Blog Pages
Force Diagram
The best way to optimize the speed at which the course was completed with as few penalties as possible was to always have a force on the ball except in the, "No Touch Zone." This allowed for easier turning as well as reducing the risk of a penalty for hitting the bowling ball with the white part of the broom. Constantly exerting a force on the ball allowed small corrections to be made, instead of trying to make the ball suddenly switch directions. Furthermore, in the start of the course, the ball was placed in the front left corner of the starting block, in order to make the first turn less sharp and be able to go around the turn at a higher velocity.
Space Station Project
The Space Station Project was overall a success as the suggested station to be built was under budget and mass, and received the most votes for money to be given. Our design took advantage of the corporate real estate option in combination with residents on the station. Although our design did not appeal to the emotions involving saving the environment and did not threaten the world with a massive weapons system, it earned the most money out of every station in the competition. The total annual income was over $49 billion, much more than the budget used.
Car and Truck Collision Blog Post
Coefficients of Friction |
Kinetic Frictional Forces |
Ff = Fn * µ
100 N = 130 N * µ µcar = 0.769 ucar * 0.7 = µtruck µtruck = 0.538 µ |
Car: Fk = µ * Fn
Fk = 0.769 * (1383.457 kg * 9.8 m/s/s) Fk = 10429 N Truck: Fk = µ * Fn Fk = 0.5384 * (7119.13 kg * 9.8 m/s/s) Fk = 37567 N Fk |
Accelerations:Car: Fnet = ma
10429 N = 1383.457 kg * a a = -7.5383 m/s/s (negative due to the object slowing down) Truck: Fnet = ma 37567 N = 7119.1322 kg * a a = -5.2769 m/s/s (negative due to the object slowing down) |
Velocities (Post - Collision)Car: Vf^2 = Vi^2 + 2ax
0 = Vi^2 + 2 * -7.5383 m/s/s * 8.2 m Vi = 11.118 m/s Truck: Vf^2 = Vi^2 + 2ax 0 = Vi^2 + 2 * -5.2769 m/s/s * 11 m Vi = 10.774 Vi |
Y-Component VelocityCar: Vi * sin(theta) = Vy
Vy = 11.118 m/s * sin(33) Vy = 6.0552 m/s Truck: Vi * sin(theta) = Vy Vy = 10.774 m/s * sin(7) Vy = 1.3130 m/s |
X-Component VelocityCar: Vi * cos(theta) = Vx
Vx = 11.118 m/s * cos(33) Vx = 9.3243 m/s Truck: Vi * cos(theta) = Vx Vx = 10.774 m/s * cos(7) Vx = 10.6936 |
Directional MomentumCar: p = mv
y: 6.0552 m/s * 1383.457 kg y = 8377.1088 kg * m/s x: 9.3242 m/s * 1383.457 kg x = 12899.7681 kg * m/s Truck: p = mv y: 1.313 m/s * 7119.132 kg y = 9347.4205 kg * m/s x: 10.6936 m/s * 7119.132 kg x = 76129.1521 kg * m/s |
Vehicle's MomentumCar initial momentum: sum of both x directional momentums post collision
8377.1088 kg * m/s + 9347.4205 kg * m/s = 17724.5293 kg * m/s Truck initial momentum: sum of both y directional momentums post collision 12899.7681 kg * m/s + 76129.1521 kg * m/s = 89028.9202 kg * m/s |
Velocity Prior to the CollisionCar: p = mv
17724.5293 kg * m/s = 1383.457 kg * v v = 12.8117 m/s Truck: p = mv 89028.92 kg * m/s = 7119.13 * v v = 12.505586 |
DiscussionsBased on the calculations done and the data collected, it seems that the truck driver, Lincoln Hawk, lied about his speed prior to the collision. He claimed to be driving only 6.7 m/s, but was actually traveling at 12.5056 m/s. Therefore, the truck driver is at fault, even though the car driver should not have pulled out in front of the truck while having a flashing red light.
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Rotational Kinematics Blog Post
Gaining an understanding of rotational kinematics through the physics of throwing a football
A stereotypically dumb football player explains the physics of throwing a football, of which he struggles to do well:
MotionAfter the football is thrown, kinematic equations can be used to calculate different quantities that describe the motion of the football in the air. For example, have the initial vertical velocity and the height at which the ball was released can be used to calculate the time that the ball will be in the air. The kinematic equations help describe the motion of the ball in the air, whether it be time in the air, final vertical velocity, or horizontal distance traveled. In this scenario, it would be beneficial for the quarterback to throw the ball slightly upwards to create an initial vertical velocity. This way, since there is a constant acceleration downwards, an initial vertical velocity will allow the ball to be in the air longer, allowing it to travel farther.
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EnergyWhile it may not seem true, the law of conservation of energy applies to the throwing of a football. Although it seems that the ball goes from rest to flying through the air with little to no energy transfer, there is still internal chemical potential energy being used to yield the kinetic energy in a throw of a football. As with a traditional energy example like a ball rolling down a hill, the higher the amount of initial potential energy that is being converted, the higher the amount of kinetic energy the ball has. A quarterback can use the law of conservation of energy to understand how much energy it will take to throw the ball at a certain velocity, as the higher the kinetic energy of an object, the higher the velocity of that object, if the mass is constant. In this scenario, the mass is constant.
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RotationWhen a football is thrown, it is not only moving translationally, but additionally is rotating. The cause of this rotation comes from the motion of the arm before the throw and a pronation of the wrist during the throw. In the wind or rain, quarterbacks are taught to pronate, or twist, their wrists in a certain manner to create a "spiral." The reason this is desired is because it reduces air resistance and allows the ball to travel faster through the air. This combination of rotational energy and kinetic energy creates not only a linear momentum in the ball, but furthermore a angular momentum. This means that the ball will not stop spinning while in the air due to its rotational inertia, meaning the ball will efficiently fly through the air to a desired target if thrown correctly.
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