Solving Total Work Problem on Rotating Object

[Pinon1977]

Please see attached sketch. I am trying to determine the total work done by this system. I am not sire where to even start. I know w=fd, but i am not certain how that applies to a rotating object. Please advise.

 

[BvU]

Hi,
Please don't erase the template:

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
In fact, filling it in could be very helpful to you in itself !

And remember: work = increase in kinetic energy.
What's the kinetic energy of a rotating object ?

 

[Pinon1977]

Well the moment of inertia is 300kgm2 x 1.57 rad/sec (15 rpms) = 471.6 Nm. Is that rhe KE?

 

[BvU]

Does it have the right dimension ?

 

[Pinon1977]

10 feet in length ...excludng the support feature of 20kg, the 100kg mass on each end has a moi of 231 and 286j.

 

[Pinon1977]

Did you did you receive my response?

 

[BvU]

DImension in the sense of units: your expression does NOT have the dimension of energy.

 

[Pinon1977]

You lost me on that one, sir. Can you ask your question in another way?

 

[jbriggs444]

You lost me on that one, sir. Can you ask your question in another way?

What are the units on...

300kgm2 x 1.57 rad/sec

Edit: In particular, pay attention to the seconds. Then compare to

471.6 Nm

What are the units on that?

 

[russ_watters]

10 feet in length ...excludng the support feature of 20kg, the 100kg mass on each end has a moi of 231 and 286j.

Do you want work done or kinetic energy?

Do you know the dfference between power and energy? How conservation of energy applies to a situation where (as described) there are no inputs or outputs?

As posed, the question is very vague and does not imply homework. Is there more/can you explain?

 

[CWatters]

Power = torque * angular velocity

At a constant speed the only torque required is that needed to overcome friction or air resistance neither of which is known.

 

[russ_watters]

Do you want work done or kinetic energy?

Do you know the dfference between power and energy? How conservation of energy applies to a situation where (as described) there are no inputs or outputs?

As posed, the question is very vague and does not imply homework. Is there more/can you explain?

@Pinon1977 I've been made aware that this thread vaguely resembles one you made in August that was closed due to its subject's resemblance to a perpetual motion machine. Is this the same topic?

Regardless, in that thread, you demonstrated a series of fundamental misunderstandings of the basic physics of rotation (torque, work/power, kinetic energy, etc.). The same misunderstandings seem to be at work in this thread. It is unfortunate: 3 months would be enough time to take some online courses in beginning physics to correct these misunderstandings. Whereever this thread goes, I recommend you do that. And along those lines, start by ensuring you can analyze more basic/typical scenarios, such as:
1. A prime mover with known torque and rpm, powering a generator; how much power is generated?
2. A basic flywheel being spun by an electric motor: how much time to spin-up? How much power to spin-up? How much kinetic energy once spun-up? How much power required to keep spinning?
3. A basic flywheel is attached to a generator. Generator draws a constant power: what happens to the flywheel?

In the end, you will come to recognize that the flywheel is irrelevant to what you really want to know, but I don't think we can help you accept that by just telling you: you'll need to do the calculations yourself and prove it.

 

[Pinon1977]

Do you want work done or kinetic energy?

Do you know the dfference between power and energy? How conservation of energy applies to a situation where (as described) there are no inputs or outputs?

As posed, the question is very vague and does not imply homework. Is there more/can you explain?

Yes, i am trying to determine how much work is done by this device at the aforementioned set up.

Here is an example ...can I use this same logic for my scenario?

Say that you have a plane that uses propellers, and you want to determine how much work the plane’s engine does on a propeller when applying a constant torque of 600 Newton-meters over 100 revolutions. You start with the work equation in terms of torque:

Plugging the numbers into the equation gives you the work:

 

[russ_watters]

Yes, i am trying to determine how much work is done by this device at the aforementioned set up.

Here is an example ...can I use this same logic for my scenario?

Say that you have a plane that uses propellers, and you want to determine how much work the plane’s engine does on a propeller when applying a constant torque of 600 Newton-meters over 100 revolutions. You start with the work equation in terms of torque:

View attachment 215396

Plugging the numbers into the equation gives you the work:

View attachment 215397

Well you didn't mention torque in your opening post scenario, so you'll have to tell us if there is one and therefore if this example applies.

