Time taken to reach 90% of terminal velocity

AI Thread Summary
To determine the time taken for a 4.5 kg metal sphere to reach 90% of its terminal velocity in a fluid with a drag coefficient of 10.5 N s^2/m^s, the terminal velocity is calculated as approximately 2.0 m/s, with 90% being 1.8 m/s. The acceleration is not constant due to the drag force, complicating the use of standard kinematic equations. A numerical solution using a spreadsheet can be effective by applying constant acceleration formulas over small time intervals. Alternatively, the problem can be approached using calculus to set up a differential equation, though this method may be more complex. Understanding both numerical and calculus-based methods is essential for solving this type of motion problem.
newageanubis
Messages
15
Reaction score
0

Homework Statement


A 4.5 kg metal sphere is released in a fluid where k = 10.5 N s^2/m^s. How long does it take to reach 90% of its terminal velocity?

Homework Equations


Force of drag = kv^2, where k is the drag coefficient
(I believe we're not considering buoyancy.)

The Attempt at a Solution


ƩF = ma
a = (mg-kv^2)/m
a = [(4.5)((9.8) - (10.5)v^2]/4.5
a = 9.8 - 2.3v^2

At terminal velocity,
ƩF = 0
mg = kv^2
v = √(mg/k)
v = √[(4.5)(9.8)/10.5]
v = 2.0 m/s [down]
90% of this is 1.8 m/s [down].

So I know I have to find the time taken for the ball to achieve a velocity of 1.8 m/s^2 [down], and a have an equation with both acceleration and velocity. However, acceleration is not constant, so all of my kinematics knowledge (the constant acceleration equations) are useless, so I don't know how to proceed. We're supposed to create a graph to get the answer, but my teacher said there's a way to do this without graphing. I'm trying to find out what this method is. Thanks!
 
Last edited:
Physics news on Phys.org
Disregard what I had here before, misread what time you were trying to find. Not sure what other way to do it.
 
Last edited:
I don't see how to get it with a graph! I would be very interested in seeing how you do it.

I did solve it numerically on a spreadsheet, with headings for t, d, v and a.
You CAN use the constant accelerated motion formulas for short time intervals (the smaller the interval the more accurate they are, and you can make the time interval between rows on the spreadsheet as small as you like; keep decreasing it until the answer no longer changes significantly).

It should be possible to solve it with calculus, Replace a with dv/dt and it becomes a differential equation. Not an easy one, though. I don't know how to solve it.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top