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What are the units on...Pinon1977 said:You lost me on that one, sir. Can you ask your question in another way?
Edit: In particular, pay attention to the seconds. Then compare toPinon1977 said:300kgm2 x 1.57 rad/sec
What are the units on that?471.6 Nm
Do you want work done or kinetic energy?Pinon1977 said:10 feet in length ...excludng the support feature of 20kg, the 100kg mass on each end has a moi of 231 and 286j.
@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?russ_watters said: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.russ_watters said: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?
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.Pinon1977 said: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
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.Pinon1977 said:work is done by this device
That is enough to compute the power, as already indicated in this post #11.Pinon1977 said:The torque being exerted in this particular instance would be 120 foot pounds of torque at 15 RPMs
If it is just a support then I would think it is stationary, so does no work.Pinon1977 said:the support mechanism is also contributing to some sort of kinetic energy
Come again? 15 rpms is a rate.Pinon1977 said:And if it took 60 seconds to do 15 rpms
What units would that have? What units do you need?Pinon1977 said:15rpms x 2pie
Joules are units of work, not power. As already explained, torque times rotation rate gives power.Pinon1977 said:15750 J
I assume you mean it maintains 15rpm for 60 seconds.Pinon1977 said:it took 60 seconds to do 15 rpms
Ok, great. You have a torque and and rpm and a sample problem that uses them to find power...Pinon1977 said:My apologies. The torque being exerted in this particular instance would be 120 foot pounds of torque at 15 RPMs.
These pieces of information were not used in your sample problem, right? So why do you think they are relevant here?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
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.Yes i would agree. However i am seeking how much work is done here-not power.
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.Would it be (120ftlbs or 167.2 Nm) x (15rpms x 2pie) = 15750 J?
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.And if it took 60 seconds to do 15 rpms then the power would be 263 watts or .35 HP?
You could say that, but nobody does because it isn't useful and can cause confusion. People use watts.Pinon1977 said: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?
Yes.Pinon1977 said:Last question. So the total amoubt if work done in a 60 second time period would be 15780 J? Correct?
Total work on a rotating object refers to the amount of energy expended to rotate the object through a certain angle or distance.
To calculate total work on a rotating object, you must multiply the applied force by the distance the force acts in the direction of rotation, and then multiply that by the cosine of the angle between the force and the direction of rotation.
The formula for total work on a rotating object is W = F * d * cos(theta), where W is the total work, F is the applied force, d is the distance the force acts, and theta is the angle between the force and the direction of rotation.
The direction of rotation does not affect the total work on a rotating object, as long as the force and distance are in the same direction. However, if the force and distance are in opposite directions, the total work would be negative, indicating that work is being done against the rotation.
The concept of total work on a rotating object can be applied in various real-life situations, such as calculating the work done by a motor in rotating a shaft, determining the energy required to rotate a wheel, or analyzing the power output of a wind turbine. It is also used in engineering and physics to understand the dynamics of rotating systems.