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A particle is in uniform circular motion about the origin of an \(x y\) coordinate system, moving clockwise with a period of \(7.00 \mathrm{~s}\). At one instant, its position vector (measured from the origin) is \(\vec{r}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}-(3.00 \mathrm{~m}) \hat{\mathrm{j}}\). At that instant, what is its velocity in unit-vector notation?

Short Answer

Expert verified
The velocity is \(-2.69 \hat{\mathrm{i}} + 1.79 \hat{\mathrm{j}} \text{ m/s}.\)

Step by step solution

01

Determine the Angular Velocity (ω)

The particle is in uniform circular motion with a given period of \( T = 7.00 \text{ s} \). The angular velocity \( \omega \) is given by the formula \( \omega = \frac{2\pi}{T} \). Calculate \( \omega \) as follows:\[ \omega = \frac{2\pi}{7.00} \approx 0.897 \text{ rad/s} \]
02

Calculate the Radius (r) of the Motion

The position vector of the particle at one instant is given by \( \vec{r} = (2.00 \mathrm{~m}) \hat{\mathrm{i}} - (3.00 \mathrm{~m}) \hat{\mathrm{j}} \). The magnitude of this vector represents the radius of the circular motion, calculated as:\[ r = \sqrt{(2.00)^2 + (-3.00)^2} = \sqrt{4 + 9} = 3.61 \mathrm{~m} \]
03

Determine the Tangential Velocity (v)

The tangential velocity \( v \) for uniform circular motion is calculated using \( v = \omega \cdot r \). Substitute the values found in the previous steps to find \( v \):\[ v = 0.897 \cdot 3.61 \approx 3.241 \text{ m/s} \]
04

Determine Direction of Velocity

The velocity vector is perpendicular to the position vector and follows the direction of the motion, which is clockwise. The direction can be determined by swapping the components of the position vector and changing the sign of the original \( x \)-component (due to clockwise motion):\[ \begin{align*} v_x &= -3.00 \text{ m} \times \frac{v}{r}, \ v_y &= 2.00 \text{ m} \times \frac{v}{r}. \end{align*} \]
05

Calculate the Velocity Components

Substitute \( v \approx 3.241 \text{ m/s} \) and \( r = 3.61 \text{ m} \) to find the components:\[ \begin{align*} v_x &= -3.00 \times \frac{3.241}{3.61} \approx -2.69 \text{ m/s}, \ v_y &= 2.00 \times \frac{3.241}{3.61} \approx 1.79 \text{ m/s}. \end{align*} \]
06

Express the Velocity in Unit-Vector Notation

Combine the calculated components to express the velocity in unit-vector notation:\[ \vec{v} \approx -2.69 \hat{\mathrm{i}} + 1.79 \hat{\mathrm{j}} \text{ m/s}. \]

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Key Concepts

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

Angular Velocity
Imagine you are twirling a ball on a string around your head. As it moves in a circle, the ball covers a certain angle every second. This rate of angle covered is known as angular velocity, denoted by \( \omega \). It describes how quickly an object traverses a circular path. The unit for angular velocity is radians per second (rad/s).

In the exercise, the particle completes a full round every 7 seconds. To find \( \omega \), we use the formula \( \omega = \frac{2\pi}{T} \). Here, \( T \) is the period — the time for one complete revolution — which is 7 seconds.

After plugging in the numbers: \[ \omega = \frac{2\pi}{7} \approx 0.897 \text{ rad/s} \] This tells us the particle moves through approximately 0.897 radians each second. Understanding angular velocity helps us predict how an object will move and when it will return to a certain point in its circular path.
Tangential Velocity
Tangential velocity is about the speed of an object moving along a circular path. Imagine stepping onto a merry-go-round. The faster you spin, the more you feel like being pushed outward — that's due to tangential velocity.

Tangential velocity, \( v \), is found using the formula: \( v = \omega \cdot r \). Here \( \omega \) is already known from our angular velocity calculation, and \( r \) is the radius of the motion. In this problem, the radius was calculated as 3.61 m based on the position vector's magnitude.

Finally, inserting the values: \[ v = 0.897 \times 3.61 \approx 3.241 \text{ m/s} \] This means that at every instant, the object moves along the circular path with a speed of about 3.241 m/s. This concept is vital in determining how quickly an object moves along its path, which is crucial for tasks involving rotating objects like wheels or planets.
Unit-Vector Notation
Unit-vector notation is a way to express vectors using components along mutually perpendicular axes, usually denoted by \( \hat{\mathrm{i}} \) and \( \hat{\mathrm{j}} \) for the x and y axes respectively. When we express vectors in this form, it becomes easier to compute their components separately.

In this exercise, the position vector of the particle is given by \( \vec{r} = (2.00 \mathrm{~m}) \hat{\mathrm{i}} - (3.00 \mathrm{~m}) \hat{\mathrm{j}} \). As the particle moves, its velocity vector is perpendicular to this position vector, following the motion's clockwise direction.

To determine the particle's velocity vector, we swap and change the sign of components, resulting in: \[ \vec{v} \approx -2.69 \hat{\mathrm{i}} + 1.79 \hat{\mathrm{j}} \text{ m/s} \] This notation helps us understand the velocity's components and direction in a clear and structured way. By using the components, one can efficiently calculate and predict motion in two-dimensional space.

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Most popular questions from this chapter

A frightened rabbit moving at \(6.00 \mathrm{~m} / \mathrm{s}\) due east runs onto a large area of level ice of negligible friction. As the rabbit slides across the ice, the force of the wind causes it to have a constant acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2}\), due north. Choose a coordinate system with the origin at the rabbit's initial position on the ice and the positive \(x\) axis directed toward the east. In unit-vector notation, what are the rabbit's (a) velocity and (b) position when it has slid for \(3.00 \mathrm{~s}\) ?

A magnetic field forces an electron to move in a circle with radial acceleration \(3.0 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} .\) (a) What is the speed of the electron if the radius of its circular path is \(15 \mathrm{~cm} ?\) (b) What is the period of the motion?

A soccer ball is kicked from the ground with an initial speed of \(19.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(45^{\circ} .\) A player \(55 \mathrm{~m}\) away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?

A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

\(A\) is located \(4.0 \mathrm{~km}\) north and \(2.5 \mathrm{~km}\) east of ship \(B\). Ship \(A\) has a velocity of \(22 \mathrm{~km} / \mathrm{h}\) toward the south, and ship \(B\) has a velocity of \(40 \mathrm{~km} / \mathrm{h}\) in a direction \(37^{\circ}\) north of east. (a) What is the velocity of \(A\) relative to \(B\) in unit-vector notation with \(\hat{\mathrm{i}}\) toward the east? (b) Write an expression (in terms of \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) ) for the position of \(A\) relative to \(B\) as a function of \(t\), where \(t=0\) when the ships are in the positions described above. (c) At what time is the separation between the ships least? (d) What is that least separation?

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