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Chapter 6: Problem 116

A particle of mass 0.50 kg starts moves through a circular path in the \(x y\) -plane with a position given by \(\overrightarrow{\mathbf{r}}(t)=(4.0 \cos 3 t) \hat{\mathbf{i}}+(4.0 \sin 3 t) \hat{\mathbf{j}} \quad\) where \(\quad r \quad\) is \(\quad\) in meters and \(t\) is in seconds. (a) Find the velocity and acceleration vectors as functions of time. (b) Show that the acceleration vector always points toward the center of the circle (and thus represents centripetal acceleration). (c) Find the centripetal force vector as a function of time.

Short Answer

Expert verified
The velocity vector is \(\overrightarrow{\mathbf{v}}(t) = (-12.0 \sin 3t) \hat{\mathbf{i}} + (12.0 \cos 3t) \hat{\mathbf{j}}\). The acceleration vector is \(\overrightarrow{\mathbf{a}}(t) = (-36.0 \cos 3t) \hat{\mathbf{i}} + (-36.0 \sin 3t) \hat{\mathbf{j}}\), and it points toward the center of the circular path, thus representing centripetal acceleration. The centripetal force vector as a function of time is \(\overrightarrow{\mathbf{F_c}}(t) = (-18.0 \cos 3t) \hat{\mathbf{i}} + (-18.0 \sin 3t) \hat{\mathbf{j}}\).

Step by step solution

01

To find the velocity vector, \(\overrightarrow{\mathbf{v}}(t)\) we need to differentiate the position vector, \(\overrightarrow{\mathbf{r}}(t)\) with respect to time. \[\overrightarrow{\mathbf{v}}(t) = \frac{d\overrightarrow{\mathbf{r}}(t)}{dt}\] Given, \(\overrightarrow{\mathbf{r}}(t)=(4.0 \cos 3 t) \hat{\mathbf{i}}+(4.0 \sin 3 t) \hat{\mathbf{j}}\), \[\overrightarrow{\mathbf{v}}(t) = \frac{d(4.0 \cos 3t) \hat{\mathbf{i}}}{dt} + \frac{d(4.0 \sin 3t) \hat{\mathbf{j}}}{dt}\] Now, differentiate the components of the position vector with respect to time: \[\overrightarrow{\mathbf{v}}(t) = (-12.0 \sin 3t) \hat{\mathbf{i}} + (12.0 \cos 3t) \hat{\mathbf{j}}\] So, the velocity vector is \(\overrightarrow{\mathbf{v}}(t) = (-12.0 \sin 3t) \hat{\mathbf{i}} + (12.0 \cos 3t) \hat{\mathbf{j}}\). #Step 2: Find the acceleration vector#

To find the acceleration vector, \(\overrightarrow{\mathbf{a}}(t)\) we need to differentiate the velocity vector, \(\overrightarrow{\mathbf{v}}(t)\) with respect to time. \[\overrightarrow{\mathbf{a}}(t) = \frac{d\overrightarrow{\mathbf{v}}(t)}{dt}\] Given, \(\overrightarrow{\mathbf{v}}(t) = (-12.0 \sin 3t) \hat{\mathbf{i}} + (12.0 \cos 3t) \hat{\mathbf{j}}\), \[\overrightarrow{\mathbf{a}}(t) = \frac{d(-12.0 \sin 3t) \hat{\mathbf{i}}}{dt} + \frac{d(12.0 \cos 3t) \hat{\mathbf{j}}}{dt}\] Now, differentiate the components of the velocity vector with respect to time: \[\overrightarrow{\mathbf{a}}(t) = (-36.0 \cos 3t) \hat{\mathbf{i}} + (-36.0 \sin 3t) \hat{\mathbf{j}}\] So, the acceleration vector is \(\overrightarrow{\mathbf{a}}(t) = (-36.0 \cos 3t) \hat{\mathbf{i}} + (-36.0 \sin 3t) \hat{\mathbf{j}}\). #Step 3: Show that the acceleration vector points toward the center of the circle#
02

