 N weight (1)
2. it is the larger force, so it needs the smaller distance from the pivot to give the same moment (1)
bCalculate Calculate the moment of the 4.0 N weight about the pivot in Fig. 4.1. Give the unit.
routine2 marks
Check answer 2 marks
1. moment = force x perpendicular distance from pivot = 4.0 × 0.80 (1)
2. = 3.2 N m (1)
cDetermine The 8.0 N weight in Fig. 4.1 is taken off and a 12 N weight is hung in its place, 0.30 m from the pivot, with the 4.0 N weight left where the figure shows it. Determine whether the beam still balances, and if it does not, state which way it turns.
demanding3 marks
Check answer 3 marks
1. anticlockwise moment of the 4.0 N weight = 4.0 × 0.80 = 3.2 N m (1)
2. clockwise moment of the 12 N weight = 12 × 0.30 = 3.6 N m (1)
3. the two moments are not equal, so the beam does not balance; the clockwise moment is the larger, so the beam turns clockwise, the 12 N end going down (1)
dExplain The beam is replaced by one of the same length that is thicker, and therefore heavier, towards its left-hand end. It rests on the same pivot at its mid-point and carries only the two weights of Fig. 4.1, the 4.0 N weight still 0.80 m from the pivot. Explain why the 8.0 N weight must now hang further from the pivot than it does in Fig. 4.1.
top of the paper4 marks
Check answer 4 marks
1. the centre of mass of the new beam lies to the left of the mid-point, and so to the left of the pivot (1)
2. the weight of the beam therefore has a turning effect about the pivot, anticlockwise, where in Fig. 4.1 it acted through the pivot and had none (1)
3. the total anticlockwise moment is now greater than 3.2 N m (1)
4. the clockwise moment must rise to match it and the force is still 8.0 N, so the distance from the pivot must increase (1)

Transfer challenge

The jib of a crane carries a load of 2000 N at a horizontal distance of 6.0 m from the tower. A counterweight hangs from the other side of the jib, 4.0 m from the tower. Calculate the counterweight needed for the jib to balance, and state one assumption you have made.

Check answer 4 marks

moment of load about the tower = 2000 × 6.0 = 12 000 N m (1)
counterweight = 12 000 / 4.0 (1)
= 3000 N (1)
assumption, any one: the jib itself is uniform so its weight acts through the tower; the distances given are horizontal distances (1)