/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A tower outside the Houses of Pa... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 16

A tower outside the Houses of Parliament in London has a famous clock commonly referred to as Big Ben, the name of its 13 -ton chiming bell. The hour hand of each clock face is \(2.7 \mathrm{m}\) long and has a mass of \(60.0 \mathrm{kg}\) Assume the hour hand to be a uniform rod attached at one end. (a) What is the torque on the clock mechanism due to the weight of one of the four hour hands when the clock strikes noon? The axis of rotation is perpendicular to a clock face and through the center of the clock. (b) What is the torque due to the weight of one hour hand about the same axis when the clock tolls 9: 00 A.M.?

Short Answer

Expert verified
Answer: The torque on the clock mechanism due to the weight of one of the hour hands at noon is 794.61 Nm, and at 9:00 AM, it is 562.925 Nm.

Step by step solution

01

Finding the weight of the hour hand

First, we find the weight of an hour hand, which is given by the following formula: \(Weight (W) = Mass (m) 脳 g\), where \(g\) is the acceleration due to gravity (\(9.81 m/s^2\)) Given mass \(m = 60.0 kg\) \(W = 60.0 kg * 9.81 m/s^2 = 588.6 N\)
02

Torque at noon

At noon, the force of gravity acts at the center of the hour hand which is halway along the length. Therefore, the perpendicular distance to the axis of rotation is half the length of the hand (\(2.7 m/2 = 1.35 m\)). Now, we can find the torque: \(Torque (\tau_{noon}) = Weight (W) 脳 \text{perpendicular distance}\) \(\tau_{noon}= 588.6 N * 1.35 m = 794.61 Nm\)
03

Torque at 9:00 AM

At 9:00 AM, the configuration of the clock changes and the perpendicular distance from the center of mass of the hour hand can be found using geometry. The hour hand forms a 45-degree angle with the vertical and the distance will be \(\frac{2.7m}{2} \cos(45^\circ)\). Using trigonometry: Perpendicular distance \(= \frac{2.7m}{2} \cos(45^\circ) = \frac{2.7m}{2} \frac{1}{\sqrt{2}} = \frac{2.7m}{\sqrt{8}} = 0.9564m\) Now, we can find the torque: \(Torque (\tau_{9AM}) = Weight (W) 脳 \text{perpendicular distance}\) \(\tau_{9AM} = 588.6 N * 0.9564 m = 562.925 Nm\)
04

Summary of the Results

a) The torque on the clock mechanism due to the weight of one of the hour hands when the clock strikes noon is \(794.61 Nm\). b) The torque due to the weight of one hour hand about the same axis when the clock tolls 9:00 AM is \(562.925 Nm\).

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Most popular questions from this chapter

A uniform door weighs \(50.0 \mathrm{N}\) and is \(1.0 \mathrm{m}\) wide and 2.6 m high. What is the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner?
Verify that the units of the rotational form of Newton's second law [Eq. (8-9)] are consistent. In other words, show that the product of a rotational inertia expressed in \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular acceleration expressed in \(\mathrm{rad} / \mathrm{s}^{2}\) is a torque expressed in \(\mathrm{N} \cdot \mathrm{m}\).
An experimental flywheel, used to store energy and replace an automobile engine, is a solid disk of mass \(200.0 \mathrm{kg}\) and radius $0.40 \mathrm{m} .\( (a) What is its rotational inertia? (b) When driving at \)22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph}),$ the fully energized flywheel is rotating at an angular speed of \(3160 \mathrm{rad} / \mathrm{s} .\) What is the initial rotational kinetic energy of the flywheel? (c) If the total mass of the car is \(1000.0 \mathrm{kg},\) find the ratio of the initial rotational kinetic energy of the flywheel to the translational kinetic energy of the car. (d) If the force of air resistance on the car is \(670.0 \mathrm{N},\) how far can the car travel at a speed of $22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph})$ with the initial stored energy? Ignore losses of mechanical energy due to means other than air resistance.

A door weighing \(300.0 \mathrm{N} \quad\) measures \(2.00 \mathrm{m} \times 3.00 \mathrm{m}\) and is of uniform density; that is, the mass is uniformly distributed throughout the volume. A doorknob is attached to the door as shown. Where is the center of gravity if the doorknob weighs \(5.0 \mathrm{N}\) and is located \(0.25 \mathrm{m}\) from the edge?
How long would a braking torque of \(4.00 \mathrm{N}\).m have to act to just stop a spinning wheel that has an initial angular momentum of $6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} ?$
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