[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...
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
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...
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...
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)
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]
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]
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
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
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
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
to convert N to kg divide by g
Regards
Hugh
9.81 m/s^2
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
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.
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).
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).