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Torque of the solid cylinder?

by kubilayyilmaz
Tags: cylinder, solid, torque
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kubilayyilmaz
#1
Sep14-08, 10:02 AM
P: 4
Hello,
I am new in this forum. I have a project about gearboxes. In the picture,How can I calculate the torque on the axle? The cylinder is turnning and it's speed is 1m/sec.Mass is 2500kg.
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kubilayyilmaz
#2
Sep14-08, 10:15 AM
P: 4
Hello,
I am new in this forum. I have a project about gearboxes. In the picture,How can I calculate the torque on the axle? The cylinder is turnning and it's speed is 1m/sec.Mass is 2500kg.
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picture.GIF  
Q_Goest
#3
Sep14-08, 10:25 AM
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You have a gearbox rotating an axle at constant speed. What external forces do you have resisting the rotation? What about the bearings? What happens to torque when axle rotation has to be accelerated?

sp1408
#4
Sep14-08, 10:42 AM
P: 17
Torque of the solid cylinder?

Didnt get your question,1m/sec is the velocity of what?
And plus if the solid cylinder rotates with constant angular velocity,the torque is zero
FredGarvin
#5
Sep14-08, 11:45 AM
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1 m/s should be the surface speed. If it's not, the number is nonsensical.

The torque to accelerate the mass is 0 at a constant speed, but due to gearbox loads and resistances, like those asked about by Q_goest, there will be a required torque to maintain a constant speed.

The OP needs to provide a lot more detail on the operating conditions before the question can be answered.
kubilayyilmaz
#6
Sep14-08, 03:51 PM
P: 4
Ok,I have to give a lot more details for my question.
There is a solid cylinder in picture. Max Surface speed is 1m/s on the cylinder. I want to calculate how much torque of gearbox for turnnig the cylinder with 1m/s? I will use motor and gearbox for turnning the cyclinder.

So I need to torque value on the axle or output of the gearbox. When I calculate it,and then I can choose gearbox model.
Q_Goest
#7
Sep14-08, 05:35 PM
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I'm assuming this is a project for school. Have you taken any basic physics courses yet?

There's a torque required to overcome forces resisting the rotation of the cylinder. There's also a torque required to increase the rotational velocity of the cylinder. There is no torque required above and beyond these two. You haven't mentioned any external forces acting on the cylinder, so I'll have to assume there are none. In that case, you only need torque to accelerate the cylinder so it's rotating at the speed you want, and then the only torque required is that needed to overcome any bearing resistance or possibly some very minor wind resistances, though from the looks of it you won't have any significant aerodynamic loading on this cylinder.

If you need to accelerate the mass at a certain rate, you need to consider how much torque is required to do that. Check one of these web pages:
http://theoryx5.uwinnipeg.ca/physics/rot/node5.html
http://hyperphysics.phy-astr.gsu.edu/Hbase/n2r.html

The forces on the cylinder at constant speed are going to be primarily from the bearings, but that resistance is very small. See this page for some information about rolling element bearing frictional resistance:
http://www.roymech.co.uk/Useful_Tabl...0Friction.html
kubilayyilmaz
#8
Sep15-08, 03:31 AM
P: 4
Thanks for your help Q_Goest.
It is not a homework. I'm a electronics eng. I have a project about the the Quality control machines. This solid cylinder is a part of the nachine.

I have looked your advised web pages and I have tried to calculate the torque with the following formulas;

Please say is it right or not ?

Torque = I (moment of inertia to the solid cylinder,rotating at Ix=Iy) x aw (a is angular acceleration)
I= 1/12.m.(3r2+h2)=1/12.2500.(3.1+2,56) = 1158,3 Nm and then

At the start and stop time
aw= a( tangential acceleration ) / r = (dV/dt)/r here I want to reach 0-1m/s in 3sec.So
aw= 1(m/s) / 1m = 1 rad/s2
Torque = 1158,3 Nm x 1 =1158,3 Nm.

Right or not?
Q_Goest
#9
Sep15-08, 07:29 PM
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check rotational inertia, looks wrong. for a solid cylinder, I = 1/2 MR^2. See this web page to calculate inertia:
http://hypertextbook.com/physics/mec...ional-inertia/

Check angular acceleration and velocity. Final velocity is .5 rad/s so angular velocity is .5/3 = .16667

plug those into the torque equation t = I (a) shown in web pages referenced above.
Analysis
#10
Sep19-08, 01:08 AM
P: 55
""If you need to accelerate the mass at a certain rate, you need to consider how much torque is required to do that""

I agree the points you said But i have one querey
TOrque=I * Alpha
While designing if we consider induction motors it will behave differently.
how to consider acceleration timings alpha to get required torque?

Prakash
Wondermotor
#11
Sep20-08, 08:39 PM
P: 4
look into moment of inertia. you use that to figure out your torque.
Analysis
#12
Sep24-08, 10:30 PM
P: 55
To... wonder motor
Reply to "Look into moment of inertia you use that to find torque"

Can't understand your reply can you brief me little?
(becoz mass moment of inertia is kg m^2 from that...?)

Sorry to my late reply
Prakash


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