Ch. 11 - Angular Momentum; General Rotation

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

Giancoli Douglas 5th edition

Ch. 11 - Angular Momentum; General Rotation

Problema 73

Capitolo 11, Problema 73

The time-dependent position of a point object which moves counterclockwise along the circumference of a circle (radius R) in the xy plane with constant speed υ is given by $\vec{r}$ = î R cos ωt + ĵ R sin ωt where the constant ω = v/R. Determine the velocity $\vec{v}$ and angular velocity $\vec{w}$ of this object and then show that these three vectors obey the relation $\vec{v} = \vec{ω} \times \vec{r}$ .

Guida verificata passo dopo passo

Start by understanding the given position vector: $ \mathbf{r}(t) = \hat{i} R \cos(\omega t) + \hat{j} R \sin(\omega t) $. This represents the position of the object in the xy-plane as a function of time, where $ R $ is the radius of the circle, and $ \omega $ is the angular velocity.

To find the velocity vector $ \mathbf{v}(t) $, take the time derivative of the position vector $ \mathbf{r}(t) $: $ \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \hat{i} (-R \omega \sin(\omega t)) + \hat{j} (R \omega \cos(\omega t)) $. Simplify this to $ \mathbf{v}(t) = -R \omega \sin(\omega t) \hat{i} + R \omega \cos(\omega t) \hat{j} $.

Next, determine the angular velocity vector $ \mathbf{\omega} $. Since the motion is counterclockwise in the xy-plane, the angular velocity vector points along the positive z-axis. Thus, $ \mathbf{\omega} = \omega \hat{k} $, where $ \hat{k} $ is the unit vector in the z-direction.

Now, verify the relationship $ \mathbf{v} = \mathbf{\omega} \times \mathbf{r} $. Compute the cross product $ \mathbf{\omega} \times \mathbf{r} $: $ \mathbf{\omega} \times \mathbf{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & \omega \\ R \cos(\omega t) & R \sin(\omega t) & 0 \end{vmatrix} $. Expand the determinant to get $ \mathbf{\omega} \times \mathbf{r} = (-R \omega \sin(\omega t)) \hat{i} + (R \omega \cos(\omega t)) \hat{j} $.

Compare the result of the cross product $ \mathbf{\omega} \times \mathbf{r} $ with the velocity vector $ \mathbf{v}(t) $. You will see that $ \mathbf{v}(t) = \mathbf{\omega} \times \mathbf{r}(t) $, confirming the given relationship.

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Circular Motion

Circular motion refers to the movement of an object along the circumference of a circle. In this context, the object moves with a constant speed, which means that while its speed remains unchanged, its direction is continuously changing. This change in direction results in an acceleration directed towards the center of the circle, known as centripetal acceleration, which is essential for maintaining circular motion.

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Intro to Circular Motion

Velocity and Angular Velocity

Velocity is a vector quantity that describes the rate of change of an object's position with respect to time, including both speed and direction. Angular velocity, on the other hand, measures how quickly an object rotates around a point or axis, expressed in radians per second. The relationship between linear velocity and angular velocity is given by the formula v = ωR, where R is the radius of the circular path.

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Escape Velocity

Cross Product of Vectors

The cross product of two vectors results in a third vector that is perpendicular to the plane formed by the original vectors. In the context of circular motion, the relationship v→ = ω→ x r→ illustrates how the linear velocity vector (v→) is derived from the angular velocity vector (ω→) and the position vector (r→). This relationship is fundamental in understanding how rotational motion translates into linear motion.

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