Terminal Velocity In Glycerin 2.10 in Classical Mechanics

In summary, the problem involves a steel ball bearing being dropped in glycerin. The dominant drag force is linear drag, given by flin = 3*pi*n*D*v, where D is the sphere's diameter, v is velocity, and n is the viscosity of the fluid. The problem asks to find the characteristic time and terminal speed of the ball bearing, as well as the time it takes to reach 95% of its terminal velocity. It also introduces the concept of Archimedes buoyant force as a third force and asks to compute the ratio of quadratic drag force to linear drag force at terminal speed.
  • #1
thomaslanny
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0
Terminal Velocity In Glycerin 2.10 in "Classical Mechanics"

Homework Statement


For a steel ball bearing (diameter 2mm and density 7.8g/cm3) dropped in glycerin (density 1.3g/cm3 and viscosity 12 N s/m2 at STP) the dominant drag force is linear drag given by flin = 3*pi*n*D*v where D is the sphere's diameter, v is velocity, n is the viscosity of the fluid
a) Find the Characteristic time and terminal speed vter. Include Archimedes buoyant force as a 3rd force.
b) How long after it is dropped from rest will the ball bearing have reached 95% of its terminal velocity
c) Use flin = 3*pi*n*D*v and fquad=kpAv2 (p being density) with k = 1/4 and compute fquad/flin at the terminal speed


Homework Equations


v(t) where t = characteristic time = 0.63vter
vter = g*t
0.95vter = 3t where t = characteristic time
Fbouyancy = (pi/6)d3p*g (p being density of fluid)


The Attempt at a Solution



So far I've gotten for part a)
(3.2672x10-8 kg)*9.8m/s = 3.202E-7 N (for gravitational force)
3*pi*n*D*v = 3.202E-7N - (pi/6)d3p*g

Is this correct? Solve for v above and that's terminal velocity? if so I'm good for the rest of the problem.
 
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  • #2


Looks A-OK. I did not check your gravitational force number.
 
  • #3


Yeah it's actually 3.2019E-4 N for the F... my mistake
 

What is terminal velocity in glycerin 2.10?

Terminal velocity in glycerin 2.10 is the maximum velocity that an object can reach when falling through a viscous liquid such as glycerin. It is the point at which the drag force from the liquid equals the force of gravity on the object.

How is terminal velocity in glycerin 2.10 calculated?

The equation for calculating terminal velocity in glycerin 2.10 is Vt = (2mg)/ρAC, where m is the mass of the object, g is the acceleration due to gravity, ρ is the density of the glycerin, A is the cross-sectional area of the object, and C is the drag coefficient.

What factors affect terminal velocity in glycerin 2.10?

The factors that affect terminal velocity in glycerin 2.10 include the mass and size of the object, the density and viscosity of the glycerin, and the shape and surface area of the object. These all impact the drag force and therefore the terminal velocity.

Can terminal velocity in glycerin 2.10 be altered?

Yes, terminal velocity in glycerin 2.10 can be altered by changing the properties of the object or the glycerin. For example, increasing the mass or surface area of the object will increase the terminal velocity, while increasing the viscosity of the glycerin will decrease the terminal velocity.

What is the significance of studying terminal velocity in glycerin 2.10?

Studying terminal velocity in glycerin 2.10 is important for understanding the behavior of objects in viscous fluids, which is relevant in many industrial and natural processes. It also helps to demonstrate the principles of drag and terminal velocity in classical mechanics.

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