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# Rube Goldberg

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## yolie nic

on 28 May 2013

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#### Transcript of Rube Goldberg

Rube Goldberg
Machine Vivy Li
Yolie Saepung
Brian Xu
Period 4
AP Physics B
Mr. What is Rube Goldberg Machine? Concept 1: Conservation
of Energy Concept 2: Momentum Concept 4: Buoyancy Concept 5: Torque Concept 3: Centripetal Force Concept 6: Circuit A complicated machine that performs a simple task
Chain reaction
Named after an American cartoonist and inventor
Became an adjective in the dictionary to describe accomplishing something simple through complex ways The car rolls down from the ramp.
The car is initially at a higher altitude, exhibiting higher gravitational potential energy. It is also at rest so the kinetic energy is zero.
Letting the height at the table top be zero, the gravitational potential energy at the table top is also zero. The speed of the car increases as it rolls down the ramp.
Going down the ramp, the kinetic energy increases as the gravitational energy decreases. (KEi+PEi = KEf+PEf+heat) The first domino falls on the second domino that is initially at rest.The momentum in a collision has to be conserved, so the momentum before the collision is equal to the momentum after the collision.
In order to show the calculation, we assume the collision is elastic When the jenga falls off the table, the weight that is holding down the balloon is removed. So, the balloon will float up.
Something will float in a fluid when the buoyant force pushing it up (Fb) is greater or equal to the force pulling it down. The force pulling it down is gravity and is a function of the mass of the object . The ruler is balanced on top of the box and and falls off the table when the balloon hits it.
Initially, the forces acting on the ruler are the gravitational force of earth on ruler and the normal force of the box on the ruler which are equal in the opposite direction.
When the balloon hits the ruler, it loses the torque equilibrium because the balloon provides an upward force so the vertical forces are no longer equal. The ruler falls off the table and the string that is attached to the ruler turns the switch on.
The switch is connected to a battery and two light bulbs in series.
For series connection, the current flowing in the circuit is SAME everywhere. I1=I2=Ibatt
The electric potentials add up together equal to the electric potential of the battery. V1+V2=Vbatt
The electric potential of the two light bulbs are 4.5 v each and the battery has 9V. The ball that is initially wound up around the poll is released and it travels in circle around the poll.
The forces acting on the ball are the gravitational force of the earth on the ball and the tension force of the string on the ball. The vertical component of the tension is equal and opposite to the gravitational force so the forces cancel.
The net force or the centripetal force that makes the ball go in circle is the horizontal component of the tension pointing towards the center of the circle at all times. Concept 7: Lens Concept 8: Nuclear Reactor In a nuclear reactor, large atoms are forced to collide with one another.
The reactions are exothermic.
Water is pumped along the nuclear reactor’s core where the reaction takes place.
The water is vaporized and rises through pipes that are connected to a steam turbine.
As the turbine spins, a generator connected to it charges. Toy car Ramp Pole String Ball Ruler String Balloon Box Initial Velocity = 0m/s
Kinetic Energy = 0J
Height = 0.32m
Mass of the car=0.033kg
Potential Energy = mgh = (0.033kg)(9.8m/s^2)(0.32m) = 0.1 J Final Height = 0m
Potential Energy = 0J
Kinetic Energy = 0.1J
Velocity = sqrt(2KE/m) = sqrt(2*0.1J/0.033kg) = 2.5m/s Lightbulb Battery Switch String Wire Calculation of Lens The light provided by the light bulb shines through the object and lens and an image is formed on the screen.
The two lens are converging lens because the waists are wider.
Focal point formula: (1/Si)+(1/So)=(1/f)
Magnification= Si/So Focal point of the thin lens:
1/21.5cm +1/19.25cm= 1/f
f=10cm
Focal point of the thick lens:
1/40cm+1/31.2cm=1/f
f=17.5cm
Magnification of the thin lens:
19.25cm /21.5cm=0.9
Magnification of the thick lens:
31.2cm/40cm=0.78 Initial Final Calculations for Momentum Since the ball is tied around the pole, the radius of the circle is increasing.
Centripetal force=mv^2/r
We know the mass and velocity of the ball, but the radius is not constant Centripetal Force (cont'd) Initial Centripetal Force To find the buoyant force, we measured the circumference of the balloon to find the radius and use it to find the volume of the balloon.
Fb = mass fluid displaced x g = weight fluid displaced
The more massive an object is the greater the force of gravity pulling it
Fb = (ρHe)(Vballoon)g = (018kg/m3)(0.01625m3)(9.8m/s2) = 0.0287 N Buoyancy (cont'd) Final Initial Initial Final Final The Thin Lens
The object is behind the focal point and the center of curvature so the image formed is inverted, real, and smaller. The Thick Lens
The object is behind the focal point and after the center of curvature so the image formed is still inverted, real , and smaller. Lens Lightbulb Protractor Screen We will do the calculations for the momentum when the toy car hits the first and second jenga.
Mass of the car = 0.033kg
Mass of jenga = 0.012kg When the car hits the first jenga;
(mcar)(vcar) = (mjenga1)(vjenga1)
(0.033kg)(2.5m/s) = (0.012kg)(vjenga1)
vjenga1 = 6.875m/s When the first jenga hits the second jenga
(mjenga1)(vjenga1) = (mjenga2)(vjenga2)
(0.012kg)(6.875m/s) = (0.024kg)(vjenga2)
vjenga2 = 3.345m/s
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