Question

A spring with spring constant 35 N/m is attached to the ceiling, and a 5.5-cm-diameter, 1.3 kg metal cylinder is attached to its lower end. The cylinder is held so that the spring is neither stretched nor compressed, then a tank of water is placed underneath with the surface of the water just touching the bottom of the cylinder. When released, the cylinder will oscillate a few times but, damped by the water, quickly reach an equilibrium position.

Requried:
When in equilibrium, what length of the cylinder is submerged?

Answer 1

Step-by-step explanation:

Let length of cylinder submerged be L .

Upward force due to restoration force of spring = k L

= 35 L .

Buoyant force by water in upward direction = Vρg

V is volume of water displaced , ρ is density of water , g is acceleration due to gravity .

V = π R²L , R is radius of cylinder

V = 3.14 x (2.75 x 10⁻²)² x L

= 23.75 x 10⁻⁴ L

Buoyant force by water in upward direction =23.75 x 10⁻⁴ L X 1000 X 9.8

= 23.275L N

Total upward force = 35 L + 23.275L

= 58.275 L

For equilibrium ,

Total upward force = weight of cylinder

58.275 L = 1.3 x 9.8

L = .2186 m

= 21.86 cm