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Chapter 8: Problem 50

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

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

Step by step solution

01

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: $$ \alpha = \frac{4.4\pi - 0}{3.20} = \frac{22\pi}{16} \, \text{rad/s}^2 $$
02

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: $$ \tau = (9.20 \times 10^{-2})\left(\frac{22\pi}{16}\right) = \frac{1014\pi \times 10^{-2}}{16} \, \text{N}\cdot\text{m} $$
03

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 \(0.444\pi \, \text{N}\).

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Most popular questions from this chapter

A stone used to grind wheat into flour is turned through 12 revolutions by a constant force of \(20.0 \mathrm{N}\) applied to the rim of a 10.0 -cm-radius shaft connected to the wheel. How much work is done on the stone during the 12 revolutions?
A bowling ball made for a child has half the radius of an adult bowling ball. They are made of the same material (and therefore have the same mass per unit volume). By what factor is the (a) mass and (b) rotational inertia of the child's ball reduced compared with the adult ball?
The angular momentum of a spinning wheel is $240 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ After application of a constant braking torque for \(2.5 \mathrm{s}\), it slows and has a new angular momentum of $115 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ What is the torque applied?
A bicycle wheel, of radius \(0.30 \mathrm{m}\) and mass \(2 \mathrm{kg}\) (concentrated on the rim), is rotating at 4.00 rev/s. After 50 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?
A hollow cylinder, a uniform solid sphere, and a uniform solid cylinder all have the same mass \(m .\) The three objects are rolling on a horizontal surface with identical transnational speeds \(v .\) Find their total kinetic energies in terms of \(m\) and \(v\) and order them from smallest to largest.
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