Question

A frog is riding on the top of a cylindrical piece of wood floating in still water.
Half of the wood, with a diameter of 4 cm and length 20 cm, is immersed in water.  
The density of water is 1 gm/cc.  
a) What is the mass of the wood along with the frog?  
b) After the frog slowly goes into the water only one third of the wood remains immersed in water. Calculate the mass of the frog.  
c) Calculate x, the distance between the water level and the center of the circular end of the wooden piece.  
d) Briefly describe the motion of the wood after the instance the frog moves into the water. Give a rough   sketch of x as a function of time.  

Answer 1

a) mass= 251.2

Answer 2

weight pushing down = Force buoyancy pushing up = weight of liquid displaced

weight of liquid displaced = volume x density x g = Pi x 2cm^2 x 10cm x 1 gm/cm^3 x g so mass of wood plus frog = weight/g = Pi x 4 x 10 x 1 gm = 125.6 gm 

Now weight of wood = weight of liquid diplaced 

Pi x 2cm^2 x 10cm x rho gm/cm^3 x g = Pi x 2cm^2 x 1 gm/cm^3 x 10/3 x g 

so rho = 1/3 gm/cm^3 

so mass of wood = Pi x 4 x 20 x 1/3 = 251.2/3 = 83. 66 gm 

subtract the two masses to get mass of froggie. 

 If 1/3 is below the surface then 2/3 must be above, so 2/3 20cm = 13.33cm