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Ch. 11 - Angular Momentum; General Rotation
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 11, Problem 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 r\(\overrightarrow{r}\) = î R cos ωt + ĵ R sin ωt where the constant ω = v/R. Determine the velocity v\(\overrightarrow{v}\) and angular velocity w\(\overrightarrow{w}\) of this object and then show that these three vectors obey the relationv=ω×r\(\overrightarrow{v}\)=\(\overrightarrow{\omega}\[\times\]\overrightarrow{r}\).

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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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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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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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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Related Practice
Textbook Question

A toy gyroscope consists of a 170-g disk with a radius of 5.5 cm mounted at the center of a thin axle 21 cm long (Fig. 11–42). The gyroscope spins at 45 rev/s. One end of its axle rests on a stand and the other end precesses horizontally about the stand. How long does it take the gyroscope to precess once around?

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Textbook Question

A merry-go-round with a moment of inertia equal to 860 kg·m² and a radius of 3.0 m rotates with negligible friction at 1.70 rad/s. A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation causing the platform to slow to 1.25 rad/s. What is her mass?

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Textbook Question

The position of a particle with mass m traveling on a helical path (see Fig. 11–48) is given by r\(\overrightarrow{r}\) = R cos (2πz/d) î + R sin (2πz/d) ĵ + zk̂ where R and d are the radius and pitch of the helix, respectively, and z has time dependence z = v𝓏t where v𝓏 is the (constant) component of velocity in the z direction. Determine the time-dependent angular momentum L\(\overrightarrow{L}\) of the particle about the origin.

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Textbook Question

A boy rolls a tire along a straight level street. The tire has mass 8.0 kg, radius 0.32 m and moment of inertia about its central axis of symmetry of 0.83 kg·m². The boy pushes the tire forward away from him at a speed of 2.1 m/s and sees that the tire leans 12° to the right (Fig. 11–49). How will the resultant torque due to gravity and the normal force FN\(\overrightarrow{F_{N}\)} affect the subsequent motion of the tire? 

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Textbook Question

Water drives a waterwheel (or turbine) of radius R = 3.0 m as shown in Fig. 11–50. The water enters at a speed v₁ = 7.0m/s and exits from the waterwheel at a speed v₂= 3.8 m/s. If the water causes the waterwheel to make one revolution every 6.0 s, how much power is delivered to the wheel?

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Textbook Question

Suppose the solid wheel of Fig. 11–42 has a mass of 260 g and rotates at 85 rad/s; it has radius 6.0 cm and is mounted at the center of a horizontal thin axle 25 cm long. At what rate does the axle precess?

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