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Anonymous Poster

Torque Required

01/13/2010 8:42 AM

hi I have been having trouble with a question and i think i've finally done it. I was wondering if there is anyone that can confirm that i've done this correct.

Here is the question: Question 3 A flywheel has a mass of 700 kg and a radius of gyration, k of 0.7 m. What is the torque required to increase the speed of the flywheel at a constant rate from 80 rev min-1 to 200 rev min-1 in a time of 20 s? Note that the moment of inertia of a body about an axis = mass*k2 where k is the radius of gyration of the body about that axis. Mass =700 Kg k = 0.7 m Current rev min-1 = 80 rev min-1 Required rev min-1 in 20 seconds = 200 rev min-1 (〖revs〗^(-1)*2π)/60= 〖rads〗^(-1) The total kinetic energy is given by the formula: Where: KE = kinetic energy in J m = mass in kg k = radius of gyration in m ωi = initial acceleration in rad/sec ωf = final acceleration in rad/sec = 63191 J Where: P = power in W E = energy in J t = time in s The power developed, P = T. Therefore T=p/ω ω = 2π (RPM/60) 2π(200-80/60)= 12.6 3159.6/12.6= 250.7 Nm so Torque Required is 250.7 Nm

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Anonymous Poster
#1

Re: Torque Required

01/13/2010 9:35 AM

?

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Guru

Join Date: Mar 2007
Location: City of Light
Posts: 3943
Good Answers: 183
#2

Re: Torque Required

01/13/2010 10:55 AM

We do not solve home work !!!

It is not in your interest to have prechewed results or approaches.

Present your detailed computations and you will get a comment and a correction.

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Anonymous Poster
#3
In reply to #2

Re: Torque Required

01/13/2010 11:26 PM

Who are "WE"?

... like you know the answer, right?!

Guest, ask lynlynch he is one of the contribuitors ... he will answer you.

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Guru

Join Date: Mar 2007
Location: City of Light
Posts: 3943
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#5
In reply to #3

Re: Torque Required

01/14/2010 3:56 AM

"We" = participants to CR4

I doubt that you will get an other reaction from thos who contribute constantly since it is a general policy of CR4 which you should know if you read the threads.

But I am willing to make some corrections: angular accelerations are measured in rad/s².

There are in the text several values not necessary. Rotating speed is in rad/s.

If it is a constant acceleration then its value is (max.speed-min speed)/time.

Acceleration is Vector/inertia where vector is either force or torque depending on the nature of the movement. It follows that vector is acceleration x inertia.

My result differs from the presented one. I think it is a computational error either at my end or at the other.

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Guru

Join Date: Jun 2006
Location: Ottawa Canada
Posts: 1975
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#6
In reply to #3

Re: Torque Required

01/14/2010 10:14 AM

That sounds like a challenge!

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Guru

Join Date: Oct 2006
Location: Surrey BC Canada
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#4

Re: Torque Required

01/14/2010 12:07 AM

In British units

N= rpm change, t in sseconds, W = lbs, K is radius of gyration in ft.

T = (WK^2 N)/(308t)

Question is to derive the equation and determine the constant in metric units.

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Anonymous Poster
#7

Re: Torque Required

01/14/2010 10:50 AM

Like Newtons Law F= m * a

There are two components of required torque, torque by fricction and torque by aceleration.

Torque = Inertia system (kg-m2) * angular aceleration (rad/seg2)

Good Luck!!!

Grettings

R Meza

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Guru

Join Date: Mar 2007
Location: City of Light
Posts: 3943
Good Answers: 183
#8
In reply to #7

Re: Torque Required

01/14/2010 1:35 PM

What is "seg"?

A new time unit?

As far as I know (but of course I do not know all the international unit for time is (or was) the second(E), seconde(F) or Sekunde(G) with the international symbol "s". Am I wrong?

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