Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

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

A flywheel of mass 182 kg has an effective radius of \(0.62 \mathrm{m}\) (assume the mass is concentrated along a circumference located at the effective radius of the flywheel). (a) How much work is done to bring this wheel from rest to a speed of 120 rpm in a time interval of 30.0 s? (b) What is the applied torque on the flywheel (assumed constant)?
A turntable must spin at 33.3 rpm \((3.49 \mathrm{rad} / \mathrm{s})\) to play an old-fashioned vinyl record. How much torque must the motor deliver if the turntable is to reach its final angular speed in 2.0 revolutions, starting from rest? The turntable is a uniform disk of diameter \(30.5 \mathrm{cm}\) and mass \(0.22 \mathrm{kg}\).
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 $6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} ?$
A merry-go-round (radius \(R\), rotational inertia \(I_{\mathrm{i}}\) ) spins with negligible friction. Its initial angular velocity is \(\omega_{i}\) A child (mass \(m\) ) on the merry-go-round moves from the center out to the rim. (a) Calculate the angular velocity after the child moves out to the rim. (b) Calculate the rotational kinetic energy and angular momentum of the system (merry-go-round + child) before and after.
A centrifuge has a rotational inertia of $6.5 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}$ How much energy must be supplied to bring it from rest to 420 rad/s \((4000 \text { rpm }) ?\)
See all solutions

Recommended explanations on Physics Textbooks

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