Flying Pig Lab
FLYING PIG LAB
Jackson Fordham Procedure
1.
Find the mass of the Pig
2.
Attach pig to ceiling with string, measuring the length of string
3.
Find the time it takes the pig to make one rotation
4.
Find the circumference by finding θ from the length of the string and the height of the pig from the ceiling
5.
Use θ to find the radius of the circle by inverse cosine
6.
Use the radius to find the circumference
7.
Find the velocity by dividing the circumference by the time it took the pig to complete one full rotation
8.
Use θ and velocity to solve for the vertical component of tension using the horizontal component Data
Mass of Pig: 0.15g
Length of string: 0.845m
Time of One Rotation :
Trial 1: 1.85s
Trial 2: 1.96s
Trial 3: 1.76s
Trial 4: 1.81s
Trial 5: 1.90s
Average time of single rotation: 1.856s
X2
X1 Vertical distance (X2) = 0.775m
Hypotenuse (Length of string= 0.845m
horizon θ = 23.5° horizontal distance (the radius, X1) = 0.34m
Calculations
Fc=mac
ac=mv^2/r θ: cos^1 θ = (0.775/0.845) θ= 23.5° r:
0.775*tan(23.5) = 0.34 m Circumference:
2ᵷ(0.34) = 2.13 m Average velocity:
2.13m / 1.856s = 1.152 m/s FTx = mv^2/r
FTx = 0.15(1.152)^2 / 0.34 = 0.585 N FT= 0.585 / sin23.5 = 1.47 ac = mv^2/r ac = (0.15)(1.152)^2 / 0.34 = 3.9 m/s^2
Jackson Fordham Procedure
1.
Find the mass of the Pig
2.
Attach pig to ceiling with string, measuring the length of string
3.
Find the time it takes the pig to make one rotation
4.
Find the circumference by finding θ from the length of the string and the height of the pig from the ceiling
5.
Use θ to find the radius of the circle by inverse cosine
6.
Use the radius to find the circumference
7.
Find the velocity by dividing the circumference by the time it took the pig to complete one full rotation
8.
Use θ and velocity to solve for the vertical component of tension using the horizontal component Data
Mass of Pig: 0.15g
Length of string: 0.845m
Time of One Rotation :
Trial 1: 1.85s
Trial 2: 1.96s
Trial 3: 1.76s
Trial 4: 1.81s
Trial 5: 1.90s
Average time of single rotation: 1.856s
X2
X1 Vertical distance (X2) = 0.775m
Hypotenuse (Length of string= 0.845m
horizon θ = 23.5° horizontal distance (the radius, X1) = 0.34m
Calculations
Fc=mac
ac=mv^2/r θ: cos^1 θ = (0.775/0.845) θ= 23.5° r:
0.775*tan(23.5) = 0.34 m Circumference:
2ᵷ(0.34) = 2.13 m Average velocity:
2.13m / 1.856s = 1.152 m/s FTx = mv^2/r
FTx = 0.15(1.152)^2 / 0.34 = 0.585 N FT= 0.585 / sin23.5 = 1.47 ac = mv^2/r ac = (0.15)(1.152)^2 / 0.34 = 3.9 m/s^2