Question

PART A) Consider a sea lion swimming horizontally in 20C sea water. It is neutrally buoyant. Its mass is 200kg, body surface area 2.1 m2, and drag coefficient Co= 0.0038. After accelerating from rest for a while (about half minute or so), the animal reaches a state of terminal velocity U_term of 1.1 m/.s. Calculate the drag and the thrust forces at that moment.

PART B) The sea lion studied in the previous question is now vertically diving. It is now negatively buoyant at W - B = 1.3N (by letting air out of its lungs), but not sinking fast enough to catch the prey that it has located below the surface. Here it must generate thrust with its fore flippers to go down faster, say with a force of Th= 3.2N. After a short while it reaches terminal velocity. What is the value of this terminal velocity?

PART C) As it sinks further down, the pressure underwater has increased to values high enough to squeeze its fur and skin and render the body even !!1Qil: negatively buoyant, now to W - B = 2.9N. At this point the sea lion decides to stop producing thrust and begin accelerating (or decelerating) to reach a new state of terminal speed. What is the value of this new terminal speed?

Answer 1

(A) We know that the drag force is given by
F_D = (1/2)C_opAV²
where p is density of sea water = 1027 kg/m³
A is area = 2.1 m²
V is terminal velocity = 1.1 m/s
Now putting all the values
F_D = (1/2)*0.0038*(1027)*2.1*(1.1)² = 4.958 N
We know at the terminal velocity , net force is zero , therefore
F_T - F_D = 0
Where F_T is thrust force
F_T = F_D = 4.958 N
(B) After applying force when it reaches the terminal velocity , FBD is shown in the figure
From FBD
F_T + W - B = 0
B = F_T + W = = 3.2 +(mg) = 3.2 +(200*9.81) = 1965.2 N
(1/2)C_o*pA*V² = 1965.2
(1/2)*0.0038*1027*2.1*V² = 1965.2
V = 22 m/s
hence the terminal velocity of lion will be 22 m/s
(C) Now
W - B = 2.9 N
(200*9.81) - 2.9 = B
B = 1959.1
(1/2)C_o*pAv² = 1959.1
(1/2)*0.0038*1027*2.1*v² = 1959.1
v = 21.86 m/s