Problem 98 - Fundamentals of Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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:\[ \omega = \frac{2\pi}{7.00} \approx 0.897 \text{ rad/s} \]

Step 2: 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:\[ r = \sqrt{(2.00)^2 + (-3.00)^2} = \sqrt{4 + 9} = 3.61 \mathrm{~m} \]

Step 3: 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 \( v \):\[ v = 0.897 \cdot 3.61 \approx 3.241 \text{ m/s} \]

Step 4: 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):\[ \begin{align*} v_x &= -3.00 \text{ m} \times \frac{v}{r}, \ v_y &= 2.00 \text{ m} \times \frac{v}{r}. \end{align*} \]

Step 5: Calculate the Velocity Components

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

Step 6: Express the Velocity in Unit-Vector Notation

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

Key Concepts

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.