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 84

Capitolo 11, Problema 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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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.

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Percorso guidato

08:12

Torque & Acceleration (Rotational Dynamics)

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.

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Percorso guidato

08:55

Net Torque & Sign of Torque

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

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