Luna Park - Answers

Luna Park - Answers

Carousel Solutions

Measurements

Period using a stopwatch. Period = 25.3 s

Questions

1. As you travel around the carousel in a circular path at a constant speed, are you accelerating? Explain making reference to Newton’s laws.

Yes. Even though your speed is constant, you are constantly changing direction, so your velocity is NOT constant. Acceleration is a measure of the change in velocity, so you are accelerating.

2.What is the direction of the acceleration? You may wish to use a vector diagram to confirm this answer.

The direction of the acceleration is inwards towards the centre. The velocity vector is along the tangent. The ‘change in velocity’, v, from v1 to v2, that is, what is added to v1 to get v2 is inwards.

3. The radius of the path of the outer horses is 7.33m. Measure the period of the Carousel and calculate the speed of the outer horse.

Using v = 2r/T, v = 1.82 m/s

4.Now determine the centripetal acceleration of the outer horse.

Using a = v2/r or a = 4r/T2, a = 0.45 m/s2

Imagine you were standing on the floor of the carousel with one of the fixed vertical rods on your inside (Note: It is not advisable to do this). You hold the rod for support. The net force on you is inwards because you are travelling in a circle.

5Draw all the forces acting on you. Label each as Force by A on You, for example, your weight force is described as the Force by Earth on You.

6.What would you feel in your arm?

The arm would feel as it was being stretched.

7.If you now let go of the rod, what would you do to stay upright?

Lean inwards. This will produce an inward horizontal frictional force on your shoes by the floor. This force provides the inward acceleration.

8.If you now moved so that the rod is on your outside, draw all the forces acting on you.

9.What would you feel in your arm? The arm would feel as it was being squashed.

Twin Dragon Answers

Ride Information Provided By LunaPark

Radius, that is, the distance from the axle to seat level:12m

Length of rubber strip on the bottom of the dragon boat:10 m

Angle between the left and right boat supports: 32º

Mass of the dragon boat:2200 kg

Max swing:60º each way

Measurements Needed

The time for a single oscillation (that is, the boat’s movement from one side, to the other, and back to where it started from) when the ride is operating at its maximum. T = 7.4 s
The length of time it took for the ride to come to a halt after it began slowing down. T = 50 s

Draw a sketch-diagram of the Twin Dragon ride and label the above information appropriately on the ride.

As the ride starts from stationary, using point-form, briefly describe how the ride reaches its maximum speed. Use concepts such as “gravitational potential energy,” “kinetic energy,” “mechanical energy,” and “friction”.

There are tyres underneath that can be pushed against the rubber strip.
When the tyres are turned by the engine, they push the rubber strip backwards, making the Twin Dragon go higher. The force by the tyres on the rubber strip is a friction force, just as with care tyres on the road. This force acts on the rubber strip along its length. This force x length equals Work Done which produces a gain in Kinetic Energy. The Twin Dragon then rises against gravity and its Kinetic Energy (KE) is converted into Gravitational Potential Energy (GPE).
As the Twin Dragon swing down and up the other side it is travelling in the opposite direction, so the tyres are come off the rubber strip. Its GPE is converted to KE as it comes down and back into GPE as it goes up the other side.
When the Twin Dragon swings back, the tyres again contact the rubber strip and push the Twin Dragon again, resulting in a gain in KE and producing a greater height and more GPE.
This cycles repeats until the Twin Dragon reaches maximum speed and maximum height.

How many people are on-board the ride? 30
The weight of an average adult person is estimated by engineers as between 65 kg and 70 kg. Using the 70kg estimate, what is the estimated total mass of the boat including the occupants? Mass = 2200 + 30 x 70 = 4300 kg
While it is swinging from side-to-side, the twin dragon can be said to be undergoing wave-like motion. What name is given to the time it takes for the boat to complete one oscillation? Period
When it is operating at its maximum, time how long it takes the ride to complete a cycle. (Note: A more accurate reading would be obtained by measuring the time for 4 or 5 complete cycles and then dividing that time by the number of cycles.) T = 7.4 s
What is the frequency of the ride’s oscillation (that is, the number of oscillations or part of an oscillation per second? Frequency = 0.135 Hz

During the course of the boats pendulum movement – that is swinging from side to side – the oscillations can be thought of as consisting of Vertical Movement and Horizontal Movement.

