Calculations
Conversion 1:
In this conversion, the ball is moving down the first ramp and accelerating.
Figure 1: First ramp with dimensions.
x = 5.52 in sinØ = y/h = 0.5/5.5 h2 = x2 + y2
y = 0.50 in sinØ = 0.0909 x = sqrt(h2 – y2)
h = 5.50 in Ø = sin-1(0.0909) x = sqrt(5.52 – 0.52)
Ø = 5.22o x = 5.52 in
100 marbles = 0.54 kg è 1 marble = 0.0054 kg = 00119 lb
· assume f (friction) is zero.
energy: (from top of counter)
Figure 2: FBD of ball (top), first ramp with dimensions from counter (bottom).
h0 = 24.5 in * (1 ft / 12 in) = 2.04 ft * (0.3048 m / 1 ft) = 0.6218 m
hf = 24 in * (1 ft / 12 in) = 2.00 ft * (0.3048 m / 1 ft) = 0.6096 m
PEi + KEi = PEf + KEf
mgh0 + 0 = mghf + 0.5mvf2
(9.81 m/s2)(0.6218 m) = (9.81 m/s2)(0.6096 m) + 0.5vf2
vf = sqrt(0.2394 m2/s2)
vf = 0.489 m/s
- The ball transforms some of its initial energy (all in potential form) into kinetic energy.
Ei = PEi = mgh0 = (0.0054 kg)(9.81 m/s2)(0.6218 m)
= 0.0329 J = 32.9 mJ
PEf = mghf = (0.0054 kg)(9.81 m/s2)(0.6096 m)
= 0.0323 J = 32.3 mJ
KEf = 0.5mvf2 = 0.5(0.0054 kg)(0.489 m/s)2
= 6.46 x 10-4 J = 646 uJ
Ef = PEf + KEf = 32.3 mJ + 646 uJ = 32.9 mJ = PEi
Conversion 2:
In this conversion, the ball is moving down the second ramp and accelerating.
Figure 3: Second ramp with dimensions.
x = 6.98 in sinØ = y/h = 0.5/7.0 h2 = x2 + y2
y = 0.50 in sinØ = 0.0714 x = sqrt(h2 – y2)
h = 7.00 in Ø = sin-1(0.0714) x = sqrt(7.02 – 0.52)
Ø = 4.10o x = 6.98 in
m = 0.0054 kg = 0.0119 lb
* Assume f (friction) is zero. Also, since the ramps are angled as shown in the figure, the marble comes to a stop before rolling down the second ramp, therefore, the velocities from the first ramp are not involved in this conversion.
Figure 4: FBD of ball (top), second ramp with dimensions (bottom).
energy: (from top of counter)
h0 = 23.5 in * (1 ft / 12 in) = 1.958 ft * (0.3048 m / 1 ft) = 0.5969 m
hf = 23 in * (1 ft / 12 in) = 1.917 ft * (0.3048 m / 1 ft) = 0.5842 m
PEi + KEi = PEf + KEf
mgh0 + 0 = mghf + 0.5mvf2
(9.81 m/s2)(0.5969 m) = (9.81 m/s2)(0.5842 m) + 0.5vf2
vf = sqrt(0.2492 m2/s2)
vf = 0.499 m/s
- The ball transforms some of its initial energy (all in potential form) into kinetic energy.
Ei = PEi = mgh0 = (0.0054 kg)(9.81 m/s2)(0.5969 m)
= 0.0316 J = 31.6 mJ
PEf = mghf = (0.0054 kg)(9.81 m/s2)(0.5848 m)
= 0.0309 J = 30.9 mJ
KEf = 0.5mvf2 = 0.5(0.0054 kg)(0.499 m/s)2
= 6.72 x 10-4 J = 672 uJ
Ef = PEf + KEf = 31.6 mJ + 672 uJ = 31.6 mJ = PEi
Conversion 3:
In this conversion, the marble falls in the cup, dropping the cup, which lifts another cup containing a large marble via a pulley.
Figure 5: Pulley setup with dimensions.
msmall = 0.0054 kg = 0.0119 lb
mlarge = 2.5 * msmall = 0.0135 kg = 0.0298 lb
A: marble falls w/ initial velocity into the cup
Figure 6: Diagrams for ball, FBD of ball, and pulley setup with dimensions.
