Name: ________________________

AQA A Level Physics

Engineering Physics 3.11.1.5 to 3.11.1.6 Class: ________________________

Date: ________________________

Time: 78 minutes

Marks: 57 marks

Q1 to Q3 to be worked through with tutor. Q4 to Q6 to be

Comments:
worked through independently.

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 A student carries out an experiment to determine the moment of inertia of a turntable. The
1 diagram shows the turntable with a small lump of plasticine held above it. An optical sensor
connected to a data recorder measures the angular speed of the turntable.

The turntable is made to rotate and then it rotates freely. The lump of plasticine is dropped from a
small height above the turntable and sticks to it. Results from the experiment are as follows.

mass of plasticine = 16.0 g

radius at which plasticine sticks to the turntable = 125 mm
angular speed of turntable immediately before plasticine is dropped = 3.46 rad s–1
angular speed of turntable immediately after plasticine is dropped = 3.31 rad s–1

The student treats the plasticine as a point mass.

(a) Explain why the turntable speed decreases when the plasticine sticks to it.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

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(b) Use the results of the experiment to determine the moment of inertia of the turntable.

Give your answer to an appropriate number of significant figures.

moment of inertia ____________________ kg m2

(3)

(c) (i) Calculate the change in rotational kinetic energy of the turntable and plasticine from
the instant before the plasticine is dropped until immediately after it sticks to the
turntable.

change in kinetic energy ____________________ J

(2)

(ii) Explain the change in rotational kinetic energy.

______________________________________________________________

______________________________________________________________
(1)
(Total 9 marks)

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 (a) A playground roundabout has a moment of inertia about its vertical axis of rotation of
2
82 kg m2. Two children are standing on the roundabout which is rotating freely at
35 revolutions per minute. The children can be considered to be point masses of 39 kg and
28 kg and their distances from the centre are as shown in the figure below.

(i) Calculate the total moment of inertia of the roundabout and children about the axis of
rotation. Give your answer to an appropriate number of significant figures.

answer = ______________________ kg m2
(3)

(ii) Calculate the total rotational kinetic energy of the roundabout and children.

answer = ______________________ J
(2)

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(b) The children move closer to the centre of the roundabout so that they are both at a
distance of 0.36 m from the centre. This changes the total moment of inertia to 91 kg m2.

(i) Explain why the roundabout speeds up as the children move to the centre of the
roundabout.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________
(2)

(ii) Calculate the new angular speed of the roundabout. You may assume that the
frictional torque at the roundabout bearing is negligible.

answer = ______________________ rad s–1

(2)

(iii) Calculate the new rotational kinetic energy of the roundabout and children.

answer = ______________________ J
(1)

(c) Explain where the increase of rotational kinetic energy of the roundabout and children has
come from.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(1)
(Total 11 marks)

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 An early form of four-stroke gas engine stores kinetic energy in a large flywheel driven by the
3 crankshaft. The engine is started from rest with its load disconnected and produces a torque
which accelerates the flywheel to its off-load running speed of 90 rev min–1.

(a) The flywheel has a moment of inertia of 250 kg m2 and takes 8.0 s to accelerate from rest
to 90 rev min–1.

(i) Show that the angular acceleration of the flywheel is 1.2 rad s–2.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(ii) Calculate the average accelerating torque acting on the flywheel.

______________________________________________________________

______________________________________________________________

______________________________________________________________

(iii) Calculate the rotational kinetic energy stored in the flywheel at the end of its
acceleration.

______________________________________________________________

______________________________________________________________

______________________________________________________________

(iv) Calculate the average useful power output of the engine during this accelerating
period.

______________________________________________________________

______________________________________________________________

______________________________________________________________
(5)

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(b) When the engine is running steadily at 90 rev min–1 off-load, the gas supply is suddenly
shut off and the flywheel continues to rotate for a further 32 complete turns before coming
to rest.

