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Ch. 05 - Using Newton's Laws: Friction, Circular Motion, Drag Forces
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 05 - Using Newton's Laws: Friction, Circular Motion, Drag ForcesProblema 62b
Capitolo 5, Problema 62b

The position of a particle moving in the xy plane is given by r→\(\overrightarrow{r}\) = (2.0m) cos [(3.0 rad/s)t ] i^\(\hat{i}\) +(2.0m) sin [(3.0 rad/s)t ] j^\(\hat{j}\), where r is in meters and t is in seconds. Calculate the velocity and acceleration vectors as functions of time.

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Start by understanding the given position vector: \( \vec{r}(t) = (2.0 \text{ m}) \cos[(3.0 \text{ rad/s})t] \hat{i} + (2.0 \text{ m}) \sin[(3.0 \text{ rad/s})t] \hat{j} \). This describes the motion of the particle in the xy-plane as a function of time.
To find the velocity vector \( \vec{v}(t) \), take the time derivative of the position vector \( \vec{r}(t) \). Use the derivative rules for trigonometric functions: \( \frac{d}{dt}[\cos(\omega t)] = -\omega \sin(\omega t) \) and \( \frac{d}{dt}[\sin(\omega t)] = \omega \cos(\omega t) \).
Apply the derivative to each component of \( \vec{r}(t) \): \( \frac{d}{dt}[(2.0 \text{ m}) \cos(3.0t)] = -(2.0 \text{ m})(3.0 \text{ rad/s}) \sin(3.0t) \) for the \( \hat{i} \)-component, and \( \frac{d}{dt}[(2.0 \text{ m}) \sin(3.0t)] = (2.0 \text{ m})(3.0 \text{ rad/s}) \cos(3.0t) \) for the \( \hat{j} \)-component.
Combine the results to write the velocity vector: \( \vec{v}(t) = -(6.0 \text{ m/s}) \sin(3.0t) \hat{i} + (6.0 \text{ m/s}) \cos(3.0t) \hat{j} \).
To find the acceleration vector \( \vec{a}(t) \), take the time derivative of the velocity vector \( \vec{v}(t) \). Use the same derivative rules for trigonometric functions. For the \( \hat{i} \)-component, \( \frac{d}{dt}[-(6.0 \text{ m/s}) \sin(3.0t)] = -(6.0 \text{ m/s})(3.0 \text{ rad/s}) \cos(3.0t) \), and for the \( \hat{j} \)-component, \( \frac{d}{dt}[(6.0 \text{ m/s}) \cos(3.0t)] = -(6.0 \text{ m/s})(3.0 \text{ rad/s}) \sin(3.0t) \). Combine these to write \( \vec{a}(t) = -(18.0 \text{ m/s}^2) \cos(3.0t) \hat{i} - (18.0 \text{ m/s}^2) \sin(3.0t) \hat{j} \).

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Position Vector

The position vector describes the location of a particle in space relative to a reference point. In this case, the position vector r→ is expressed in terms of its components along the x and y axes, using trigonometric functions to indicate circular motion. Understanding the position vector is crucial for deriving other motion-related quantities such as velocity and acceleration.
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Final Position Vector

Velocity Vector

The velocity vector represents the rate of change of the position vector with respect to time. It is calculated by taking the derivative of the position vector r→ with respect to time t. In this scenario, the velocity will also exhibit a periodic nature due to the trigonometric functions involved, reflecting the particle's circular motion in the xy plane.
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Acceleration Vector

The acceleration vector indicates the rate of change of the velocity vector with respect to time. It is obtained by differentiating the velocity vector. For a particle in circular motion, the acceleration can be both tangential and centripetal, and understanding its components is essential for analyzing the dynamics of the particle's motion in the xy plane.
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