The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water is supplied through the hose at 0.006 m3/s and is ejected through the four nozzles into the atmosphere. Determine the constant angular velocity of the arms. Neglect friction at the vertical axis.

the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec

So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

V_single = 29.8566 m/s

Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

ω_single = Vt / r

from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

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