(i) Estimate the average retarding torque acting on the flywheel.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(ii) Assuming that the average retarding torque on the flywheel is the same when the
engine is running, estimate the average power which the engine must develop when
running to keep the flywheel turning at 90 rev min–1 off-load.

______________________________________________________________

______________________________________________________________

______________________________________________________________
(5)
(Total 10 marks)

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 The graph below shows how the moment of inertia I of a diver performing a reverse dive varies
4 with time t from just after he has left the springboard until he enters the water.

The diver starts with his arms extended above his head (position 1), and then brings his legs
towards his chest as he rotates (position 2). After somersaulting in mid-air, he extends his arms
and legs before entering the water (position 3).

(a) Explain how moving the legs towards the chest causes the moment of inertia of the diver
about the axis of rotation to decrease.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

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(b) (i) Explain in terms of angular momentum why the angular velocity of the diver varies
during the dive.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________
(2)

(ii) Describe how the angular velocity of the diver varies throughout the dive.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________
(1)

(c) At time t = 0 the angular velocity of the diver is 4.4 rad s–1 and his moment of inertia about
the axis of rotation is 10.9 kg m2.

With reference to the graph above calculate the maximum angular velocity of the diver
during the dive.

angular velocity ____________________ rad s–1

(3)
(Total 8 marks)

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 The turntable of a microwave oven has a moment of inertia of 8.2 × 10–3 kg m2 about its vertical
5
axis of rotation.

(a) With the drive disconnected, the turntable is set spinning. Starting at an angular speed of
6.4 rad s–1 it makes 8.3 revolutions before coming to rest.

(i) Calculate the angular deceleration of the turntable, assuming that the deceleration is
uniform. State an appropriate unit for your answer.

angular deceleration ____________________ unit ___________

(4)

(ii) Calculate the magnitude of the frictional torque acting at the turntable bearings.

torque ____________________ N m
(1)

(b) The turntable drive is reconnected. A circular pie is placed centrally on the turntable. The
power input to the microwave oven is 900 W, and to cook the pie the oven is switched on
for 270 seconds. The turntable reaches its operating speed of 0.78 rad s–1 almost
immediately, and the friction torque is the same as in part (a)(ii).

(i) Calculate the work done to keep the turntable rotating for 270 s at a constant angular
speed of 0.78 rad s–1 as the pie cooks.

work done ____________________ J

(2)

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 (ii) Show that the ratio

is of the order of 105.

(2)
(Total 9 marks)

The following figure shows a motor-driven winch for raising loads on a building site. As the motor
6 turns the cable is wound around the drum, raising the load.

The drum, axle and other rotating parts have a moment of inertia about the axis of rotation of
7.4 kg m2, and the mass of the load is 85 kg. The drum has a radius of 0.088 m.

The load is accelerated uniformly from rest to a speed of 2.2 m s–1. When it is accelerating it
rises through a height of 3.5 m. It then continues at the constant speed of 2.2 m s–1.

(a) Show that the drum turns through 40 rad as the load accelerates.

(1)

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(b) Calculate the angular speed of the drum when the load is moving at 2.2 m s–1.

angular speed ____________________ rad s–1

(1)

(c) (i) Show that for the time that the load is accelerating the total increase in energy of the
load and the rotating parts is about 5400 J.

(3)

(ii) A constant frictional torque of 5.2 N m acts at the bearings of the winch.

Calculate the total work done by the motor to accelerate the load.

Give your answer to an appropriate number of significant figures.

total work done ____________________ J

(3)

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(d) Calculate the maximum power developed by the motor.

maximum power ____________________ W

(2)
(Total 10 marks)

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Mark schemes

(a) Law of conservation of angular momentum applies and I1 ω1 = I2ω2

1
OR Law of conservation of angular momentum applies and angular momentum = I ω ✓
(because no external torque acts)

Adding plasticine increases I ✓

So ω must decrease to maintain I ω constant / to conserve angular momentum ✓

3

(b) I × 3.46 =(I + 0.016 × 0.1252) × 3.31 ✓

I = 0.00552 kg m2 ✓ 3 sf ✓
Useful: mr2 = 2.5 × 10−4
Sig fig mark s an independent mark
If method correct but incorrect conversion of g to kg or mm to m,
award 1 mark out of first 2 marks
3