The acceleration vector we found is \(\overrightarrow{\mathbf{a}}(t) = (-36.0 \cos 3t) \hat{\mathbf{i}} + (-36.0 \sin 3t) \hat{\mathbf{j}}\). Let's compare this to the position vector. Observe that the position vector \(\overrightarrow{\mathbf{r}}(t)\) and the acceleration vector \(\overrightarrow{\mathbf{a}}(t)\) are proportional and have opposite signs in their components: \[\overrightarrow{\mathbf{a}}(t) = -9 \overrightarrow{\mathbf{r}}(t)\] This means that the acceleration vector points in the opposite direction of the position vector, i.e., it points toward the center of the circular path. Thus, the acceleration vector represents centripetal acceleration. #Step 4: Find the centripetal force vector as a function of time#

We are given the mass of the particle, \(m = 0.50 \ kg\). We can find the centripetal force vector by using the centripetal acceleration vector and the mass: \[\overrightarrow{\mathbf{F_c}}(t) = m \overrightarrow{\mathbf{a}}(t)\] \[\overrightarrow{\mathbf{F_c}}(t) = (0.50) (-36.0 \cos 3t) \hat{\mathbf{i}} + (0.50) (-36.0 \sin 3t) \hat{\mathbf{j}}\] So, the centripetal force vector as a function of time is \(\overrightarrow{\mathbf{F_c}}(t) = (-18.0 \cos 3t) \hat{\mathbf{i}} + (-18.0 \sin 3t) \hat{\mathbf{j}}\).

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Key Concepts

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

Velocity Vector
When a particle moves through space, its velocity vector represents the direction and speed at any given moment. The velocity vector is a crucial concept in physics, as it allows us to understand how objects are moving. In the context of circular motion, the velocity vector is always tangential to the path of the object and it changes over time, even if the speed remains constant, because the direction of motion is continually changing. When we have a position vector, such as \(\overrightarrow{\mathbf{r}}(t)=(4.0 \cos 3 t) \hat{\mathbf{i}}+(4.0 \sin 3 t) \hat{\mathbf{j}} \), the velocity vector is simply the rate of change of the position. Hence, by differentiating the position vector with respect to time, we find how fast and in what direction the particle is moving at any instant.
Acceleration Vector
Acceleration is the rate at which the velocity of an object changes with time. The acceleration vector, thereby, gives us information about how the velocity changes. Not to be confused with speed, which is a scalar quantity, acceleration focuses on changes in velocity, which includes changes in both magnitude and direction. In the case of uniform circular motion, while the speed may not change, the velocity does due to changes in direction, resulting in an acceleration towards the center of the circle. By differentiating the velocity vector with respect to time, we obtain the acceleration vector \(\overrightarrow{\mathbf{a}}(t) = (-36.0 \cos 3 t) \hat{\mathbf{i}} + (-36.0 \sin 3 t) \hat{\mathbf{j}}\). This demonstrates the crucial principle that in any circular motion, there is always an inward acceleration, indicating the continuous change in direction of the object.
Circular Motion
Circular motion refers to the movement of an object along a curve that is at a constant distance from a central point, creating a circular path. In circular motion, even if the speed (the magnitude of the velocity) of the object remains constant, the object is still accelerating because the direction of the velocity vector is continually changing. This is a non-intuitive concept since we often associate acceleration only with changes in speed. In our exercise, the motion described by the position vector \(\overrightarrow{\mathbf{r}}(t)\) reflects such uniform circular motion, which can be depicted mathematically and graphed to illustrate how the particle moves in the xy-plane.
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path and is directed towards the center around which the object is moving. It is not a force in itself, but rather the result of other forces that are directed towards the circle's center, such as gravity, tension, or friction. The formula \(\overrightarrow{\mathbf{F_c}}(t) = m \overrightarrow{\mathbf{a}}(t)\) relates centripetal force to the acceleration vector and the mass of the particle. Since the acceleration vector points inward, so does the centripetal force. It's important to recognize that without a net centripetal force, an object cannot continue to follow a circular path, and understanding this force vector as a function of time is critical in analyzing the motion of objects in circular paths.

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Most popular questions from this chapter

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \(\overrightarrow{\mathbf{F}}\) so that the swing ropes are \(30.0^{\circ}\) with respect to the vertical. (a) Calculate the tension in each of the two ropes supporting the swing under these conditions. (b) Calculate the magnitude of \(\overrightarrow{\mathbf{F}}\)

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