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  • A moment of inertia, I, equal to (1/2)*M*R^2
  • M = mass = 80 kg
  • R = radius = 0.0.5 meters

Part 1

Find the amount of work that must be done to bring the wheel from rest to an angular velocity of 8.38 radians/sec

Solution:

Recall from a previous module that the rotational kinetic energy for a rotating object is given by

Ks = (1/2)*I*w^2

  • where Ks represents the kinetic energy for the system
  • I represents the rotational inertia for the system
  • w represents the angular velocity of the system

We could rewrite this equation as

deltaKs = (1/2)*I*(w0 - wf)^2

where

  • deltaKs represents the change in kinetic energy
  • w0 represents the initial kinetic energy
  • wf represents the final kinetic energy

However, since the initial kinetic energy value is zero, that would simplycomplicate the algebra. Therefore, we will stick with the original equation .

We either have, or can calculate values for all of the terms in this equation. Substituting the values given above gives us

Ks = (1/2)*I*w^2 , or

Ks = (1/2)*((1/2)*M*R^2)*w^2 , or

Ks = (1/2)*((1/2)*80kg*(0.5m)^2)*(8.38 radians/sec)^2

Entering this expression into the Google calculator gives us

Ks = 351 joules

This is the amount of work that must be done to bring the wheel from rest to an angular velocity of 8.38 radians/sec

Part 2

If the motor that drives the wheel delivers a constant torque of 10 N*m during this time, how many revolutions does the wheel turn in coming up to speed.

Solution:

We know how to relate the displacement angle and the work for a constant torque using the equation in Figure 2 .

W = T*A

where

  • W represents the work done by a constant torque
  • T represents the constant torque
  • A represents the angle of displacement measured in radians resulting from the application of the constant torque

In this case, we know the amount of work and the value of the torque and need to find the angle. Therefore,

A = W*joules/T*n*m

However, this gives us the angular displacement in radians. We need to scale to convert it to revolutions.

A = (W*joules/T*n*m)/2*pi, or

A = (351joules/10newton meters)/(2*pi), or

A = 5.59 revolutions

This is the number of revolutions that the wheel turns in coming up to speed.

Do the computations

I encourage you to repeat the computations that I have presented in this lesson to confirm that you get the same results. Experiment withthe scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

Housekeeping material
  • Module name: Angular Momentum -- Torque, Work and Energy
  • File: Phy1330.htm
  • Revised: 10/02/15
  • Keywords:
    • physics
    • accessible
    • accessibility
    • blind
    • graph board
    • protractor
    • screen reader
    • refreshable Braille display
    • JavaScript
    • trigonometry
    • force
    • torque
    • work
    • energy
Disclaimers:

Financial : Although the openstax CNX site makes it possible for you to download a PDF file for the collection that contains thismodule at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should beaware that some of the HTML elements in this module may not translate well into PDF.

You also need to know that Prof. Baldwin receives no financial compensation from openstax CNX even if you purchase the PDF version of the collection.

In the past, unknown individuals have copied Prof. Baldwin's modules from cnx.org, converted them to Kindle books, and placed them for sale on Amazon.com showing Prof. Baldwin as the author.Prof. Baldwin neither receives compensation for those sales nor does he know who doesreceive compensation. If you purchase such a book, please be aware that it is a copy of a collection that is freelyavailable on openstax CNX and that it was made and published without the prior knowledge of Prof. Baldwin.

Affiliation : Prof. Baldwin is a professor of Computer Information Technology at Austin Community College in Austin, TX.

-end-

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
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Sherica
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Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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is it 3×y ?
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J, combine like terms 7x-4y
Bridget Reply
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Asali
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what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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what's the easiest and fastest way to the synthesize AgNP?
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China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Accessible physics concepts for blind students. OpenStax CNX. Oct 02, 2015 Download for free at https://legacy.cnx.org/content/col11294/1.36
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