Question

A smooth ball that has a diameter of 100 mm is dropped in a tank filled with oil of density 900 kg/m and dynamic viscosity 0.036 Pa-s. The ball travels 70 cm in 2 seconds. If it is assumed that terminal speed has been reached. What is the density of the ball?

Answer 1

The density of the ball is 902.31
`\pmb{\cfrac{kg}{m^3}}`

Step-by-step explanation:

In order to determinate the density of the ball, under the assumption that terminal speed has been reached, Stoke's Law for terminal velocity can be used. But first we need to convert all units to SI units so we can find the terminal velocity.

Given information in SI units.

Diameter of the ball.

`d=100\,mm= 0.1 \, m`

Radius of the ball.

`r=\cfrac d2 \\ r =\cfrac{0.1}{2}\\ r=0.05 \, m`

Oil density.

`\rho_f = 900  \,\cfrac{kg}{m^3}`

Dynamic viscosity of the oil.

`\mu= 0.036 Pa\cdot s = 0.036 \cfrac{Ns}{m^2}`

Distance traveled

`x=70\, cm = 0.7 \, m`

Time

`t=2\,s`

Finding the terminal velocity.

Under the assumption that the ball has reached terminal velocity in that time then the average velocity will give us the terminal velocity that is

`v=\cfrac xt \\ v= \cfrac{0.7}{2} \\ v= 0.35\, \cfrac ms`

Finding density of the ball.

The Stoke's law for terminal velocity of sphere falling in a fluid is given by

`v = \cfrac 29 \cfrac{(\rho_b-\rho_f)}{\mu }g r^2`

where
`\rho_b` is the density of the ball.

We can solve for it and we get

`9v\mu = 2(\rho_b -\rho_f)gr^2`

Getting the density in one side.

`\cfrac{9v\mu}{2gr^2 }= \rho_b -\rho_f \\ \rho_b =\rho_f +\cfrac{9v\mu}{2gr^2 }`

Replacing values we get

`\rho_b =900 +\cfrac{9(0.35)(0.036)}{2(9.8)(0.05)^2 }`

`\boxed{\rho_b=902.31 \, \cfrac{kg}{m^3}}`

The density of the ball is 902.31
`\pmb{\cfrac{kg}{m^3}}`