Download Metropolissolutions

Luna Park
Physics
This ride uses cables to initially
haul two connected 4-person carts
up a track then allows gravity to
take over its motion. The riders
experience passages of free fall and
both linear and centripetal
accelerations.
DATA
Maximum height = 13.5m
Radius of spiral = 7.0m
g = 9.8 m/s2
Mass of carriages carrying 8 people
= 1500 kg (approx)
1. The track is approximately 500m long. Time one lap.
Distance = 500m
Time = 120 sec.
Average speed of carts = 4.2 m/s = 15 km/h
2. Initially the carriages are towed up a steep incline from a height of 0.50m to a height of
13.5m.
(a) How much work is done on the carriages in towing the 1500 kg it up this incline?
Work = gain in P.E. = mgh = 1500 x 9.8 x 13
1.91 x 105 J
(b) This takes the engine 16 seconds. What power is required to do this?
P = Energy / time = 1.91 x 105 J / 16sec = 11,940W
11.9 kW
(c) If the engine is only 60% efficient, what is the engine’s power output?
0.60 x Output = 11.9 => Output = 11.9kW / 0.6
-1-
19.9 kW
3. The first spiral section:
Data:
Distance along track A to B
= 90m
A
Vertical drop A to B = 5.2m
Speed A = 7.0 m/s
Speed B = 10 m/s
Radius = 7.0 m
B
Track banked at 380
(a) What is the carriages’ decrease in P.E. from A to B as it drops 5.2m in height?
PE = mgh = 1500 x 9.8 x 5.2 = 76,440
7.64 x 104 J
(b) Assuming no losses, what is the carriages’ increase in K.E. in dropping from A to B?
7.64 x 104 J
Gain in K.E. = Loss in P.E.
(c) Use the values of velocity at A and B in the data to calculate the real increase in K.E
from A to B:
38,250 J
KE = KEB – KEA = ½ x 1500 x (102 – 72) = 38,250
(d) How much K.E. has been lost? What form of energy has it been converted to?
KE lost is converted to heat by friction.
Loss = Theoretical KE gain – actual KE gain = 76,400J – 38,250J
3.82 x 104 J
(e) How much work is done by the friction force of the track as the carriages move from
A to B?
3.82 x 104 J
(f) Work done = Force x Distance. What is the size of the friction force between the
track and the carriages over this 90m section of track?
Friction Force = Work done / displacement = 3.82 x 104 J / 90m
(g) At point B, calculate the size of the your centripetal acceleration:
acent. = v2/R = 10.02/7.0
-2-
420
14.3
N
m/s2
(h) Divide this value by 9.8 to calculate how many “g’s” this is: (that is, how many times
the normal acceleration due to gravity you experience here).
1.46
g’s
14.3 / 9.8 = 1.46
(therefore you would “feel” about 1 ½ times as heavy)
(i) The track in this section is banked, what is the advantage of doing this?
The normal reaction force now has a component
directed towards the centre of the circle that supplies
the centripetal force necessary to force the carriage
into circular motion.
(j) Calculate the size of the forces at B:
(i) The vertical component of the track’s normal
reaction force on the carriages’ NV
NV = W = mg = 1500 x 9.8 = 14,700N
NH
N
NV
(ii) The horizontal component of this force, NH:
NH = NV tan 380 = 14,700 tan380
= 11,490N
380
380
(iii) The total reaction force N of the track on the carriage:
NTotal = (14,7002 + 11,4852)1/2 = 18,650N
W
FCar. on Track
(iv) What is the magnitude of the force that the carriages’ apply to the track.
Draw this vector on the diagram and label it FCarriage. on Track
FCarriage. on Track = FTrack on Carriage = 18,650N
4. The second drop (the largest):
X (vx = 3.0 m/s)
DATA
Speed at X = 3.0m/s
Vertical drop = 9.9 m
Length of track from
X to Y = 20.5m
12.5m
Y
2.6m
-3-
(a) Assuming no losses, what should be the total K.E. of the carriages at the point Y?
KEY = KEX + PE = ½ x 1500 x 32 + 1500 x 9.8 x (12.5 – 2.6)
1.52 x 105
J
(b) Use your answer for the track’s friction force from 3(f) to calculate the energy lost
due to friction from X to Y (20.5 m):
8,610 J
Energy lost = friction x distance = 420N x 20.5m
(c) Taking this loss into account, what is the total K.E. of the carriages at Y?
1.5228 x 105 – 8,610 = 143,670J
1.44 x 105 J
(d) Use this value to calculate your speed at point Y: How does this value compare with
the measured value of about 50km/h?
½ x 1500 x v2 = 143,670
v
13.8 m/s
2  143,670
1500
49.8 km/h
The radius of the circular dip at Y is 12m.
(e) What is the magnitude of the centripetal
acceleration on the carriages at this point?
N
12m
ac = v2 / R = 13.82 / 12 = 15.9 m/s2
F
(f) What is the magnitude of the Net force, F,
on the carriage’s at Y?
F = m x ac = 1500kg x 15.9m/s2 = 2.38 x 104
v
Y
W
(g) Calculate the magnitude of the Reaction force, N, on the carriages at Y:
F = N – W, =>
N = F + W = 2.38 x 104 + 1500 x 9.8 = 38,500N
(h) For the 1500kg carriages on the diagram above draw vectors for:
(i) The weight force W
(ii) The Normal Reaction force N of the track on the carriages
(iii) The Net force, F = FCENTRIPETAL of the track on the carriages = N - W
(iv) The force of the carriages on the track FCart. on Track
-4-