CTF Photo Project - By Brendan Foley

Gravity: Statue on Block Pedestal

(Natural)

            In this photo, we place a coveted “World’s Greatest Boss” trophy on a colorful wooden-block pedestal. The statue was acquired from a friend, who gave it to my brother, in reference to the “like a boss” meme, back when it was relevant (now it’s more vintage). On a scale, the trophy had a mass of .24 kg, and the platform it rests on was 3kg. When you factor in gravity’s push on them, the trophy pushes on the ground with a force of 2.35 N, and the pedestal 29.4 N. Combined, they exert a force of 31.75 N.

Projectile Motion: Independent X and Y Velocities: Marble Rolling Off Table

(Natural)

            This photo shows a marble being rolled off a table, keeping it’s X launch velocity constant throughout its flight. The only acceleration it experiences is from gravity on its way down to the ground. The marble is the same marble in the previously mentioned “marble in a bowl experiment,” so it has a mass of 20 g, or .02 kg. The table it falls from is 67.3 cm

(.67 m) tall, so we can solve its time and Y-velocity using the equation:

Vyf^2=Vyi^2 + 2ag (ΔY)

Vyf^2= (0) + 2(9.8 m/s^2) (.67 m)

Vyf^2= 13.132

Vyf= 3.63 m/s

T = Vyf/ag

T = 3.62 m/s   =  .37 seconds

     9.8 m/s^2

Because the initial Y-Velocity was zero, it made finding the time pretty easy. Next, we’ll find the X-velocity of the marble. Because the marble fell .2 m from the launch point, we can use this equation, utilizing the ΔX and T:

Vx= ΔX

          T

Vx= .2 m   = .54 m/s

         .37s

To get the final velocity it was going though, we have to add the velocities together. For this, we can use the Pythagorean Theorem.

.54^2 m/s + 3.63^2 m/s= V^2

.29 m/s + 13.18 m/s= V^2

13.47 m/s= V^2

3.67 m/s= V

Centripetal Force: Marble in a Bowl
(Contrived)

            In this photo, there’s a medium sized marble in a Tupperware bowl, finishing up a loop around it. When it picks up speed, normal force keeps the marble suspended against the edge of the bowl, instead of the bottom. If spun around the bowl too quickly (and not carefully enough), the marble will exit the bowl, tangential to the curve. While spinning at top speeds, the marble laps the 19.2 cm bowl 204 times per minute (204 RPMs). To solve the centripetal force of the spinning marble, we take its mass (20 grams) and multiply it by the velocity. To change our 204 RPM into a standard velocity, we take the radius of our bowl (9.6 cm), multiply by 2Π (we’ll use Π=3.14) and the RPMs (204), and divide by 60, to convert into cm/second.

So…

(2(3.14) x 9.6 cm x 204 RPM)   = 204.97 cm/ s  = 2.05 m/s
               60 sec

With our exact velocity, we can solve for the centripetal force by squaring our force (2.05^2, multiplying by the mass (20 grams =.02kg), and dividing by the radius (.096 m).

Fc= (2.05^2 m/sec x .02 kg)     =  .88 N
                 .096 m

Although it’s going very quickly, it doesn’t have much force behind it.

Gravity: Angles effecting Suspension

(Contrived)

            This photo shows how gravity pulling down on the center of an object suspended on string takes much more force to suspend evenly, because gravity works against the holder. Pulling outwards on the string will never overcome the downwards force of gravity on the basket. The basket with blocks in it is .28 kg, and the angle of the basket pulling down on the string is 133°. The reference angle of 133 is 47°, and 23.5° is the deviation from horizontal. The sin of 23.5° is .4. To solve the force necessary to pull in order to hold the basket at this angle is:

Sin(23.5) (Ft)= .28 x 9.8

(.4)(Ft) = 2.74

Ft = 6.86 N

As you can see, it doesn’t take a lot of force to suspend this at such a steep angle. But let’s say I wanted to hold it very close to level, as close as possible.

Sin(1)= .0174

.274/.0174 = 157. 47 N

157.47 N/ 9.8 m/s^2= 15.3 kg

Just to hold this .28 kg basket very nearly level, I would have to pull with a force equivalent to lifting 15.3 kg with each hand horizontally.

Rocket Project Write-Up -By Brendan Foley

Materials:

·       2 Liter Bottles- ($4)

·       Paper Nosecone ($.03)

·       Duct Tape ($3.50)

·       Card stock folders($1 apiece)

·       Foam Packaging (Acquired)

·       Bubble Wrap (Acquired)

·       Trash Bag ($.5)

·       Yarn ($1)
Procedure:
First, we assembled out materials, and planned what we were going to do.

We decide to to cut one of the bottles down, and cut a nose cone for it, making the cockpit for our eggonaut.
Unsure about how our parachute would deploy, we decide to make an emergency capsule made of foam and bubble wrap to ensure eggonaut survival.
Next, we cut the card stock folder into wings, and attach them to the untouched bottle.
Then, we use duct tape to attach the bottles, and reinforce the wings.
For our parachute mechanism, we used a large trash bag, and punched holes through it, then laced yarn through it.
We would then insert the emergency foam capsule into the cockpit of the rocket, followed by the parachute, then put on the fuel cell, and the nosecone put atop it all.
Results:
Our rocket only traveled a bit more than 35 feet and stayed in the air for 2.7 seconds. On the test run, the parachute deployed, but it never deployed for the real launch. This could have been due to a parachute packing malfunction, the nosecone being worn out after the test, or we were just lucky the first time it deployed. Fortunately, the egg easily survived the fall, possibly because it was only a 35ft drop, but also due to the foam capsule and bubble wrap. Unfortunately, this capsule could easily have been a double edged sword, and its weight could have negatively effected the height. The nosecone was also imperfect, and could have been tighter fit and sharper, which would have caused more balance, and less drag.
Conclusion:
I believe having the backup eggonaut survival capsule was the right way to go. What would have been best would have been a reliable parachute deploying mechanism, but we couldn't figure that out. Having a straighter nosecone, and possibly a longer rocket could have helped a lot with the height deficiency. 

Calculation:
We need to solve for height, so I'll use the formula: Δy= Vit +1/2agt^2
Δy=0+1/2(9.8)(2.7)^2
Δy= 35.7 ft
This calculation may be more accurate than Mr, Yav's estimate against the side of Sobroto, although it may be a little low, as this takes the initial velocity as 0. If it's greater on launch, then it could be the 50ft Mr. Yav initially estimated.
To Solve for Initial Velocity, we can use the handy hatchet equation, using our new height, and by saying that the Vf at the apogee is 0, and Gravity=-9.8m/s^2
0=Vi^2+(2)(-9.8m/s^2)(35.7)
0=Vi^2-699.72
699.72=Vi^2
Vi=26.45

FBDs: