Question

You have a glass ball with a radius of 2.00 mm and a density of 2500 kg/m3. You hold the ball so it is fully submerged, just below the surface, in a tall cylinder full of glycerin, and then release the ball from rest. Take the viscosity of glycerin to be 1.5 Pa s and the density of glycerin to be 1250 kg/m3. Use g = 10 N/kg = 10 m/s2. Also, note that the drag force on a ball moving through a fluid is: Fdrag = 6πηrv . (a) Note that initially the ball is at rest. Sketch (to scale) the free-body diagram of the ball just after it is released, while its velocity is negligible. (b) Calculate the magnitude of the ball’s initial acceleration. (c) Eventually, the ball reaches a terminal (constant) velocity. Sketch (to scale) the free-body diagram of the ball when it is moving at its terminal velocity. (d) Calculate the magnitude of the terminal velocity. (e) What is the magnitude of the ball’s acceleration, when the ball reaches terminal velocity? (f) Let’s say that the force of gravity acting on the ball is 4F, directed down. We can then express all the forces in terms of F.

Answer 1

(a) check attachment

(b)5 m/s²

Step-by-step explanation:

Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv

(a) for answer check attachment.

(b) For the magnitude of the balls initial acceleration:

Initial net force(f) = mg - upthrust

=
`mg - ((m)/(p) )pg.g`

acceleration (a) =
`Acceleration(a)=(f)/(m)\\=g - ((pg)/(p))g\\=g(1-(pg)/(p) )\\=10(1-(1250)/(2500) )\\a=10(1-0.5)\\a=5 m/s²`

c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)

`mg=6πnrv + upthrust`

d.) For the magnitude of terminal velocity:

`mg=6πnrv + ((m)/(p))pg.g\\\\((4)/(3)πr^(3) p)g=6πnrv +(4)/(3)πr^3pg.g\\\\V = (2)/(9).((2*10^(-3))^(2)*(2500-1250)*10)/(1.5)\\\\=0.79cm/s`

e.) when the ball reaches terminal velocity, the acceleration is zero (0)