Vx = VsinØ = (0.499 m/s)sin(4.10o) = 0.0357 m/s V = 0.499 m/s
Ø = 4.10o
Vy = VcosØ = (0.499 m/s)cos(4.10o) = 0.498 m/s from before
h0 = 23 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.5848 m
hf = 21 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.5334 m
energy: (from top of counter) * as before, some energy is converted
into kinetic energy
PEi + KEi = PEf + KEf
mgh0 + 0.5mvf2 = mghf + 0.5mvf2
(9.81 m/s2)(0.5848 m) + 0.5(0.498 m/s)2 = (9.81 m/s2)(0.5334 m) + 0.5vf2
5.731 m2/s2 + 0.124 m2/s2 = 5.233 m2/s2 + 0.5mvf2
vf = sqrt(1.244 m2/s2)
vf = 1.115 m/s
PEi = mgh0 = (0.0054 kg)(9.81 m/s2)(0.5848 m)
= 0.0310 J = 31.0 mJ
KEf = 0.5mvf2 = 0.5(0.0054 kg)(0.498 m/s)2
= 6.70 x 10-4 J = 670 uJ
PEf = mghf = (0.0054 kg)(9.81 m/s2)(0.5334m)
= 0.0336 J = 33.6 mJ
KEf = 0.5mvf2 = 0.5(0.0054 kg)(1.115 m/s)2
= 0.0036 J = 3.36 mJ
Ei = PEi + KEi = 31.0 mJ + 670 uJ = 31.7 mJ
Ef = PEf + KEf = 28.3 mJ + 3.36 mJ = 31.7 mJ
B marble falls into cup, and continues falling down to box top
* assume, since cup is counter-weighted, the weight of the cup is a non-factor in the calculations, also, only gravity works on the ball.
Figure 7: Diagram for ball falling into pulley cup with dimensions.
h0 = 21 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.5334 m
hf = 17.5 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.4445 m
energy: (from top of counter) * as before, some energy is converted
into kinetic energy
PEi + KEi = PEf + KEf
mgh0 + 0.5mvf2 = mghf + 0.5mvf2
(9.81 m/s2)(0.5334 m) + 0.5(1.115 m/s)2 = (9.81 m/s2)(0.4445 m) + 0.5vf2
5.233 m2/s2 + 0.622 m2/s2 = 4.361 m2/s2 + 0.5mvf2
vf = sqrt(2.988 m2/s2)
vf = 1.729 m/s
PEi = mgh0 = (0.0054 kg)(9.81 m/s2)(0.5334 m)
= 0.0282 J = 28.2 mJ
KEf = 0.5mvf2 = 0.5(0.0054 kg)(1.115 m/s)2
= 0.00336 J = 3.36 mJ
PEf = mghf = (0.0054 kg)(9.81 m/s2)(0.4445 m)
= 0.0235 J = 23.5 mJ
KEf = 0.5mvf2 = 0.5(0.0054 kg)(1.729 m/s)2
= 0.0807 J = 8.07 mJ
Ei = PEi + KEi = 28.2 mJ + 3.36 mJ = 31.6 mJ
Ef = PEf + KEf = 28.5 mJ + 8.07 mJ = 31.6 mJ
C the cup is pulled up by the counter-weight across the pulley and the large marble rolls out
* assume, the net effect of the pulley is to pull the marble straight up before releasing it.
Figure 8: Pulley system diagram (left) and ball FBD (right).
h0 = 6.5 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.1651 m
hf = 7.5 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.1905 m
energy: (from top of counter) * as before, some energy is converted
into kinetic energy
PEi = mgh0 = (0.0135 kg)(9.81 m/s2)(0.1651 m)
= 0.0219 J = 21.9 mJ
PEf = mghf = (0.0135 kg)(9.81 m/s2)(0.1905 m)
= 0.0252 J = 25.2 mJ
Since PEi != PEf there must be energy added to this system by the external force since PEf > PEi:
PEi + Eadd = PEf
Eadd = PEf – PEi = 25.2 mJ – 21.9 mJ
Eadd = 3.3 mJ
* From Conversion 3: Part B, we notice that PEi from above is the same as the KEi from the ball, thus confirming that energy is added to the system, and the source is the ball’s initial kinetic energy.
Conversion 4:
In this conversion, the ball is released from the cup by the pulley, and rolls down the ramp to engage the tripping mechanism for the next section.
Figure 9: Ramp diagram with dimensions.
x = 11 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.2794 m h2 = x2 + y2
y = 3 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.0762 m h = 0.290 m
* Assume f (friction) is zero.
Figure 10: FBD of ball moving down the ramp.
energy: (from top of counter)
h0 = 15 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.381 m
hf = 12 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.3048 m
PEi + KEi = PEf + KEf
mgh0 + 0 = mghf + 0.5mvf2
(9.81 m/s2)(0.381 m) = (9.81 m/s2)(0.3048 m) + 0.5vf2
vf = sqrt(1.496 m2/s2)
vf = 1.223 m/s
- The ball transforms some of its initial energy (all in potential form) into kinetic energy.