(c) (i) ΔE = ½I ω12 − ½(I +mr2)ω22

= [½ × 5.52 × 10−3 × 3.462] ‒
[½ × 5.77 × 10−3 × 3.312] ✓
= 1.39 × 10−3 J ✓
CE for I of turntable or I of plasticine from 2b
Answers will vary depending on rounding e.g. accept 1.43 × 10−3
2

(ii) Work done against friction / deforming plasticine as it collides with turntable / to
move or acclerate plasticine ✓
Allow heat loss on collision
Do not allow energy to sound
1
[9]

(a) (i) I = 82 + 39 × 0.902 + 28 × 0.502 (1)

2
= 120 kg m2 (1) to 2 sig figs (1)
3

(ii) ω = 35 × 2π/60 (1) = 3.7 rad s–1

E = ½I ω2 = 0.5 × 120 × 3.72 = 820 J (1)

(accept 800 to 821 J depending on sf carried through)
2

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 (b) (i) angular momentum must be conserved (1)

so if I decreases ω must increase (1)

2

(ii) 120 × 3.7 = 91 × ω2 (1)

ω2 = 4.9 rad s–1 (1)

2

(iii) E = 0.5 × 91 × 4.92 = 1100 J (1090 J) (1)

(give CE only if correct I value used)

accept 1050 – 1100 J
1

(c) work done or energy transferred as children move towards

the centre (1)

or work done as centripetal force moves inwards (1)

1
[11]

3
(a) (i) = 9.4 (rad s–1)

(= 1.18 rad s–2)(1)

(ii) T(= Iα)= 250 × 1.18 = 295 or 300 N m (1)

(iii) k.e. × 250 × 9.42 = 1.1 × 104 J (1)

(iv) power = 1.4 kW (1)

(5)

(b) (i) θ = 32 × 2π = 200(rad) (1)

α= = 0.221(rad s–2) (1)

T (= Iα) = 55 N m (1)

(ii) P (= Tω) = 55 × 9.4 = 520 W (1)

(5)
[10]

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 (a) Use of I = Σmr2 or expressed in words
4
With legs close to chest, more mass at smaller r, so I smaller
2

(b) (i) Angular momentum is conserved / must remain constant OR no external

torque acts √
WTTE

as I decreases, ω increases and vice versa to maintain I ω constant

OR as I varies, ω must vary to maintain I ω constant
2

(ii) (Angular velocity increases initially then decreases (as he straightens up

to enter the water)).
No mark for just ang. vel starts low then
increases then decreases, i.e. for
describing ω only at positions 1,2 and 3.

With one detail point e.g.

• Angular velocity when entering water is greater than at time t = 0

s.

• Angular velocity increases, decreases, increases, decreases

• Maximum angular velocity at t = 0.4 s

• Greatest rate of change of ang. vel. is near the start

• Angular velocity will vary as inverse of M of I graph

1

(c) angular. momentum = 10.9 × 4.4 = 48 (N m s)

(ωmax occurs at minimum I )

Allow 6.3 to 6.5. If out of tolerance e.g. 6.2
give AE for final answer

minimum I = 6.4 kg m2 (at 0.4 s)

6.4 × ωmax = 48 leading to

ωmax = 7.5 rad s−1

3
(Total 8 marks)

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 (a) (i) 8.3 rev = 8.3 × 2 rad (= 52 rad )
5
Use of ω22 = ω12 + 2αθ

0 = 6.42 + 2 × α × 52
If eqtn(s) of motion used correctly with θ = 8.3 (giving α = 2.5 ), give
2 out of first 3 marks.

