Question

Consider water being drained from a cylindrical container of diameter D through a hole in the cap, of diameter d, as shown below. Let A be a point on the surface of the water and let B be a point right at the hole. The level of water is h above the hole.

Water flowing into bucket

If the height of the water level is h = 24 cm, what is the value of vB2−vA2, in SI units?

If the diameters are d = 0.8 cm,and D = 11 cm,, what is the ratio of the speeds, v_B/v_A?

Find equation 3! Then compute the acceleration of the surface for diameters d = 1.1 cm and D = 10 cm, in m/s²? Do not forget the sign!

Numerical check: What is the drainage time, in s, for a container with diameters D = 12 cm and d = 0.6 cm, that is initially filled to a height of h = 16 cm above the draining hole?

In the lab, we will be dealing with a bottle with a diameter around 9 cm. A typical diameter of the drainage hole is 11 mm. In other words, D is much larger than d, and therefore the “-1” in the expression above is quite irrelevant. What is the percentage error that we introduce if we ignore the “-1” in the expression above?

Could you please answer Questions 3 - 7?

Answer 1

Question 3 :

V_B² - V_A² = 2 gh

V_B² - V_A² = 2 x 9.8 x 0.24

V_B² - V_A² = 4.704

Question 4 :

v_B/v_A = D²/d²

v_B/v_A = (11/0.8)²

v_B/v_A = 189.0625

Question 5 :

- 2 a y = 2gh / ((D/d)⁴ - 1)

since y = h

- 2 a h = 2gh / ((D/d)⁴ - 1)

a = - g / ((D/d)⁴ - 1)

a = - 9.8 / ((10/1.1)⁴ - 1)

a = - 0.00144 m/s²

Question 6 :

v = v_A + at

0 - v_A = a t

- sqrt(2gh/((D/d)⁴ - 1)) = - (g / ((D/d)⁴ - 1)) t

2gh/((D/d)⁴ - 1) = (g² / ((D/d)⁴ - 1)²) t²

2h = (g / ((D/d)⁴ - 1)) t²

t = sqrt(2 h ((D/d)⁴ - 1)/g)

t = sqrt(2 (0.16) ((12/0.6)⁴ - 1)/9.8)

t = 72.3 sec