Miks15
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| posted on 21/1/10 at 08:10 PM |
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a little maths help:)
ive been given this maths problem:
A flywheel having moment of inertia I = 0.5kgm^2 and diameter 0.5m is rotatingat a rate of 1000rpm when a constant tangential friction force of 1kN is
applied to its rim. Calculate the time in minutes required for the flywheel to come to rest.
Now i found some formula saying: Torque = moment of inertia x angular accelaration
I worked the angular acceleration to be 500 rads/ second
and the angular velocity intially to be 104.72 rads/sec (1000rpm x 2pi / 60)
Giving me a time to come to rest of 0.21 seconds. Is this right or have i done something wrong?
Cheers guys
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NeilP
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| posted on 21/1/10 at 08:17 PM |
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Bugger if I can remember my mechanical engineering but I think you'd use conservation of energy to work it out?...
[pause] - Nope. You'd still need to do a calc on the speed of the wheel over the time to standstill so over to the brighter / younger / better
memoried on here...
If you pay peanuts...
Mentale, yar? Yar, mentale!
Drive it like you stole it!
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Toltec
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| posted on 21/1/10 at 09:56 PM |
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I think you have either used the wrong calculation for your angular acceleration or messed up the units.
Note the applied torque is in Nm.
I may be wrong, it has been about 28 years...
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Miks15
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| posted on 21/1/10 at 09:59 PM |
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It is the formula i got off the internet, with some good working.
I have used 1000 N for the force, with 0.25m as the radius of the flywheel, giving 250Nm for the torque.
Which when divided by 0.5 (the moment of inertia) i get 500 rads per second (being the SI unit for angular accelaration)
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Toltec
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| posted on 21/1/10 at 10:05 PM |
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The moment of inertia is in kgm2, the torque in Nm, does that sound right 
Edit - Actually is does, I thought they were being sneaky by mixing units, but kgm is the torque applied by a kilo in Earth's gravity.
Dimensionally.
N = kgm/s2
Inertia = kgm2
Ang Acc = 1/s2
So you are right.
[Edited on 21/1/10 by Toltec]
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matt_claydon
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| posted on 21/1/10 at 10:26 PM |
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It looks right to me. Given they asked for time in minutes I wonder if the question was meant to be 1N rather than 1kN?
Angular accel is rad/sec^2 though.
[Edited on 21/1/10 by matt_claydon]
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Miks15
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| posted on 21/1/10 at 10:41 PM |
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o yeh good point matt, had the square in my working
well i can only give the answer to the question given i guess! cheers
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rachaeljf
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| posted on 21/1/10 at 11:29 PM |
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You need to divide the braking torque (Nm) by g (9.81 m/s^2) to get your units (dimensions) consistent. You end up taking 2.1 seconds to bring the
flywheel to a halt, still pretty quick!
It's always a good idea to do a dimension check on your formulae and equations; get everything expressed in m, kg and s and you will be ok.
For example, torque or moment in Nm becomes kg.m.s^-2.m (from F = ma). When you try angular accel'n a = T/J you will see the units drop out to
s^-2 , which are indeed the units of radians/s^2.
It does look like they meant 1 N rather than 1 kN. A 100 kg friction load would seem quite large.
Just to be the pedantic engineer, SI quantities should be expressed as XX-space-units, e.g. 50 kN, not 50kN!
Cheers R
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matt_gsxr
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| posted on 22/1/10 at 12:58 AM |
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You had the answer nearly right first time (except maybe a dodgy unit, and arguably the sign on your angular acceleration which technically is a
vector).
The answer isn't 0.21s
It is 0.0035 minutes (they asked for the answer in minutes).
Matt
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Miks15
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| posted on 22/1/10 at 06:20 AM |
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quote:
Originally posted by rachaeljf
You need to divide the braking torque (Nm) by g (9.81 m/s^2) to get your units (dimensions) consistent. You end up taking 2.1 seconds to bring the
flywheel to a halt, still pretty quick!
It's always a good idea to do a dimension check on your formulae and equations; get everything expressed in m, kg and s and you will be ok.
For example, torque or moment in Nm becomes kg.m.s^-2.m (from F = ma). When you try angular accel'n a = T/J you will see the units drop out to
s^-2 , which are indeed the units of radians/s^2.
It does look like they meant 1 N rather than 1 kN. A 100 kg friction load would seem quite large.
Just to be the pedantic engineer, SI quantities should be expressed as XX-space-units, e.g. 50 kN, not 50kN!
Cheers R
I do understand the fact that it needs to be homogenious on both sides... but i cant see why your would have to divide by g? Surely g has nothing to
do with any of the calculation above?
And point taken on the space, will remember that for the future.
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hughpinder
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| posted on 22/1/10 at 09:12 AM |
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to convert N to kg divide by g
Regards
Hugh
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Lightning
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| posted on 22/1/10 at 09:40 AM |
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9.81 m/s^2
Steve
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Miks15
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| posted on 22/1/10 at 02:02 PM |
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but i dont need to convert N to kg, the formula i ahve stated is already homogenious.
Torque = Moment of Inertia * angular accelaration
Kg.m.s^-2 * m = kgm^2 * s^-2
So its kgm^2s^-2 = kgm^2s^-2
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Liam
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| posted on 22/1/10 at 05:21 PM |
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quote:
Originally posted by rachaeljf
For example, torque or moment in Nm becomes kg.m.s^-2.m (from F = ma). When you try angular accel'n a = T/J you will see the units drop out to
s^-2 , which are indeed the units of radians/s^2.
Indeed, indeed. So wasn't he correct to use his Nm torque value and get 500 rads^-2 for the angular acceleration and 0.21 seconds for the time
to stop?
Liam
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rachaeljf
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| posted on 22/1/10 at 06:10 PM |
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No Liam, you can't use N on one side and kg on the other, g has to get involved somewhere. 1 Nm = 1/9.81 kgm^2s^-2.
By definition, 1 kg = 9.81 N (more or less).
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Toltec
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| posted on 23/1/10 at 06:20 PM |
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quote:
Originally posted by rachaeljf
No Liam, you can't use N on one side and kg on the other, g has to get involved somewhere. 1 Nm = 1/9.81 kgm^2s^-2.
By definition, 1 kg = 9.81 N (more or less).
That was what I thought initially, however torque is force X distance and Newtons are a unit of force and kg is mass. The conversion of Nm to kgm is
done as the force applied by a one kg mass in one g is used for the torque value. To put it another way kgm and lbft are only valid when referenced
to a gravitational field, i.e Earths, Nm does not contain that assumption.
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