Log out Go to App
Go to App

Learning Materials

Features

Discover

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: \(Weight (W) = Mass (m) × g\), 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: \(Torque (\tau_{noon}) = Weight (W) × \text{perpendicular distance}\) \(\tau_{noon}= 588.6 N * 1.35 m = 794.61 Nm\)
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: \(Torque (\tau_{9AM}) = Weight (W) × \text{perpendicular distance}\) \(\tau_{9AM} = 588.6 N * 0.9564 m = 562.925 Nm\)
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 \(562.925 Nm\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The angular momentum of a spinning wheel is $240 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ After application of a constant braking torque for \(2.5 \mathrm{s}\), it slows and has a new angular momentum of $115 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ What is the torque applied?
A Ferris wheel rotates because a motor exerts a torque on the wheel. The radius of the London Eye, a huge observation wheel on the banks of the Thames, is \(67.5 \mathrm{m}\) and its mass is \(1.90 \times 10^{6} \mathrm{kg} .\) The cruising angular speed of the wheel is $3.50 \times 10^{-3} \mathrm{rad} / \mathrm{s} .$ (a) How much work does the motor need to do to bring the stationary wheel up to cruising speed? [Hint: Treat the wheel as a hoop.] (b) What is the torque (assumed constant) the motor needs to provide to the wheel if it takes 20.0 s to reach the cruising angular speed?
A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber's waist \(15 \mathrm{cm}\) to the right of his center of gravity. If the climber weighs \(770 \mathrm{N},\) find (a) the tension in the rope and (b) the magnitude and direction of the contact force exerted by the wall on the climber's feet.
A house painter is standing on a uniform, horizontal platform that is held in equilibrium by two cables attached to supports on the roof. The painter has a mass of \(75 \mathrm{kg}\) and the mass of the platform is \(20.0 \mathrm{kg} .\) The distance from the left end of the platform to where the painter is standing is \(d=2.0 \mathrm{m}\) and the total length of the platform is $5.0 \mathrm{m}$ (a) How large is the force exerted by the left-hand cable on the platform? (b) How large is the force exerted by the right-hand cable?
A crustacean (Hemisquilla ensigera) rotates its anterior limb to strike a mollusk, intending to break it open. The limb reaches an angular velocity of 175 rad/s in \(1.50 \mathrm{ms} .\) We can approximate the limb as a thin rod rotating about an axis perpendicular to one end (the joint where the limb attaches to the crustacean). (a) If the mass of the limb is \(28.0 \mathrm{g}\) and the length is \(3.80 \mathrm{cm}\) what is the rotational inertia of the limb about that axis? (b) If the ex-tensor muscle is 3.00 mm from the joint and acts perpendicular to the limb, what is the muscular force required to achieve the blow?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Thermodynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electricity and Magnetism

Read Explanation

Fundamentals of Physics

Read Explanation

Physics of Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free