 

[haruspex]

work is done by this device

The diagram does not suffice. It just shows two 100 kg masses at the ends of an arm rotating at a constant rate. If that is all there is to it then no work is being done.
There is apparently another force "20kg", but 20kg is a mass not a force, and it is described as a "support", so it is not at all clear what this is or what it is doing.

 

[Pinon1977]

My apologies. The torque being exerted in this particular instance would be 120 foot pounds of torque at 15 RPMs. The overall travel diameter is approximately 10 feet with the only real Mass being the 100-kilogram out on each end. However I'm sure the support mechanism is also contributing to some sort of kinetic energy but let's just leave it out of the equation for right now if it's easier. Please advise

 

[haruspex]

The torque being exerted in this particular instance would be 120 foot pounds of torque at 15 RPMs

That is enough to compute the power, as already indicated in this post #11.

the support mechanism is also contributing to some sort of kinetic energy

If it is just a support then I would think it is stationary, so does no work.

 

[Pinon1977]

Yes i would agree. However i am seeking how much work is done here-not power. Would it be (120ftlbs or 167.2 Nm) x (15rpms x 2pie) = 15750 J? And if it took 60 seconds to do 15 rpms then the power would be 263 watts or .35 HP?

 

[jbriggs444]

And if it took 60 seconds to do 15 rpms

Come again? 15 rpms is a rate.

 

[haruspex]

15rpms x 2pie

What units would that have? What units do you need?

15750 J

Joules are units of work, not power. As already explained, torque times rotation rate gives power.

it took 60 seconds to do 15 rpms

I assume you mean it maintains 15rpm for 60 seconds.

 

[russ_watters]

My apologies. The torque being exerted in this particular instance would be 120 foot pounds of torque at 15 RPMs.

Ok, great. You have a torque and and rpm and a sample problem that uses them to find power...

The overall travel diameter is approximately 10 feet with the only real Mass being the 100-kilogram out on each end. However I'm sure the support mechanism is also contributing to some sort of kinetic energy but let's just leave it out of the equation for right now if it's easier. Please advise

These pieces of information were not used in your sample problem, right? So why do you think they are relevant here?

Yes i would agree. However i am seeking how much work is done here-not power.

It doesn't seem like you understand how work and power are related. Please state your understanding of the relationship so we can tell for sure.

Would it be (120ftlbs or 167.2 Nm) x (15rpms x 2pie) = 15750 J?

Please pay more attention to your units. An "rpm" has per unit time (minutes) in it. A joule does not. So these units do not match. However, if you are looking for Joules expended in one minute, that's the correct answer.

And if it took 60 seconds to do 15 rpms then the power would be 263 watts or .35 HP?

Again, you expressed it wrong, but the answer is correct for what you probably really meant. Here's how to say it: 15 rpm is 15 revolutions in one minute, or 60 seconds. So 15750 / 60 is 263 joules per second or 263 watts.

 

[Pinon1977]

Great. I appreciate the feedback. I apologize if my verbiage or terminology isn't 100% up up to speed. Okay, so that being said, would 263 joules per second would produce 15780 joules over 60 seconds? Would the units be joules per minute at that point?

 

[russ_watters]

Great. I appreciate the feedback. I apologize if my verbiage or terminology isn't 100% up up to speed. Okay, so that being said, would 263 joules per second would produce 15780 joules over 60 seconds? Would the units be joules per minute at that point?

You could say that, but nobody does because it isn't useful and can cause confusion. People use watts.

 

[Pinon1977]

Last question. So the total amoubt if work done in a 60 second time period would be 15780 J? Correct?

 

[haruspex]

Last question. So the total amoubt if work done in a 60 second time period would be 15780 J? Correct?

Yes.

 

[Pinon1977]

Thank you for yalls patience. It is appreciated.

 

[CWatters]

If the "torque being exerted" is the torque required to keep it turning at a constant 15rpm then the input power is indeed about 263W or 15780 Joules per minute.

If there is no load connected to the system 263W is also the same power lost to friction and/or air resistance.