Question

A marble is dropped into a graduated cylinder that is filled with water. Use the words

Answer 1

Answer:

v_t = $\frac{2 g \ ( \rho_b - \rho_l ) r^2 }{9 \ \eta }$

Explanation:

Let's describe the motion and forces on the marble when it falls into the graduated cylinder filled with water.

Initially, the marble in the air falls under the influence of its weight (gravity force), when it enters the water, a friction force appears that INCREASES with SPEED and opposes the movement, this force is called Stokes force.

             

              Fr = 6π R η v

where η is the viscosity of water

also because of this in a liquid it has an upward thrust, given by Archimedes' law, this thrust is CONSTANT

              B = ρ_l g V_l

in that case the volume of liquid(V_l)  is equal to the volume of the marble.

Constant force of gravity

               W = mg

using the concept of density

                ρ_body = m / V

                W = ρ_body g V_body

we apply Newton's second law

                Fr + B - W = ma

the friction force increases and the other two are constant, so acceleration decreases until reaching zero, at this point the forces are in BALANCE

                Fr + B - W = 0

and the speed remains constant, being called TERMINAL SPEED

                v_t = $\frac{2 g \ ( \rho_b - \rho_l ) r^2 }{9 \ \eta }$

from this point on, the displacement is proportional to the time