OR use of θ = ½(ω1+ ω2)t leading to t = 16.25 s and ω 2 = ω 2 + αt

α = (−) 0.39 rad s−2

Accept: s−2
Unit mark is an independent mark
4

(ii) T = Iα

= 8.2 × 10−3 × 0.39 = 3.2 × 10−3 N m

Give CE from a i
1

(b) (i) (W = Tθ or W = Tωt) where θ = 0.78 × 270 √ (= 210 rad)

= 3.2 × 10−3 × 210 = 0.67 J

Give CE from a ii
2

(b) (ii)

CE from b i. Must be in the form: number × 105 with number

calculated correctly.
900 × 270 or 2.4(3) × 105 or equivalent must be seen for 1stmark
1 mark for only writing 3.6 × 105
2
(Total 9 marks)

6 (a) = 6.3 rev

6.3 × 2π = 39.8 rad or 40 rad ✓

OR

= 39.8 or 40 rad ✓
If correct working shown with answer 40 rad give the mark
Accept alternative route using equations of motion
1

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(b) ω = v / r = 2.2 / 0.088 = 25 rad s−1 ✓
1

(c) (i) E = ½Iω2 + ½mv2 + mgh

= (0.5 × 7.4 × 252)
+ (0.5 × 85 × 2.22)
+ (85 × 9.81 × 3.5)
= 2310 ✓
+ 206 ✓
+ 2920 ✓
( = 5440 J or 5400 J )
CE from 1b
½ I ω2 + ½mv2 = 2310 + 210 = 2520 J
½ I ω2 + mgh = 2310 + 2920 = 5230 J
½mv2 + mgh = 210 + 2920 = 3130 J
Each of these is worth 2 marks
3

(ii) Work done against friction = Tθ

= 5.2 × 40 = 210J ✓
Total work done = W = 5400 + 210
= 5600J ✓ 2 sig fig ✓
CE if used their answer to i rather than 5400J
Accept 5700 J (using 5440 J)
Sig fig mark is an independent mark
3

(d) Time of travel = distance / average speed = 3.5 / 1.1 = 3.2s ✓

Pave = = 1750 W

Pmax = Pave × 2 = 3500 W ✓

OR accelerating torque = T = W / θ
= 5600 / 40 = 140 N m ✓
P = T ωmax = 140× 25 = 3500 W ✓
CE from ii
1780 W if 5650 J used
2
[10]

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Examiner reports
The majority of students were able to see that part (a) was a question requiring the application of
1 conservation of angular momentum, even though angular momentum was not mentioned in the
text of the question. Some, however, thought the conservation of rotational kinetic energy
applied. Nearly all students scored the 1 mark for realising that the plasticine increased the
moment of inertia of the turntable.

In part (b) answers that failed to get the physics (and numbers) correct initially could only score
the third mark, and then only if they wrote any answer to 3 significant figures. Those students
who made an error in converting g to kg or mm to m (or both) lost one of the two marks for the
calculation. A common error was to use mr rather than mr2 for the moment of inertia of the
plasticine.

Part (c)(i) was reasonably well-answered. Examiners used ‘carried error’ where incorrect values
of moments of inertia for the turntable and plasticine from part (b) were used appropriately in the
calculation.

Part (c)(ii) elicited some very weak answers. Some just referred to loss of energy because the
turntable slows down, or ‘energy after is less than energy before’, or the extra mass makes it
slow down, without realising that there would be energy loss (or work done in accelerating the
plasticine) during the collision. If this had been set as a collision in a linear dynamics context,
they would probably have done better.

This question was set in the context of a rotating playground roundabout and (a) (i) was
2
answered correctly by a majority of candidates. Most correctly applied I = Σmr2 to the children but
some then forgot to add the moment of inertia of the roundabout itself. A few candidates did not
square the radii. The data in the question were given to two significant figures and a mark was
awarded for giving the final answer to two significant figures. Many candidates failed to gain
credit by not doing this.

In (a) (ii) most were able to convert 35 revolutions per minute to rad s–1 and then go on
successfully to calculate the rotational kinetic energy.

