Question

A lawn sprinkler has horizontal arms of radius \( a = 10\,\text{cm} = 0.1\,\text{m} \). Water exits through jets of area \( A = 25\,\text{cm}^2 = 2.5\times10^{-3}\,\text{m}^2 \) with velocity \( V = 1\,\text{m/s} \). Each jet leaves orthogonal to the arm and makes a \( 60^\circ \) angle with the horizontal. Find the torque (N·m) required to hold the sprinkler stationary. (Round off to two decimal places).

Hint:

Only the horizontal component of jet velocity produces torque. For two arms, multiply the torque by 2.

Answer 1

Correct Answer: 0.24

Mass flow rate is:
\[ \dot{m} = \rho A V = 1000(2.5 \times 10^{-3})(1) = 2.5\,\text{kg/s} \] Horizontal velocity component:
\[ V_h = V\cos 60^\circ = 1 \times 0.5 = 0.5 \, \text{m/s} \] Force due to momentum change:
\[ F = \dot{m} V_h = 2.5 \times 0.5 = 1.25\,\text{N} \] Torque from one arm:
\[ \tau_1 = F a = 1.25 \times 0.1 = 0.125\,\text{N·m} \] Since the sprinkler has two identical arms:
\[ \tau_{\text{total}} = 2 \times 0.125 = 0.25\,\text{N·m} \]