Horizontal Movement

At what point(s) during the boat’s motion is the horizontal velocity 0 ms-1 ?At either end of the swing
At what point(s) during the pendulum motion is the horizontal velocity a maximum? At the bottom of the swing

Vertical Movement

At what point(s) during the boat’s motion is the vertical velocity 0 ms-1 ?Both at the bottom of the swing and either end of the swing
At what point(s) during the boat’s motion is the vertical velocity a maximum? On the way up just after the tyres pull away from the rubber strip

Momentum

At what point in the boat’s swinging movement is horizontal momentum a maximum? At the bottom of the swing, when the velocity is a maximum.
At what point in the boat’s swinging movement is horizontal momentum 0 kgms-1 ?At either end of the swing
When is vertical momentum 0 kgms-1 ?Both at the bottom of the swing and either end of the swing.
When is vertical momentum at a maximum? On the way up just after the tyres pull away from the rubber strip

The

Coney

Island

Top

Drop

Data: Mass of 12 person carriage = 1150 kg. Include units in your answers.

1. On the diagrams below, draw a vector (when needed) to represent the direction of the net force on a rider in the chair.

a) Stationary at X at the topb) in free fall between X and Y c) braking between Y and Z

Free fall 3.5 m

a = 9.8 m/s2

Braking 0.8m

2.Now on each of the diagrams draw in a vector for the weight force (when needed), labeled F by Earth on rider

3.Now on each of the diagrams draw in a vector for the reaction force by the chair on the rider (when needed), labeled F by chair on rider. Consider the size as well as the direction of this force.

4. Energy

(a) How much work is done by the cables to raise the chair and riders from Z to X?

PE = mgh = 1150 x 9.8 x (0.8 + 3.5) = 4.85 x 104 J

(b) What is the decrease in Gravitational Potential Energy of the carriage when full (assume your body mass) as it falls from position X to Y?

PE = mgh = 750 x 9.8 x 3.5 = 3.94 x 104 J

Gain in KE = Loss in PE = 3.94 x 104 J

(d) Use this to find out how fast the carriage and its occupants are falling at the end of free fall, Y?



5. Kinematics and Dynamics

(a) Use your answer from 4(d) to calculate how long you are in free fall from stationary at X to the point Y, 3.5 m lower:

v = u + at  t = (v-u)/a = (8.3 – 0)/ 9.8 = 0.85 sec

(b) From Y to Z the seat decelerates under braking from the speed calculated in 2(d) to zero. Assume this deceleration is constant, calculate its size:

v2 – u2 = 2ax  a = (v2-u2)/2x = (8.32 – 02)/(2 x 0.8) = 42.9m/s2

(c) How many “gees” is this acceleration in 5b that the riders experience? (That is, how many times is this greater than the acceleration due to gravity, 9.8 m/s2).

No.of g’s=42.9m/s2 / 9.8 = 4.4

(d) What force does the cushioning have to supply to bring the carriage to rest over this distance?

F = ma = 750 x 42.9 = 3300 N

(e)You “feel” your weight through the reaction force acting on you, i.e. the force by chair on rider. This is called your apparent weight. Determine your apparent weight at

i) Position Xii) In free falliii) During deceleration from Y to Z

(i) W = N = mg = 80kg x 9.8 = 784N for an 80kg person, here Wapp = WReal

(ii) In free fall, N = 0 and Wapp= 0 (you “feel” weightless). Your real weight is still = m x g

(iii)F = N – W  N = F + W = ma + mg = 80 x 42.9 + 80 x 9.8 = 7,350 N (This is 7,350 / 784 = 9.4 times normal body weight).

Silly Serpent Answers

Drawings

1) Draw the shape of the track

a)Label the points on the track where you would experience;

b)The highest speed

c)The highest acceleration

d)The highest gravitational potential energy

e)The highest kinetic energy

2)Draw a diagram of you sitting in a carriage at the highest point on the track during the ride. Draw in all the forces acting on you at this point. . Label the forces as “Force by A on You”.

Weight down (F by Earth on You) &

Normal reaction up (F by Seat on You)forward push from the seat (F by seat on You)

3.On the drawing of the first two cars below draw in the forces acting on the Serpent car as the ride starts off.

4.Using another coloured (red) pen draw in the net force acting on a passenger in the carriage at this point.

Estimations

Your mass 70 kg. The mass of the empty carriage 700 kg

Height of rail at the lowest point 0.50 m. Maximum height of the rail 4.0 m

Number of curves 2x180 and 1x 360 The radius of each curve 2x5m and 1x3m.

