Physics Forces help (terminal velocity / upthrust )?

Question

Question goes like this: A solid steel sphere is held just below the surface of water and released. Its terminal velocity, v, is measured. If the experiment was repeated in places with different values of g(acceleration of free fall), a graph of v against g would most closely resemble,
A. graph starts from origin, is always increasing but increasing at a decreasing rate.
B. graph starts from origin, is always increasing but increasing at a increasing rate.
C. Graph is linear, but does not start from origin.
D. Graph is linear, starts from origin.
E. Graph is a constant value of v, ie. Horizontal line.

Can any kind soul explain please?? Thanks in advance :D

Answer 1

The answer is D.

Note that Soham assumes friction proportional to velocity^2. For experiments of steel balls falling in water a better approximation is Stoke's formula for a sphere: friction = 6 pi eta r v).





When terminal velocity is reached, the net sum of forces must be zero.



Three forces at work:

gravity (m g, downward))

buoyancy (rho g V, Archimedes' principle, upward)

drag (C v , assuming friction to be linear in velocity, a very good approximation in these types of experiments), upward)



So



C v = m g - rho g V

or

v = some constant * g



In words: when terminal velocity is reached, drag (proportional to velocity) balances effective weight (weight minus buoyancy) which is proportional to g. So v proportional to g.

Answer 2

The answer is option (A)



Let us understand this step by step.



We know that the expression for terminal velocity is;





Mathematically, terminal velocity—without considering buoyancy effects—is given by

V = root( 2mg/pAC)

where

V= terminal velocity,

m= mass of the falling object,

g= acceleration due to gravity,

C= drag coefficient,

p= density of the fluid through which the object is falling, and

A= projected area of the object.



In the above expression we see that v proportional to sqrt(g).



When g=0 , i.e , there is no acceleration due to gravity , the ball would not fall through the liquid and hence will have a terminal velocity of 0. Hence the graph will start from the origin.



Now we see that v proportional to sqrt(g) hence the graph will be corresponding to a natural graph of root(x).Hence it will be increasing but at a decreasing rate.



Hope this helped. :)

Answer 3

the answer is option D.



when a sphere starts to fall it will xperience two upward forces ad one downward force



upward ... upthrust and viscous force,while downward .... only weight .



at terminal velocity these forces balance each other, i.e F(v) + U = W

use stokes law to find out viscous force =6.pi.eta.radius.terminal velocity

while u can easily find W-U

u see that v is directly related to g... v = constant*g

so the graph will be linear and passing through origin at some angle, which depends upon radius and the densities...

Answer 4

The deal is, that air drag is commonly proportional to the SQUARE of velocity. This is right for any type of drag that's ruled by way of have an effect on strain (believe air and water at top speeds). This isn't actual for drag that's ruled by way of viscosity (believe oils at gradual speeds). At terminal velocity, the drag drive is by way of definition same to the burden of the falling item. That is the one method that the item can fall at a regular velocity, consequently the title. Weight is m*g Drag at terminal velocity: D_term = m*g By the character of the drive of drag, it's proportional to the rectangular of velocity. Make a false method for drag, with a main regular (K) in entrance of velocity squared. We can connect items to it, of Newton-seconds^two/meter^two, in the event you wish. We can provide K a thought, however that's an issue for a different day. D = K*v^two We have an interest within the case, while D = D_term/two. And we wish to understand what velocity that occurs at. Given: D_term = K*v_term^two Construct the case of: D_term/two = K*v^two Solve first equation for K: K = D_term/v_term^two Solve moment equation for v: v = sqrt(D_term/(two*K)) Plug in K: v = sqrt(D_term/(two*(D_term/v_term^two))) Simplify: v = v_term/sqrt(two) Given v_term = 60 m/s^two Answer: v = forty two m/s