Candidates who were confident about the conservation of angular momentum wrote correct and
concise answers to (b) (i). However, many candidates answered in terms of conservation of
rotational kinetic energy, which cannot be applied here. It was very surprising to see the
conservation of angular momentum applied perfectly correctly in the calculation for (b) (ii) even
though (b) (i) had been incorrectly answered.

Part (b) (iii) was usually correctly answered.

In part (c) relatively few candidates realised that the increase in kinetic energy came from the
work done by the children in moving to the centre. The best answers from candidates also
included reference to centripetal force.

Candidates were very confident with rotational dynamics questions of this type and there were
3 many completely correct, or nearly correct, answers to both parts of the question. The only error
which occurred frequently was in part (a)(iv), where candidates used the expression P = Tω for
the power, with the wrong value of ax Those candidates who divided kinetic energy by time
invariably obtained the correct answer.

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 One very odd aspect of this question was that part (a)(iii) was very often answered to 5. 6. 7 and,
once or twice, 10 significant figures, even by candidates who were clearly meticulous in rounding
off their answers everywhere else. Excessive significant figures are now very rarely seen in this
module. Large numbers of candidates were penalised for this mistake and many of these were
also unable to give the correct unit of torque in part (iii) - N m.

This question on moment of inertia and angular momentum was set in the context of a diver
4
somersaulting during a dive. In part (a) (i) most candidates quoted I = Σmr2 but fewer were able
to explain that by bringing his legs closer to the chest, the diver altered the way the mass was
distributed about the axis of rotation. It was not enough to say that because the ‘radius’ (without
saying what they meant by radius here) had been reduced, the moment of inertia must decrease.

Part (b) (i) was generally answered well. The examiners were looking for: conservation of angular
momentum, angular momentum = Iω (or in words) and that if I varies ω must vary to keep Iω
constant.

In part (b) (ii) the question asked for a description of how the angular velocity varied throughout
the dive, so some evidence that candidates had studied the graph was needed. The most
common way of gaining the mark was to write that the angular velocity increased, decreased,
increased slightly then decreased. A significant number of candidates failed to get the mark
because they referred only to the angular velocity in positions 1, 2 and 3.

The angular momentum calculation in part (c) was correctly carried out by the majority of
candidates. One error, made by relatively few candidates, was to read the lowest moment of
inertia from the graph as 6.2 kg m2 instead of 6.4 kg m2, and this cost them a mark. A very small
number of candidates calculated an answer by mistakenly thinking that conservation of rotational
kinetic energy applied here.

Candidates were asked to apply their knowledge of rotational dynamics to the motion of a
5 turntable in a microwave oven. The calculations were generally performed well, with those who
fell at the first hurdle (not converting revolutions to radians) being penalised only once, with a
carried error given in subsequent parts provided the rest of their working was correct. In part (a)
(i), nearly all candidates gained the mark for the correct unit.

In parts (a) (ii) and (b) (i), most candidates had a good understanding of the part played by
frictional torque.

In part (b) (ii), some candidates made the mistake of dividing power by work.

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 Parts (a) and (b) were answered well by the majority of candidates, but a significant minority
6 chose a correct but very circuitous route using equations of motion for both linear and angular
motion. All that was needed in part (a) was the use of s = rθ or 2π x height raised divided by
circumference.

Most candidates were able to pick up some marks in part (c)(i) but fewer than half of the
candidates realised that there were three energies involved - gravitational potential, translational
kinetic and rotational kinetic. The energy most frequently ignored was the increase in ½mv2 of
the rising load. In part (c)(ii) all candidates had to do was calculate Tθ for the work done against
friction and add it to the answer for part (c)(i), but all too many started this part from scratch as if
it had no relation to part (c)(i) at all, or simply just left the work done against friction as the total
work done, i.e. the final answer. These candidates could score two marks, however, as one mark
was for giving an answer to two significant figures.

Two methods were possible for part (d); those who used work done divided by time taken (i.e.
average power) did not then multiply by 2 to get the maximum power. Many who used P = Tωmax
were given a carried error if their wrong answer to part (c)(ii) was used correctly.

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