Data and Calculations
This table shows the series of trials
to find the average time (in seconds)
that it takes for the socket to revolve
around ten times
to find the average time (in seconds)
that it takes for the socket to revolve
around ten times
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Measurable Givens:
Gravity = (9.8 m/s2)
Mass of Socket = (.03 Kg)
Radius = (.72m)
Mass = (.15Kg through .5Kg)
Time = (varies by weight)
Gravity = (9.8 m/s2)
Mass of Socket = (.03 Kg)
Radius = (.72m)
Mass = (.15Kg through .5Kg)
Time = (varies by weight)
Line of best fit:
Y = 15.147x + 3.5026
Y = 15.147x + 3.5026
Line of best fit:
Y = 15.147x + 3.5026
Y = 15.147x + 3.5026
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This table show the breakdown of
everything we do know or have
conducted trials to calculate
everything we do know or have
conducted trials to calculate
To be able to calculate the velocity of
the rotating socket for each case is a
simply case of setting up equations. To
do so we need to set up a few free
body diagrams.
the rotating socket for each case is a
simply case of setting up equations. To
do so we need to set up a few free
body diagrams.
In diagram number one there is gravity and the tension of the fishing line
adding to become the sum of the forces. Since the weight is hanging in the
air the two forces are equal. Mg = T.
In the second Diagram there is again two forces acting on the socket. The
centrifugal force pulling the socket outwards and the tension in the fishing
line pulling the object in. Since the socket remains in a rotation the two
forces are in equilibrium and thus they equal each other. Fc = T.
The tension in the fishing line is the same for both diagrams and because
of that we can set the two equations equal to each other. Fc = Mg
Fc is the centrifugal force acting on an object that is spinning. Fc = (Mv2)/r
Therefore Mg = (Mv2)/r
adding to become the sum of the forces. Since the weight is hanging in the
air the two forces are equal. Mg = T.
In the second Diagram there is again two forces acting on the socket. The
centrifugal force pulling the socket outwards and the tension in the fishing
line pulling the object in. Since the socket remains in a rotation the two
forces are in equilibrium and thus they equal each other. Fc = T.
The tension in the fishing line is the same for both diagrams and because
of that we can set the two equations equal to each other. Fc = Mg
Fc is the centrifugal force acting on an object that is spinning. Fc = (Mv2)/r
Therefore Mg = (Mv2)/r
1
2
Using the new equation Mg = (Mv2)/r. We can no solve for the velocity in each of the
given mass trials. We know the value of every term in the equation except v. Solve for v
in each of the mass trials. Then graph velocity verses mass on a scatter plot and find a
best fit line.
given mass trials. We know the value of every term in the equation except v. Solve for v
in each of the mass trials. Then graph velocity verses mass on a scatter plot and find a
best fit line.
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Since we do not know the mass of the mystery
mass we cannot use the above equations to solve for the velocity.
The alternative to this method is to use the length of the radius
and total time for ten revolutions to calculate the velocity. This
method is the only way to find the velocity of the socket as it
spins when we do not know the mass of the hanging object. The
first step is to find the period of revolution. Period =
time/revolutions. (P = t/Rev). After we have done that we compute
the circumference of the circle. Circumference = two x pi x radius
(C= 2πr). Once both the period and the circumference has been
compiled, we can solve for the velocity. Velocity = Circumference
/ Period (V = C/P).
Solving for mystery mass: The velocity: part 1
Time = 4.03
Revolutions = 10
Period = 4.03/10 = .403
Radius = .72m
C = 2πr
C =1.44π
V= C/P
V= 1.44π / .403
V= 11.22 m/s
Solving for mystery mass: The velocity: part 1
Time = 4.03
Revolutions = 10
Period = 4.03/10 = .403
Radius = .72m
C = 2πr
C =1.44π
V= C/P
V= 1.44π / .403
V= 11.22 m/s
Unknowns:
Velocity
Mystery Mass
Velocity
Mystery Mass
Estimating the
Mass:
Using the line of best fit from the graph we made earlier we can guess what the mass would be. Y is the Velocity and X is the Mass
11.22 = 15.147(X) + 3.5026
Y = .972 Kg
Using the line of best fit from the graph we made earlier we can guess what the mass would be. Y is the Velocity and X is the Mass
11.22 = 15.147(X) + 3.5026
Y = .972 Kg
Solving for the mystery mass: the Mass: part
2
Now that we know the velocity we can use the original equation to solve for the mass of the mystery mass.
Mg = (mv2)/r
M = (.03(125.89)) / ((.72) x (9.8)
M = .54 Kg
When we weighed the mystery mass we found it to weigh .6 Kg
Continue to Results
Now that we know the velocity we can use the original equation to solve for the mass of the mystery mass.
Mg = (mv2)/r
M = (.03(125.89)) / ((.72) x (9.8)
M = .54 Kg
When we weighed the mystery mass we found it to weigh .6 Kg
Continue to Results