3. DETAILS MY NOTES PRACTICE ANOTHER A cylindrical log of radius 0.4 meter, 7 meters in length, and with a mass of 100 kilograms is placed vertically in a lake so that it is free to bob up and down. Assume that there is no water resistance. A weight of 100 kilograms of negligible volume is attached to the bottom of the log so that it remains vertical (so the total mass of the log and weight together is 200 kilograms). The mass density of water is 1000 kilograms per cubic meter. (For convenience, assume that the acceleration due to gravity is g= 10 meters per sec?). There are two forces acting on the log: gravity and the buoyant force of the water. The buoyant force can be computed from Archimedes' principle: An object that is completely or partially submerged in a fluid is acted on by an upward (buoyant) force equal to the weight (weight = mass *g) of the displaced fluid. Let t be time in seconds and let d(t) denote the depth (in meters) of the bottom of the log. Compute the depth deg of the log in its equilibrium position, i.e. when the magnitude of the buoyant force is exactly equal the combined weight (in Newtons) of the log plus the mass. deg= Write down a differential equation for d(t). d" + Now let y(t) - d(t) - deg be the displacement of the log from its equilibrium position. Assuming that y(0) - 1 meters and y'(0) - 0 meters/sec, write down an initial value problem for y y" + y(0) - 1 and y'(0) - 0 y = 0 Solve the initial value problem you wrote down in part (c). y(t) - If you are looking at the log from h(t) - boat on the surface, you don't see the bottom of the log. Write a formula for the h(t), the height of the log above the surface of the lake. Bymbolic formatting help
3. DETAILS MY NOTES PRACTICE ANOTHER A cylindrical log of radius 0.4 meter, 7 meters in length, and with a mass of 100 kilograms is placed vertically in a lake so that it is free to bob up and down. Assume that there is no water resistance. A weight of 100 kilograms of negligible volume is attached to the bottom of the log so that it remains vertical (so the total mass of the log and weight together is 200 kilograms). The mass density of water is 1000 kilograms per cubic meter. (For convenience, assume that the acceleration due to gravity is g= 10 meters per sec?). There are two forces acting on the log: gravity and the buoyant force of the water. The buoyant force can be computed from Archimedes' principle: An object that is completely or partially submerged in a fluid is acted on by an upward (buoyant) force equal to the weight (weight = mass *g) of the displaced fluid. Let t be time in seconds and let d(t) denote the depth (in meters) of the bottom of the log. Compute the depth deg of the log in its equilibrium position, i.e. when the magnitude of the buoyant force is exactly equal the combined weight (in Newtons) of the log plus the mass. deg= Write down a differential equation for d(t). d" + Now let y(t) - d(t) - deg be the displacement of the log from its equilibrium position. Assuming that y(0) - 1 meters and y'(0) - 0 meters/sec, write down an initial value problem for y y" + y(0) - 1 and y'(0) - 0 y = 0 Solve the initial value problem you wrote down in part (c). y(t) - If you are looking at the log from h(t) - boat on the surface, you don't see the bottom of the log. Write a formula for the h(t), the height of the log above the surface of the lake. Bymbolic formatting help
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Transcribed Image Text:3.
DETAILS
MY NOTES
PRACTICE ANOTHER
A cylindrical log of radius 0.4 meter, 7 meters in length, and with a mass of 100 kilograms is placed vertically in a lake so that it is free to bob up and down. Assume that there is no water resistance. A
weight of 100 kilograms of negligible volume is attached to the bottom of the log so that it remains vertical (so the total mass of the log and weight together is 200 kilograms). The mass density of
water is 1000 kilograms per cubic meter. (For convenience, assume that the acceleration due to gravity is g= 10 meters per sec?).
There are two forces acting on the log: gravity and the buoyant force of the water. The buoyant force can be computed from Archimedes' principle:
An object that is completely or partially submerged in a fluid is acted on by an upward (buoyant) force equal to the weight (weight = mass *g) of the displaced fluid.
Let t be time in seconds and let d(t) denote the depth (in meters) of the bottom of the log.
Compute the depth den of the log in its equilibrium position, i.e. when the magnitude of the buoyant force is exactly equal the combined weight (in Newtons) of the log plus the mass.
deg =
Write down a differential equation for d(t).
d"
Now let y(t) = d(t) - deg be the displacement of the log from its equilibrium position. Assuming that y(0) - 1 meters and y'(0) = 0 meters/sec, write down an initial value problem for y.
y" +
y(0) = 1 and y'(0) = 0
y = 0
Solve the initial value problem you wrote down in part (c).
y(t) =
If you are looking at the log from a boat on the surface, you don't see the bottom of the log. Write a formula for the h(t), the height of the log above the surface of the lake.
h(t) -
symbolic formatting help
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