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

Giancoli Douglas 5th edition

Ch. 11 - Angular Momentum; General Rotation

Problem 84

Chapter 11, Problem 84

A radio transmission tower has a mass of 76 kg and is 12 m high. The tower is anchored to the ground by a flexible joint at its base, but it is secured by three cables 120° apart (Fig. 11–52). In an analysis of a potential failure, a mechanical engineer needs to determine the behavior of the tower if one of the cables breaks. The tower would fall away from the broken cable, rotating about its base. Determine the speed of the top of the tower as a function of the rotation angle θ. Start your analysis with the rotational dynamics equation of motion d $\vec{L}$ /dt = ${\vec{τ_{e x t}}}_{}$ . Approximate the tower as a tall thin rod.

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Start by identifying the rotational dynamics equation: $ \frac{d\mathbf{L}}{dt} = \mathbf{\tau}_{\text{ext}} $, where $ \mathbf{L} $ is the angular momentum and $ \mathbf{\tau}_{\text{ext}} $ is the external torque. The external torque in this case is due to the gravitational force acting on the center of mass of the tower.

Model the tower as a tall thin rod. The moment of inertia $ I $ of a thin rod rotating about one end is given by $ I = \frac{1}{3} M L^2 $, where $ M $ is the mass of the rod (76 kg) and $ L $ is its length (12 m).

The torque $ \tau $ due to gravity is calculated as $ \tau = M g \frac{L}{2} \sin(\theta) $, where $ g $ is the acceleration due to gravity, $ L/2 $ is the distance from the pivot to the center of mass, and $ \sin(\theta) $ accounts for the angle of rotation.

Using the rotational dynamics equation, relate the angular acceleration $ \alpha $ to the torque: $ \tau = I \alpha $. Substituting $ I $ and $ \tau $, we get $ \frac{1}{3} M L^2 \alpha = M g \frac{L}{2} \sin(\theta) $. Simplify to find $ \alpha = \frac{3 g \sin(\theta)}{2 L} $.

To find the speed of the top of the tower as a function of $ \theta $, use the relationship between angular velocity $ \omega $ and angular acceleration: $ \omega^2 = 2 \int \alpha \, d\theta $. Substituting $ \alpha $, integrate $ \omega^2 = \int \frac{3 g \sin(\theta)}{L} \, d\theta $. Finally, the linear speed of the top of the tower is $ v = \omega L $.

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Key Concepts

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

Rotational Dynamics

Rotational dynamics involves the study of the motion of objects that rotate about an axis. It is governed by Newton's laws applied to rotational motion, where torque (τ) is the rotational equivalent of force. The equation dL/dt = τ_ext relates the change in angular momentum (L) to the external torque acting on the system, which is crucial for analyzing the behavior of rotating objects like the tower.

Torque

Torque is a measure of the rotational force applied to an object, calculated as the product of the force and the distance from the pivot point (lever arm). In the context of the tower, the torque generated by the tension in the cables will influence its rotational motion when one cable fails. Understanding how to calculate and apply torque is essential for predicting the tower's behavior during rotation.

Angular Velocity and Acceleration

Angular velocity refers to the rate of change of the angle of rotation, while angular acceleration is the rate of change of angular velocity. These concepts are critical for determining the speed of the top of the tower as it rotates after a cable breaks. By relating angular displacement to angular velocity and acceleration, one can derive the motion of the tower as a function of the rotation angle θ.

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Rotational Velocity & Acceleration

Related Practice

Textbook Question

A particle of mass m uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius R: $\vec{r}$ = î R cos θ + ĵ R sin θ with θ = ω₀t + (1/2)αt² , where the constants ω₀ and α are the initial angular velocity and angular acceleration, respectively. Determine the object’s tangential acceleration $\vec{a}$ _tan and determine the torque acting on the object using $\vec{τ} = \vec{r} \times F$ .

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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 $\vec{F_{N}}$ 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

A baseball bat has a sweet spot where a ball can be hit with almost effortless transmission of energy. A careful analysis of baseball dynamics shows that this special spot is located at the point where an applied force would result in pure rotation of the bat about the handle grip. Determine the location of the sweet spot, xₛ, of the bat shown in Fig. 11–53. The linear mass density of the bat is given roughly by (0.61 + 3.3x²) kg/m, where x is in meters measured from the end of the handle. The entire bat is 0.84 m long. The desired rotation point should be 5.0 cm from the thin end where the bat is held. [Hint: Where is the cm of the bat?]

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