Distance traveled by the Serpent in one second about 3.0 m.

The total circumference of each curve added to give the total distance around the track ½ (2  5)x2 + 2x3 + three straight lengths 5m + 6m + 8m= 69m m.

Measure the time for one rotation of the serpent = 15seconds.

Stopping distance = 4.0 m. Stopping time = 2.0 s.

Calculations (include units)

1) Using your estimation for the total distance around the track calculate your average velocity as you complete one circuit.

V= d/t = 69/15 = 4.6 m/s

2) Using the average velocity and the radius for each curve, calculate the centripetal acceleration on you.

a= v2/r = 4.6x4.6/5 = 4.m/s/s and the 360 curve a = v2/r = 4.6 x 4.6 /3= 7m/s/s

3) Using the formula for gravitational potential energy, work out your change in potential energy as the serpent ride lifts you from the lowest point to the highest point.

PE = mgh = 70 x 9.8 x (4- 0.5)= 2.4x103 Joules

What is your total energy at this point in the ride?

Energy = PE + KE = 2.4 x103 + 1/2mv2 = 2.4 x103 + ½70x4.62 = 3.1 x103 J

4) Count the maximum possible number of passengers and the number of carriages . Calculate the maximum total mass of passengers when the ride is full. Use this with the average velocity to calculate the maximum momentum of the Serpent ride.

20 people with an average mass of say 50kg (more little kids than adults normally momentum = mass x velocity = 20 x 50 x 4.6 = 4.6 x103 kg m/s. this is the momentum for the passengers. There are 5 carriages with a mass of 2x 500kg and 3 x 300kg Carriage momentum = Mv = 1900 x 4.6 = 8.7 x103

Total momentum 8.7 x103+ 4.6 x103 = 1.3 x104kgm/s

5) When the ride stops what happens to this momentum? transferred to the Earth

6) Calculate the braking force required to stop the whole ride.

change in momentum is 1.3 x104 = Impulse

= Force x time = Force = impulse /time = 1.3 x104 /2seconds = 6.7 x103 Newtons

7) Work out the total kinetic energy of the serpent ride before it stops.

KE = ½ mv2 = 0.5x2900 x4.6x 4.6 = 3.1 x104 Joules

8) What happens to this energy when the Serpent stops?

Kinetic energy changed into heat & sound

Pharoah’s Curse: Answers

Equipment needed: stopwatch, calculator

The Pharoah’s Curse has two gondolas for passengers which swing in opposite directions. As the gondolas swing backwards and forwards, the angle increases each time until the gondolas go over the top. The length of the arms is 9.0 metres.

Full rotation

1.When the gondolas reach full rotation, measure the time for one rotation, calculate the average speed of rotation. (v=2r/T)T = 8.2 sec, speed = 6.9 m/s

2.Assuming this speed is constant, what is the size of the centripetal acceleration?

(a = v2/r, a = 42r/T2) a= 5.8 m/s2

3.Draw, on the figure of you in the diagram, arrows to represent:

a) the centripetal acceleration, a

b) your weight force (labelled as Force by Earth on You FEarth on You) and

c) the reaction force by the seat on you

(labelled as FSeat on You).

Consider both the size and direction of these two forces.

Gondola at the bottom of its swing

4.Estimate the kinetic energy of each gondola as it rotates in a full rotation.? The mass of each gondola is 2100kg, estimate the total mass of the passengers.

Mass with passengers = 2100 + 8 x 70 = 2660 kg. KE = ½ x 2660 x 6.92 = 63kJ

5.Looking at your answer to Q’n 4, can you suggest why the drive mechanism is hydraulic rather than mechanical? Consider the effect of gears and sudden changes in speed or power loss. A crunching of gears would lead to a massive deceleration with drastic consequences for the structure and the occupants. With an hydraulic operation the oil can absorb sudden changes and slow down the impact on the structure.

Feeling Lighter at the Top

These three diagrams show the forces acting on you at the top for three different speeds.

The Net F = ma equation is also included as well as a description of what you would experience.