Ei = PEi = mgh0 = (0.0054 kg)(9.81 m/s2)(0.381 m)
= 0.0505J = 50.5 mJ
PEf = mghf = (0.0054 kg)(9.81 m/s2)(0.3048 m)
= 0.0404 J = 40.4 mJ
KEf = 0.5mvf2 = 0.5(0.0054 kg)(1.223 m/s)2
= 0.0101 J = 10.1 mJ
Ef = PEf + KEf = 40.4 mJ + 10.1 mJ = 50.5 mJ
Conversion 5:
In this conversion, the marble engages the mechanism to spill water from a bottle, spinning a turbine, which winds a string, lifting a ping pong ball filled with 6 BB pellets, a finally, a lever spills the BBs onto a track.
Figure 11: Diagrams for the mechanism to lift and spill BBs from the ping pong ball.
* assume nearly all energy is transferred from the large marble, thru the device, to the mechanism lifting the ping pong ball filled with BBs.
From Conversion 4, Ei for the beginning of this conversion is 50.5 mJ.
* since lifting BBs, energy will be “added” to the BB lift mechanism,
mBB = 0.2 g = 2x10-4 kg
hf = 2 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.0508m
BBs: PEi + KEi + Eadd = PEf + KEf
0 + 0 + Eadd = PEf + 0 = mghf
Eadd = (6 * 2x10-4 kg)(9.81 m/s2)(0.0508 m)
Eadd = 5.98x10-4 J = 598 uJ
So, 598 uJ is required to lift and spill the BBs onto the track. With 50.5 mJ going into the system (from the large marble), the energy lost in the system is:
Eloss = Ei – Ef = 50.5 mJ – 598 uJ = 0.0499 J = 49.9 mJ
This can be attributed to rough surfaces, poor building materials (cheap items from the parts bin), faulty construction (i.e.: not enough stability in the system), and water power being wasted.
Conversion 6:
In this conversion, the BBs roll down the ramp after being spilled from the ping pong ball.
Figure 12: Elevator mechanism for ping pong ball with dimensions.
x = 5.98 in sinØ = y/h = 0.5/6.0 h2 = x2 + y2
y = 0.50 in sinØ = 0.083 x = sqrt(h2 – y2)
h = 6.00 in Ø = sin-1(0.083) x = sqrt(6.02 – 0.52)
Ø = 4.78o x = 5.98 in
* Assume f (friction) is zero, and the collision with the washer at the bottom of the ramp absorbs nearly all the energy from the BBs, effectively causing the washer to fall straight down “from rest” (Vx = 0).
energy: (from top of counter)
h0 = 10.25 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.260 m
hf = 9.75 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.248 m
Figure 13 : FBD for BBs rolling down the ramp.
* Assume that when BBs fall onto ramp, they begin moving down the ramp from rest.
PEi + KEi = PEf + KEf
mgh0 + 0 = mghf + 0.5mvf2
(9.81 m/s2)(0.260 m) = (9.81 m/s2)(0.248 m) + 0.5vf2
vf = sqrt(0.2354 m2/s2)
vf = 0.485 m/s
Now, assume Ef here is just enough to knock off the washer.
Ei = PEi = mgh0 = (0.0002 kg)(9.81 m/s2)(0.260 m)
= 0.000510 J = 510 uJ
PEf = mghf = (0.0002 kg)(9.81 m/s2)(0.248 m)
= 0.000489 J = 489 uJ
KEf = 0.5mvf2 = 0.5(0.0002 kg)(0.485 m/s)2
= 0.0000235 J = 23.5 uJ
Ef = PEf + KEf = 489 uJ + 23.5 uJ = 512 uJ
Conversion 7:
In this conversion, the BBs hit the washer, causing it to fall straight down onto the mouse trap, setting it off.
Figure 14: Diagram for washer falling off ramp onto mouse tramp with dimensions.
mwasher = 3*mBB = 0.6g = 6x10-4 kg
energy: (from top of counter)
h0 = 9.75 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.2478 m
hf = 1.5 in * (1 ft / 12 in) * (0.3048 m / 1 ft) = 0.0381 m
PEi + KEi = PEf + KEf
mgh0 + 0 = mghf + 0.5mvf2
(9.81 m/s2)(0.2478 m) = (9.81 m/s2)(0.0381 m) + 0.5vf2
vf = sqrt(4.114 m2/s2)
vf = 2.03 m/s
PEf = mghf = (0.0006 kg)(9.81 m/s2)(0.0381 m)
= 2.243x10-4 J = 224 uJ
KEf = 0.5mvf2 = 0.5(0.0006 kg)(2.03 m/s)2
= 6.09x10-4 J = 609 uJ
Ef = PEf + KEf = 224 uJ + 609 uJ = 833 uJ
So, Ei for the entire system : Ei = 32.9 mJ
Ef for the entire system : Ef = 833 uJ
Ei = Ef + Eloss
Eloss = Ei – Ef = 21.9 mJ – 833 uJ = 32.1 mJ
Percent Energy Loss = Eloss / Ei x 100 = 32.1 mJ / 32.9 mJ x 100 = 97.6 % energy absorbed into system or lost to environment.