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\(\bullet\) A certain drawbridge can be modeled as a uniform \(15,000 \mathrm{N}\) bar, 12.0 \(\mathrm{m}\) long, pivoted about its lower end. When this bridge is raised to an angle of \(60.0^{\circ}\) above the horizontal, the cable holding it suddenly breaks, allowing the bridge to fall. At the instant after the cable breaks, (a) what is the torque on this bridge about the pivot and (b) at what rate is its angular momentum changing?

Short Answer

Expert verified
Torque and rate of angular momentum change is 77,942.29 Nm.

Step by step solution

01

Understanding the Problem

We need to calculate the torque on the drawbridge about the pivot point when it is at a 60-degree angle and the cable holding it breaks. Additionally, we need to determine the rate at which its angular momentum is changing at that instant.
02

Identifying Given Values

The length of the bridge is 12.0 m and its weight is 15,000 N. The bridge is raised at an angle of 60 degrees above the horizontal.
03

Calculating the Effective Force due to Gravity

The force due to gravity acts at the center of mass of the bridge, which is at half its length, i.e., at 6 m from the pivot. The vertical component of this gravitational force, causing torque, is given by: \( F_{g} = 15,000 \times \sin(60^{\circ}) \).
04

Computing Torque about the Pivot

Torque is the product of force and the perpendicular distance from the pivot. Thus, torque \( \tau \) is calculated by: \( \tau = F_{g} \times \text{distance} = 15,000 \times \sin(60^{\circ}) \times 6 \).
05

Finding the Rate of Change of Angular Momentum

According to Newton's second law for rotation, the rate of change of angular momentum \( \frac{dL}{dt} \) is equal to the torque: \( \frac{dL}{dt} = \tau \). Hence, \( \frac{dL}{dt} \) is the same value as computed for torque.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drawbridge Physics
A drawbridge is a type of movable bridge that can be raised, lowered, or moved to allow boats to pass underneath. In physics, understanding how a drawbridge like this one functions requires examining forces and how they change during movement. When the drawbridge is raised, it's like a long bar pivoted at one end. The pivot or support acts as a fulcrum, allowing the bridge to rotate about this point. When the cable holding the bridge breaks, gravity becomes the main force acting on the bridge, making it rotate downward. The length and weight of the bridge, as well as the angle at which it is held, are crucial factors in determining the forces at play. During this transition, it's the torque and angular momentum that describe how the movement progresses.
Angular Momentum
Angular momentum is a key concept in rotational dynamics and has similarities to linear momentum in translational motion. - **Definition**: In simple terms, angular momentum is the rotational equivalent of linear momentum. It's the product of an object's rotational inertia and its angular velocity. - **For Drawbridges**: When the bridge starts to fall, it gains angular momentum. This is because it starts rotating around the pivot point. The rate at which angular momentum changes relates directly to torque, as per Newton's second law for rotation. The greater the torque, the faster this change occurs. This is why the breaking cable instantly causes an immediate rotational motion of the bridge.
Gravitational Force
Gravitational force is the attractive force that pulls objects towards the center of the Earth. This force is crucial in the operation of a drawbridge. - **Effect on the Bridge**: On the drawbridge, gravity acts at its center of mass, which is located halfway along its length. This is why calculations often use this point when determining forces and torques. When the cable breaks, gravity causes the bridge to rotate around the pivot point. The component of gravitational force that acts perpendicular to the length of the bridge creates a torque. This can be found by multiplying the gravitational force by the sine of the angle at which the bridge is raised. This component solely contributes to the torque that determines the rotational motion of the bridge.
Newton's Second Law for Rotation
This law is an extension of Newton's second law, which states that the rate of change of linear momentum is equal to the net force acting on an object. - **For Rotational Motion**: In terms of rotation, it states that the rate of change of angular momentum is equal to the net torque applied. This can be expressed mathematically as \( \frac{dL}{dt} = \tau \), where \( L \) represents angular momentum and \( \tau \) the torque.In the context of the drawbridge, once the cable breaks, the torque due to gravitational force isn't counteracted by any other force, meaning the bridge will begin rotating downward with a rate of change in angular momentum directly proportional to the torque exerted by gravity. This principle is what allows us to understand how quickly the bridge begins its rotational fall.

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Most popular questions from this chapter

\(\bullet\) A thin, light string is wrapped around the rim of a 4.00 kg solid uniform disk that is 30.0 \(\mathrm{cm}\) in diameter. A person pulls on the string with a constant force of 100.0 \(\mathrm{N}\) tangent to the disk, as shown in Figure \(10.49 .\) The disk is not attached to anything and is free to move and tum. (a) Find the angular acceleration of the disk about its center of mass and the linear acceleration of its center of mass. (b) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the accelerations in part (a)?

Supporting an injured arm: I. A 650 N person must have her injured arm supported, with the upper arm horizontal and the fore- arm vertical. (See Figure \(10.76 .\) ) According to biomedical tables and direct measurements, her upper arm is 26 \(\mathrm{cm}\) long (measured from the shoulder joint), accounts for 3.50\(\%\) of her body weight, and has a center of mass 13.0 \(\mathrm{cm}\) from her shoulder joint. Her forearm (including the hand) is 34.0 \(\mathrm{cm}\) long, makes up 3.25\(\%\) of her body weight, and has a center of mass 43.0 \(\mathrm{cm}\) from her shoulder joint. (a) Where is the center of mass of the person's arm when it is supported as shown? (b) What weight \(W\) is needed to support her arm? (c) Find the horizontal and vertical components of the force that the shoulder joint exerts on her arm.

\(\bullet\) A 750 gram grinding wheel 25.0 \(\mathrm{cm}\) in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the center.) When it is in use, it turns at a constant 220 \(\mathrm{rpm}\) about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 45.0 s with constant angular acceleration due to friction at the axle. What torque does friction exert while this wheel is slowing down?

\(\cdot\) A woman with mass 50.0 \(\mathrm{kg}\) is standing on the rim of a large disk that is rotating at 0.50 \(\mathrm{rev} / \mathrm{s}\) about an axis perpendicular to it through its center. The disk has a mass of 110 \(\mathrm{kg}\) and a radius of 4.0 \(\mathrm{m}\) . Calculate the magnitude of the total angular momentum of the woman-plus-disk system, assuming that you can treat the woman as a point.

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{N},\) with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

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