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Chapter 9: Problem 2

In San Francisco a very simple technique is used to turn around a cable car when it reaches the end of its route. The car rolls onto a turntable, which can rotate about a vertical axis through its center. Then, two people push perpendicularly on the car, one at each end, as shown in the drawing. The turntable is rotated one-half of a revolution to turn the car around. If the length of the car is \(9.20 \mathrm{~m}\) and each person pushes with a \(185-\mathrm{N}\) force, what is the magnitude of the net torque applied to the car?

Short Answer

Expert verified
The net torque applied to the car is \(1702 \ \mathrm{Nm}\).

Step by step solution

01

Determine the Forces' Points of Application

Since the car is pushed perpendicularly at each end, we need to identify where the forces are applied. Each force acts at a distance of half the car's length from the center of the turntable. Thus, the distance is \(\frac{9.20 \mathrm{~m}}{2} = 4.60 \mathrm{~m}\).
02

Calculate the Torque from One Force

Torque is calculated using the formula \( \tau = F \cdot r \cdot \sin(\theta) \), where \( F \) is the force, \( r \) is the perpendicular distance from the axis of rotation, and \( \theta \) is the angle between the force direction and the lever arm. Here, \( \theta = 90^\circ \), therefore, \( \sin(\theta) = 1 \). With \( F = 185\, N \) and \( r = 4.60\, m \), the torque from one force is \( \tau = 185 \cdot 4.60 = 851 \ \mathrm{Nm} \).
03

Calculate the Total Net Torque

Since there are two people applying the same torque on opposite ends and both are contributing rotationally in the same direction, the total net torque is the sum of the torques from both ends. Thus, the net torque is \( 2 \times 851 = 1702 \ \mathrm{Nm} \).

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Key Concepts

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

Rotation
Rotation is a fundamental concept in physics, describing the circular movement of an object around an axis or a point. In the case of the cable car, it rotates around a vertical axis located at the center of the turntable. Understanding rotation involves looking at several factors:
  • Axis of rotation: This is the imaginary line that the object spins around.
  • Rotational motion: The circular path that the object follows as it rotates.
  • Revolution: A complete turn around the axis, reaching the initial start position again.
In practical scenarios, like turning a cable car, rotation helps change orientation or direction without linear movement. The objects involved - like the cable car and people pushing it - have forces acting that cause the rotation.
Torque Formula
Torque is a measure of the turning force on an object, defined by the formula: \[ \tau = F \times r \times \sin(\theta) \]Where:
  • \( \tau \) is the torque.
  • \( F \) is the magnitude of the force applied.
  • \( r \) is the distance from the axis of rotation to where the force is applied.
  • \( \theta \) is the angle between the force direction and the lever arm.
In this problem, both forces are applied perpendicularly, simplifying the formula since \( \sin(90^\circ) = 1 \). This eliminates the sine component in the calculation, making it straightforward. The torque calculation depends heavily on correctly understanding where and how the force is applied in relation to the axis.
Perpendicular Force
A perpendicular force refers to a force applied at a 90-degree angle relative to a surface or direction, and is crucial in calculating torque. In the scenario with the cable car, each person pushes perpendicularly to ensure maximum efficiency in inducing rotation.Key points about perpendicular force include:
  • Optimal for creating rotation because it maximizes the torque produced.
  • Any deviation from a 90-degree angle reduces the torque as \( \sin(\theta) \) becomes less than 1, reducing the effective force in turning the object.
  • Ensures that no part of the force is wasted in trying to move the object linearly.
Understanding this helps to maximize the effectiveness of the force applied as seen in mechanical systems or everyday scenarios such as opening doors, using wrenches, or in our case, turning a cable car.
Axis of Rotation
The axis of rotation is a vital element in understanding rotational motion. It is the line around which everything rotates, remaining fixed while the object spins around it. For the cable car on the turntable:
  • The axis of rotation is vertical and passes through the turntable's center, around which the entire car rotates.
  • Understanding this axis is important because it determines the orientation and path of rotation.
  • All calculations regarding rotational motion, like torque, use the distance from this axis as a core part.
This concept helps in visualizing and calculating movements within rotational systems, helping you determine how forces will influence motion around this axis. In any mechanical or real-world application where rotation is involved, identifying and understanding the axis of rotation is crucial.

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Most popular questions from this chapter

A particle is located at each corner of an imaginary cube. Each edge of the cube is \(0.25 \mathrm{~m}\) long, and each particle has a mass of \(0.12 \mathrm{~kg}\). What is the moment of inertia of these particles with respect to an axis that lies along one edge of the cube?

A thin, rigid, uniform rod has a mass of \(2.00 \mathrm{~kg}\) and a length of \(2.00 \mathrm{~m} .\) (a) Find the moment of inertia of the rod relative to an axis that is perpendicular to the rod at one end. (b) Suppose all the mass of the rod were located at a single point. Determine the perpendicular distance of this point from the axis in part (a), such that this point particle has the same moment of inertia as the rod. This distance is called the radius of gyration of the rod.

Interactive Solution 9.35 at presents a method for modeling this problem. MultipleConcept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{~m}\), an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of \(6.40 \mathrm{~s}\). (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

A thin uniform rod is rotating at an angular velocity of \(7.0 \mathrm{rad} / \mathrm{s}\) about an axis that is perpendicular to the rod at its center. As the drawing indicates, the rod is hinged at two places, one-quarter of the length from each end. Without the aid of external torques, the rod suddenly assumes a "u" shape, with the arms of the "u" parallel to the rotation axis. What is the angular velocity of the rotating "u"?

A baggage carousel at an airport is rotating with an angular speed of \(0.20 \mathrm{rad} / \mathrm{s}\) when the baggage begins to be loaded onto it. The moment of inertia of the carousel is 1500 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Ten pieces of baggage with an average mass of \(15 \mathrm{~kg}\) each are dropped vertically onto the carousel and come to rest at a perpendicular distance of \(2.0 \mathrm{~m}\) from the axis of rotation. (a) Assuming that no net external torque acts on the system of carousel and baggage, find the final angular speed. (b) In reality, the angular speed of a baggage carousel does not change. Therefore, what can you say qualitatively about the net external torque acting on the system?
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