Force Diagrams

Very fast, VFast, vSlow, v

Net F = ma Equations

N: Force by seat on you, down:No force from seatN: Force by restraint on you, up

N + mg = Net Force = mV2/rN = 0, mg = mv2/rmg – N = m v2/r

You would feel …

Experiences

… pressure on your backside weightless, floating pressure on your shoulders

from the seatfrom the restraint

6.Determine the speed of the gondola at the top if you were to feel weightless?

mg = mv2/r, g = v2/r, v = square root (gr) = speed = 9.4 m/s

Stationary at the top

7.Why don’t the passengers fall out of the gondola at the top when it is at rest? The bar across their thighs exerts an upward force to counter the weight force

8.For the position when the gondola is stationary at the top, draw, on the figure of you in the diagram, arrows to represent:

a) your acceleration, a (accel’n = zero)

b) your weight force (labelled as Force by Earth on You)

FEarth on You) and

c) the reaction force by the gondola on you (labelled as FGondola on You).

If all the gondola’s gravitational potential energy at the top was converted into kinetic energy at the bottom, how fast would the gondola be travelling at its lowest point? (radius = 9.0 m)

mgh = ½ mv2, v = sq root (2gh) = sq root (2 x 9.8 x 2 x 9.0) = 18.8 m/s

10.Determine the centripetal acceleration at this lowest point. 39.2 m/s2

11.Determine the size of the net force that would be on you at this point (estimate your mass).

F = ma = 2940 N on 75 kg person

12.At this point, the Reaction force (N) – weight force (mg) = Net force = ma, N – mg = ma. Determine the size of the reaction force. How many times greater than your weight is this?

Reaction Force = 2940 + 75 x 10 = 3690 N. 3690 / 750 = 4.9 times

The reaction force is what you feel, the ratio indicates how many times heavier you feel. At about a ratio of 4, pilots experience reduced vision leading to blackout. For this reason there is a hydraulic drag as the gondolas come down from the stationary position at the top.

Measure the time from stopped at the top to the bottom. Calculate the average speed. Assume the speed at the bottom is twice the average. How does this compare with Q’n9.

Time = 8.2 s, speed = dist / time = r/time = 3.4 m/s, which is much slower.

Dodgem Cars

Newton’s First Law

When two dodgem cars collide, the car may stop or change direction after colliding, but the drivers’ body will continue in their initial direction.

1.Describe your experiences and feelings when you collide:

a)Head-on with an oncoming dodgem car

You feel as if you are “thrown” forward.

b)Head-on with a stationary dodgem car

You feel as if you are “thrown” forward but with not as much force, since stationary car will then move with you in your initial direction

c)Head-on with a guardrail

You feel as if you were “thrown” forward and you may experience/feel a greater force. The guardrail is rigid and does not “move” as easily as an oncoming or stationary dodgem car.

d)at an angle with a dodgem car

You feel as if you are “thrown” forward and you should move off at an angle after the collision. You may experience a “turning” feeling or torque.

Investigating Impacts

2.Determine the approximate maximum speed of the car, by measuring the time to travel an estimated distance.

Avg. speed, v (m/s) = distance (m) / time (s) =5.0 / 3.0 = 1.7 m/s

3.The car has a mass of 600 kg. Calculate its momentum and kinetic energy at maximum speed.

Momentum (kgm/s): p = mv where v is avg. speed from above

and m = mass of car (600kg) + mass of driver = 600 + 70 = 670 kg

KE max (J) = ½ mv2 = ½ x 670 x 1.72 = 930 J

4.Observe the car crashing into the guard rail at maximum speed. Estimate how much the bumper bar on the car is compressed. scar = 4.0 cm = 0.040 m

5.The car comes to rest after the impact. Use the data in Questions 2 and 4 to determine the acceleration of the car and the duration of the impact.

Work = Fs = change in KE and F=ma

F.s = ma.s = ½ mv2 = ½ mu2, where v = 0 and u = 1.7 m/s

So magnitude of ma.s = ½ mu2 and a = u2/2s. Find a. a = 1.72 / (2 x 0.04) = 35 m/s2

Impulse = change in momentum, F.t = p. So, t = m(v - u)/ma. Simplifying t = u/a. Find t. t = 1.7 / 35 = 0.05 sec.

6.After the impact the car’s momentum and kinetic energy are zero. Where have they gone?

Momentum: Momentum is always conserved. Guardrail/Earth has moved.

Kinetic Energy: KE lost to heat, sound, in crumpling the car and guardrail.

7.List the features of the car design intended to protect the driver in an impact. For each feature, use physics concepts to explain how it works.

Seat belt: Increases the duration of slowing down and so decreases the force.

Rubber bumper bar: Increases the distance of slowing down and so decreases the force.

What makes the cars move?

8.Observe the nature of the floor and the footwear of the operator and suggest a mechanism for this ride?