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Buoancy, old exam question

  1. Jul 25, 2004 #1
    Hello people, first of all thanks to all who reads this and attempts this problem or more or less helps me see the underlying principles. A little background...i am a first semester 23 yr. old summer school physics student who has an exam tomorrow, monday night. The exam covers buoyancy, fluids, density& pressure and rotational kinetic energy. Good stuff i know....here is a problem from an old exam last semester. It reads exactly.....

    A 1.0 kg hollow ball of radius .10 m, filled with air, is released from rest at the bottom of a 2.0 m deep pool of water. How high above the surface of the water does the ball rise? Neglect all frictional effects and the changing force on the ball when it is partially submerged. Volume of a sphere = 4/3 pi r^3; rho_air (denisty of air) = 1.29 kg/m^3; rho_water (density of water) = 1000 kg/m^3.

    I am on a time crunch this morning so i'll try to condense as much as I know about this problem as I can with the time constraints i am on and will come back around 5 and read what people have to say. I really want to understand what actually is happening, because that is where physics is difficult for me.

    This problem is an application to what Bernouli and Archimede's studied. I believe the height value i'm looking for is located in the gravitational potential energy term of Bernouli's equation, rho*g*y (rho = density). The velocity terms in the equation drop out because the ball starts from rest and finishes at rest. If anything i'm saying is incorrect or if my thinking is skewd, please let me know.

    Thanks....i gotta go work a few hours then i'll be back :)
  2. jcsd
  3. Jul 25, 2004 #2
    Does anyone have any input on this problem?
  4. Jul 25, 2004 #3


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    Staff Emeritus
    Science Advisor

    You know the density of water so you know the mass of the water the sphere displaces. That, presumably, is less than the actual mass of the sphere so that the sphere WILL go up. The upward (bouyancy) force is mg where m is mass of water displaced - mass of sphere. Using that, you can calculate the upward speed of the sphere when it reaches the surface of the water. After that, gravity takes over. Using the speed at the surface of the water as "initial speed" you can calculate the kinetic energy (taking potential energy to be 0) and calculate the maximum height the sphere will reach (where all the kinetic energy has been converted to potential energy)
  5. Jul 25, 2004 #4
    Ok, i see that rho=m/V. Now when i use the Volume of water displaced to find the mass of the water displaced, do i find volume using the equation of a cylinder or of a sphere? Because the shape of the displaced water is cylindrical, however, the object is a hollow spherical object so I'm not sure which volume equation to use to calculate the masses. I didn't think about using Energy to calculate the height when the ball comes out. I knew that the ball would be in static equilibrium, thus the bouyant force and the weight of water displaced by the object would be equal, but i could not get a way to calculate the height above the water.

    Thanks for your help....could you explain the way to calculate the masses if the way i thought to calculate them using density and volume was wrong, how do you get the masses?
  6. Jul 26, 2004 #5

    Doc Al

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    Staff: Mentor

    Where did you get the idea that the shape of the displaced water is cylindrical? The ball is sphere, thus it displaces a spherical volume of water.
    What makes you think that the submerged ball is in static equilibrium? If it were, then it would just sit there, not shoot out of the water and rise in the air.

    When the ball is submerged there are two forces acting on it: its weight (down) and the buoyant force (up). Find the net force (which will be up) and then the resulting acceleration of the ball (from Newton's 2nd law). Use kinematics of accelerated motion to find what the final speed will be when the ball reaches the surface. Once the ball leaves the water with the speed just calculated, then it's just an ordinary projectile. Find how far it rises, given the acceleration is -g.

    If you wish to use energy methods (which is what I would do) it's even easier:
    Under the water you have the net force on the ball; let's call that F. That force acts over a distance D (2 m in this case) and thus does work on the ball giving it some KE as it leaves the water: KE = FD. To find out how high the ball rises in the air, set that initial KE to the PE at the top of the motion: KE = FD = mgH. Now solve for H.
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