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At the instant shown, the watersprinkler is rotating with an angular speed \(\dot{\theta}=2 \mathrm{rad} / \mathrm{s}\) and an angular acceleration \(\ddot{\theta}=3 \mathrm{rad} / \mathrm{s}^{2}\). If the nozzle lies in the vertical plane and water is flowing through it at a constant rate of \(3 \mathrm{~m} / \mathrm{s}\), determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, \(r=0.2 \mathrm{~m}\).

Short Answer

Expert verified
The magnitude of the velocity of the water particle as it exits the sprinkler is calculated to be 3.04 m/s and the acceleration is found to be 1 m/s².

Step by step solution

01

Calculate the Tangential Velocity

The tangential velocity \(v_t\) is given by the product of the rotation speed \(\dot{\theta}\) and the radius \(r\). So, we can plug in the given values to get \(v_t = \dot{\theta} \cdot r = 2 \mathrm{rad/s} \cdot 0.2 \mathrm{m} = 0.4 \mathrm{m/s}\).
02

Calculate the Radial Velocity

The radial velocity is simply the flow rate of water, given as 3 m/s.
03

Find the Resultant Velocity

The total velocity is the vector sum of the tangential and radial velocities. Because they are perpendicular, we can use the Pythagorean theorem to find their sum: \(v = \sqrt{(v_r)^2 + (v_t)^2} = \sqrt{(3 \mathrm{m/s})^2 + (0.4 \mathrm{m/s})^2} = 3.04 \mathrm{m/s}\).
04

Calculate the Radial Acceleration

The radial acceleration \(a_r\) is given by \(r \cdot (\dot{\theta})^2\). Substituting the given values, we get \(a_r = 0.2 \mathrm{m} \cdot (2 \mathrm{rad/s})^2 = 0.8 \mathrm{m/s}^2\).
05

Calculate the Tangential Acceleration

The tangential acceleration \(a_t\) is given by the product of the angular acceleration and the radius \(\ddot{\theta} \cdot r\). Plugging in the given values, we get \(a_t = 3 \mathrm{rad/s}^2 \cdot 0.2 \mathrm{m} = 0.6 \mathrm{ m/s}^2\).
06

Find the Resultant Acceleration

Like the velocity, the total acceleration is the vector sum of the tangential and radial accelerations. We can again use the Pythagorean theorem to find it: \(a = \sqrt{(a_r)^2 + (a_t)^2} = \sqrt{(0.8 \mathrm{m/s}^2)^2 + (0.6 \mathrm{m/s}^2)^2} = 1 \mathrm{m/s}^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Motion
Angular motion is all about how an object rotates around a specific point or axis. It's like watching a merry-go-round spin. Just like linear motion has speed and acceleration, angular motion has angular speed and angular acceleration. The angular speed, often denoted \(\dot{\theta}\), tells us how fast something is spinning. For instance, in our water sprinkler scenario, the angular speed is 2 rad/s, meaning it rotates 2 radians in one second.

Angular acceleration, represented as \(\ddot{\theta}\), shows how quickly the speed of the spin is changing. In the exercise, this is 3 rad/s², which means every second, the speed of the spin increases by 3 rad/s. Understanding these two concepts, angular speed and acceleration, is crucial in predicting how things rotate and change over time.
Tangential Velocity
Tangential velocity relates to the speed of an object moving along a circular path. Imagine holding onto a spinning roundabout at a playground; the tangential speed is how fast you are moving along the edge.

It's determined by the formula:
  • \(v_t = \dot{\theta} \cdot r\)
Where \(v_t\) is the tangential velocity, \(\dot{\theta}\) is the angular velocity, and \(r\) is the radius.

For the sprinkler, it has a radial distance of 0.2 m and an angular speed of 2 rad/s. So, the tangential velocity is calculated as 0.4 m/s. This means the water droplets are moving at 0.4 m/s along the path that follows the edge of their circular trajectory.
Radial Acceleration
Radial acceleration, also called centripetal acceleration, keeps an object moving in its circular path. It points toward the center of the rotation, acting like a leash that prevents the object from flying off.

The formula to calculate radial acceleration is:
  • \(a_r = r \cdot (\dot{\theta})^2\)
For the sprinkler nozzle, with a radius (r) of 0.2 m and an angular velocity (\(\dot{\theta}\)) of 2 rad/s, the radial acceleration is determined as 0.8 m/s². This is the force keeping the water droplets moving in their circular paths, ensuring they follow the spray trajectory instead of shooting straight off into the distance.
Pythagorean Theorem in Physics
The Pythagorean theorem is a powerful tool not just in geometry, but also in physics, especially when dealing with vectors. It helps us find the resultant of two perpendicular quantities, such as forces, velocities, or accelerations.

For instance, to find the total velocity or acceleration in the case of a water particle exiting a sprinkler, you sum the tangential and radial components. Since they form a right-angle vector triangle, the Pythagorean theorem applies:
  • \(v = \sqrt{(v_r)^2 + (v_t)^2}\)
  • \(a = \sqrt{(a_r)^2 + (a_t)^2}\)
Here, \(v\) and \(a\) represent the total velocity and acceleration, while \(v_r\), \(v_t\), \(a_r\), and \(a_t\) are the radial and tangential components respectively. It allows us to compute these resultant values— 3.04 m/s for velocity and 1 m/s² for acceleration—showing how the effects of spinning and flowing work together.

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Most popular questions from this chapter

The race car has an initial speed \(v_{A}=15 \mathrm{~m} / \mathrm{s}\) at \(A\). If it increases its speed along the circular track at the rate \(a_{t}=(0.4 s) \mathrm{m} / \mathrm{s}^{2},\) where \(s\) is in meters, determine the time needed for the car to travel \(20 \mathrm{~m}\). Take \(\rho=150 \mathrm{~m}\).

A man can row a boat at \(5 \mathrm{~m} / \mathrm{s}\) in still water. He wishes to cross a 50 -m-wide river to point \(B, 50 \mathrm{~m}\) downstream. If the river flows with a velocity of \(2 \mathrm{~m} / \mathrm{s}\) determine the speed of the boat and the time needed to make the crossing.

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The velocity of a particle is \(\mathbf{v}=\\{3 \mathbf{i}+(6-2 t) \mathbf{j}\\} \mathrm{m} / \mathrm{s},\) where \(t\) is in seconds. If \(\mathbf{r}=\mathbf{0}\) when \(t=0\), determine the displacement of the particle during the time interval \(t=1 \mathrm{~s}\) to \(t=3 \mathrm{~s}\)

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