Problem 122 The position of a particle is gi... [FREE SOLUTION]

Chapter 6: Problem 122

The position of a particle is given by $$ \overrightarrow{\mathbf{r}}(t)=A(\cos \omega t \hat{\mathbf{i}}+\sin \omega t \hat{\mathbf{j}}), \quad \text { where } \quad \omega \quad \text { is } \quad \mathrm{a} $$ constant. (a) Show that the particle moves in a circle of radius $A$. (b) Calculate $d \overrightarrow{\mathbf{r}} / d t$ and then show that the speed of the particle is a constant $A_{\omega^{*}}$ (c) Determine $d^{2} \overrightarrow{\mathbf{r}} / d t^{2}$ and show that $a$ is given by $a_{\mathrm{c}}=r \omega^{2}$ Calculate the centripetal force on the particle. [Hint: For (b) and (c), you will need to use $(d / d t)(\cos \omega t)=-\omega \sin \omega t$ and $(d / d t)(\sin \omega t)=\omega \cos \omega t$

Short Answer

Expert verified

The particle moves in a circular path of radius A, as shown by the equation $x^2+y^2=A^2$. The speed of the particle is constant and equal to $A\omega$. The acceleration vector is given by $a = A\omega^2$, and the centripetal force acting on the particle is $F_c = mA\omega^2$.

Step by step solution

Understanding the given position vector

The position vector $\overrightarrow{\mathbf{r}}(t) $ is given by: $\overrightarrow{\mathbf{r}}(t) = A(\cos \omega t \hat{\mathbf{i}} + \sin \omega t \hat{\mathbf{j}})$ Both coordinates are functions of t, and $\omega$ is a constant. We can analyze the motion in two dimension separately to see if it moves in a circle or not.

Proving that the path is a circle

We know that a circle with a radius A can be defined as follows: $x^2 + y^2 = A^2$ Here, $x = A \cos \omega t$ and $y = A \sin \omega t$ from the position vector. Squaring and adding the equations, we get: $(A \cos \omega t)^2 + (A \sin \omega t)^2 = A^2 (\cos^2 \omega t + \sin^2 \omega t)$ Since $\cos^2 \omega t + \sin^2 \omega t = 1$, we can simplify the equation to: $A^2 = A^2$ And hence this proves that the particle moves in a circle with a radius A.

Differentiating position vector

To calculate the speed, we need to calculate the first derivative of the position vector: $\frac{d \overrightarrow{\mathbf{r}}}{dt} = \frac{d}{dt} (A(\cos \omega t \hat{\mathbf{i}} + \sin \omega t \hat{\mathbf{j}}))$ Using the given hints: $\frac{d \overrightarrow{\mathbf{r}}}{dt} = A(-\omega\sin \omega t \hat{\mathbf{i}} + \omega\cos \omega t \hat{\mathbf{j}})$ And the speed of the particle is given by the magnitude of this: $v = A\omega$ This shows that the speed of the particle is constant and equal to $A\omega$.

Calculating the second derivative of the position vector

To calculate the acceleration, we need the second derivative of the position vector: $\frac{d^2\overrightarrow{\mathbf{r}}}{dt^2} = \frac{d}{dt}(A(-\omega\sin \omega t \hat{\mathbf{i}} + \omega\cos \omega t \hat{\mathbf{j}}))$ Differentiating again, we get: $\frac{d^2\overrightarrow{\mathbf{r}}}{dt^2} = A\omega^2(-\cos \omega t \hat{\mathbf{i}} - \sin \omega t \hat{\mathbf{j}})$ And the magnitude of this acceleration vector is: $a = A\omega^2$

Calculating the centripetal force

Since we have validated that the particle moves in a circle of radius A, the centripetal force Fc can be calculated using the following equation considering the mass of particle to be m: $F_c = m a_c = m r \omega^2$ Substituting the radius value (A) and the calculated acceleration (A蠅虏), we get: $F_c = mA\omega^2$ This gives us the centripetal force acting on the particle in circular motion.

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Key Concepts

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

Centripetal Force

Centripetal force is the invisible tether that keeps an object moving in a circular path, always pointing towards the center of the circle. In our exercise, the centripetal force is what maintains the particle's circular motion.

To understand this force better, let's look at its equation, which is derived from Newton's second law: $ F = m a $. For an object moving in a circle, the acceleration that appears in this equation is not a linear acceleration, but a centripetal (center-seeking) acceleration. This means that the force does not act to move the object along the path of the circle but keeps pulling it inward, preventing it from moving off in a straight line.

In mathematical terms, the centripetal force ($ F_c $) acting on an object of mass (m) moving at a velocity (v) along a path with radius (r) is given by:\[ F_c = m \frac{v^2}{r} \]Alternatively, when you have the angular velocity ($ \omega $), the formula becomes:\[ F_c = m r \omega^2 \]
Thus, in the context of the exercise, once you calculate the object鈥檚 centripetal acceleration, multiplying it by the object's mass gives us the centripetal force. It鈥檚 essential to remember that this force doesn't exist on its own but is the result of other forces, like tension or gravity, depending on the context of the problem.

Uniform Circular Motion

Uniform circular motion describes the movement of an object along a circular path with a constant speed. It is 'uniform' because the speed does not change, although the object's velocity is constantly changing direction. In the exercise, when we show the speed of the particle remains constant ($ A\omega $), we confirm the motion is uniform circular motion.

Uniform does not mean unchanging in all aspects. An interesting paradox lies at the heart of uniform circular motion: even though the speed is constant, the velocity is not. This is because velocity is a vector quantity 鈥� it has both magnitude and direction. Since the direction is constantly changing due to the particle's circular path, the velocity is not constant.

Uniform circular motion is a great example of a concept that appears simple on the surface but holds complex, foundational ideas in physics. Understanding this concept fully means recognizing the role of velocity, acceleration (specifically centripetal acceleration), and the forces that give rise to this sort of motion.

Angular Velocity

Angular velocity, denoted by ($ \omega $), is a vector quantity that represents how fast an object rotates or revolves relative to another point 鈥� in this case, the center of the circular path. It's measured in radians per second (rad/s).

In these sorts of problems, angular velocity provides a link between the linear dimensions of motion (like speed and radius) and the rotational aspect. The equation connecting linear and angular velocities is:\[ v = r\omega \]where (v) is the linear speed, (r) is the radius of the circular path, and ($ \omega $) is the angular velocity. This equation comes in handy when solving for the speed of an object in circular motion, as we did in our exercise.

Understanding angular velocity is critical, especially in tasks such as calculating the rotational kinetic energy of an object, determining the period of revolution, and even in complex engineering where it鈥檚 necessary to predict the behavior of rotating machinery and systems. Remember, angular velocity is all about rotation鈥攅very point on the object will have the same angular velocity, but not the same linear speed unless the points are equidistant from the axis of rotation.

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Short Answer

Step by step solution

Understanding the given position vector

Proving that the path is a circle

Differentiating position vector

Calculating the second derivative of the position vector

Calculating the centripetal force

Key Concepts

Centripetal Force

Uniform Circular Motion

Angular Velocity

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