Torque and Power Required

No power at all once it's moving except that required to overcome frictional losses in the bearings etc* (which you've told us nothing about)

Starting and stopping it are an entirely different matter.

e.g If it has really good bearings, say maglev? and is in a vacuum it will keep spinning for ages.

Del

(I reserve the right to be entirely wrong on this...)

Appreciate your observations.

Power & torque is required to start the object from standstill to 1 rpm in about 10 Seconds and then to 10 rpm in 30 secs. Pl do not consider the frictional losses.

I dunno...

Don't mind him. He's just a cat.

He can't help you because cat's don't do homework.

At least not other people's homework.

And you'd need to know how it is supported, bearing type, size, etc, just as the cat explained.

Please explain the difference between torque and power.

Not doing any homework for you. However, you don't need anything more than what is written above.

Once it is up-to-speed, the only power needed is that to overcome bearing resistance and atmospheric effects. However, that is not the whole story, as there is no conceptual need for a cylinder 7000kg rotating at 1-10rpm and nothing else going on in it. Whatever is going on will need some power too, and that has not been shared.

The time it takes to get up-to-speed will determine to a large extent the power needed. Use the equations above once the meaning of the variables has been stated.

'....Use the equations above once the meaning of the variables has been stated....'

.

Providing the equations to calculate moment of inertia and from there torque, is enough of a push in the right direction.

If everything were spelled out, allowing the values to simply be plugged in, what good would that do?

Besides, those formulas should look familiar to a student asked to answer that type of questions, so it shouldn't need to be spelled out.

.

BTW I'm pretty sure the reason we don't know the rest of the problem is because there isn't any more to it....that was as much detail as was given by the instructor.

Hence the fishing expedition in #6↑.

ahhhh. Thank you.

Is the load balanced?

A heck of a lot more torque may be required for an unbalanced load, depending on the degree, than for a balanced one.

Dear Mr. gharemp,

In my opinion, your question should contain some more details - such as Load to be driven, the data relating to Gd^2 Value etc. From Rest - to rotate Force is required to over come frictional resistance and accelerate to the required speed, in a given time. Short time means more torque required and viseversa.

Pl. indicate the working configurations.

DHAYANANDHAN.S

Please go to this site and have a look at the explanations:

http://bolvan.ph.utexas.edu/~vadim/Classes/2011s/linang.pdf

This will clear a lot of the missing knowledge and inform you on how to get your answers.

KISS, first principles.

Calculate the diameter at which half the mass is inside this diameter, and half is outside this diameter.

Dx * Dx = (Din * Din + Dout * Dout) / 2

Calculate rim velocity at 1 RPM at diameter Dx.

Vx = pi * Dx / 60

at 10 RPM

Vx =pi * Dx / 6

Tape a wire to the drum, add weights on the end until the drum begins to turn. This is the approximate frictional force to overcome.

Ff = M * 9.81 at a diam Dout, or

Ff = M * 9.81 * Dout / Dx at diam Dx

Acceleration to get to Vx in 60 secs

a = Vx / 60

Fa = Force to accelerate 7000 kg

Fa = 7000 * a

Total force required = Ff + Fa

Torque T = (Ff + Fa) * Dx / 2

Power = 2 * n * T / 60 Watts

Do same for 10 RPM and choose the higher value calculated. Should be the one at 1 RPM.

This is good!

Except: for 1 rpm, he wants it reached in 10sec while for 10rpm he wants 30sec

therefore, the higher values will be for the 10 rpm (faster acceleration: to be the same, he would need the time to be 100 sec).

One way of looking at it is the principle of Conservation of Energy.

Calculate the amount of energy needed to get the object spinning at the two different spin rates relative to its rest position. The power needed is the greater of:

• Energy of spin speed 1 minus energy of rest state, all divided by the acceleration time to get to spin speed 1
• Energy of spin speed 2 minus energy of spin state 1, all divided by the acceleration time to get to spin speed 2 from spin speed 1.

Simples!ξ

Definitely the best answer.

It is elegant, simple, and avoids the pitfalls of the other responses.

Close to the complete answer (and no, I didn't follow the link), but Gharemp did not state whether the 10 RPM speed has to be achieved from a standstill, or from the 1 RPM speed.

Ergo, we still don't know enough, unless we want to calculate it for both scenarios. And that WOULD be doing homework for him.

Not a cat, but not enough time to do homework anymore, either. Except what Mrs Micahd wants me to do.

350 lb/ft torque - 454HP

As good a guess as any given the amount of data in the question.

Zero, nil and nothing. At least according to the wording of the question. Did the OP understood HIS problem before trying to formulate it? I do doubt it very much.

IF is problem is spinning a satellite, My answer damn well nails it.

IF his problem is more pedestrian terrestrial application, mundane stuff like friction and aerodynamic braking enters the picture.

How do we supposed to comment on a primitive homework picture?!?

Well, Friends, thank you for your time & efforts in giving your valuable feedback.

Following calculations incorporate most of your relevant suggestions (by still keeping out frictional losses) --

Moment of Inertia for cylindrical body has been worked out to be 31718.75 kg-sq meters. {1/2 x mass x (square of r1 + square of r2)}

torque required to rotate the object to 1 rpm in 10 seconds =

31718.75 x 2 x 3.14/(60 x 10) = 332 kg m

Torque required to rotate the object to 10 rpm from standstill =

31718.75 x 2 x 3.14 x 10 / (60 x 30) = 1106.63 kg m

Power = Torque in N-m x rotational speed in radian /sec

= 1106.63 x 9.81 x 2 x 3.14 x 10 / 60 = 11362.6 Watts.

Power has been calculated for highest torque requirement, assuming torque required to rotate the cylinder from 1 rpm to 10 rpm in 20 Sec will be less than torque required to rotate the cylinder from standstill to 10 rpm in 30 seconds.

Now, if I have to select the gear box and then dc motor with speed control, what are the factors I need to consider?

No, you're a factor g out. MoI (kg.m²)*angular acc (rad/s²) = torque in N.m, not kg.m. You can check this from the dimensions.

I don't know much about DC motors, but I assume you have a continuous-running motor speed in mind, and your gearbox ratio clearly has to = that/30 rpm.

Also your calculation (when done correctly) gives the average torque. But the motor gives max torque at starting and falls off as the speed rises, so you might need to do some estimates and iterations. Also check that the motor is happy taking 30 secs to get to full speed. If not you might need to consider eg a fluid or powder coupling.