Question

A lawn sprinkler has three spouts that spray water, each \(15.0 \mathrm{cm}\) long. As the water is sprayed, the sprinkler turns around in a circle. The sprinkler has a total rotational inertia of $9.20 \times 10^{-2} \mathrm{kg} \cdot \mathrm{m}^{2} .\( If the sprinkler starts from rest and takes \)3.20 \mathrm{s}$ to reach its final speed of 2.2 rev/s, what force does each spout exert on the sprinkler?

Short answer

Answer: The force exerted by each spout on the sprinkler is approximately \(0.444\pi \, \text{N}\).

Step-by-step solution

Find the angular acceleration

We are given the initial angular velocity (\(\omega_i\)), final angular velocity (\(\omega_f\)), and the time it takes to reach the final angular velocity (t). We can use the formula for angular acceleration, which is: $$ \alpha = \frac{\omega_f - \omega_i}{t} $$ The initial angular velocity is 0 (starts from rest), and the final angular velocity is 2.2 rev/s. To convert rev/s to rad/s, we multiply by \(2\pi\): $$ \omega_f = 2.2 \times 2\pi \, \text{rad/s} =4.4\pi \, \text{rad/s} $$ Now, we can find the angular acceleration: $$ \alpha = \frac{4.4\pi - 0}{3.20} = \frac{22\pi}{16} \, \text{rad/s}^2 $$

Find the total torque acting on the sprinkler

Now that we have the angular acceleration, we can find the torque acting on the sprinkler, using the formula: $$ \tau = I\alpha $$ where \(\tau\) is the torque, \(I\) is the rotational inertia (\(9.20 \times 10^{-2} \mathrm{kg} \cdot \mathrm{m}^{2}\)), and \(\alpha\) is the angular acceleration. Plugging in the values, we have: $$ \tau = (9.20 \times 10^{-2})\left(\frac{22\pi}{16}\right) = \frac{1014\pi \times 10^{-2}}{16} \, \text{N}\cdot\text{m} $$

Calculate the force exerted by each spout on the sprinkler

Since the torque is applied by the three spouts, we can find the force exerted by each spout (\(F\)) by using the radius of the sprinkler (\(r = 0.15\,\text{m}\)), and the total torque on the sprinkler, using the formula: $$ \tau = F \times r $$ Now, we rearrange the formula to solve for the force exerted by each spout: $$ F = \frac{\tau}{3r} $$ Plugging in the values, we have: $$ F = \frac{\frac{1014\pi \times 10^{-2}}{16}}{3 \times 0.15} = \frac{1014\pi \times 10^{-2}}{7.2} = 0.444\pi \, \text{N} $$ Thus, the force exerted by each spout on the sprinkler is approximately \(0.444\pi \, \text{N}\).