Work Energy Method for Rotational Motion

In summary: In this case, it would be ω2-ω02=2*α*Δθ which is equal to 2*π*Δθ or 8π. So in summary, at the end of the 8th revolution, the omega of the wheel is at 8π.
  • #1
freshbox
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0

Homework Statement


Hi guys, I would like to ask for this particular question, The GPE is 0 for all position is because I set datum as the original position, GPE = 0 and when the string got detached from the axle, there is no GPE for datum 2. Am I right?
Thanks
 

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  • #2
Doesn't the height of mass B change?
 
  • #3
You mean there is GPE? o_0

Because i see the working for E2 and E3 equation the GPE is 0 and they got the answer
 
  • #4
freshbox said:
You mean there is GPE? o_0
Again, what matters is change in GPE. And since mass B changes height, there is a change in GPE regardless of what you choose as your reference point.
 
  • #5
freshbox said:
Because i see the working for E2 and E3 equation the GPE is 0 and they got the answer
Not sure what you mean.
 
  • #6
Since you say height changes = there is GPE, how come the GPE in the working for E1 and E2 are 0 and able to get the answer.
 

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  • #7
freshbox said:
Since you say height changes = there is GPE, how come the GPE in the working for E1 and E2 are 0 and able to get the answer.
For that part of the problem (the slowing down of A after the mass B has detached) only the energy of A is relevant. And the GPE of A doesn't change. (The height of B changes as it falls, not the height of A.)

And since you are given the angular acceleration of A as the mass is falling, there's no need to bother with the changing GPE of B even for the first part of the problem.
 
  • #8
GPE of A doesn't change because it's a wheel.

So the "E" equation is all about wheel A?
 
  • #9
freshbox said:
GPE of A doesn't change because it's a wheel.
The GPE doesn't change because its center of mass doesn't change height.
So the "E" equation is all about wheel A?
Right. They are equating the work done on A to its change in energy.
 
  • #10
Then for other questions how am I going to know whether to lnclude the GPE or not??

This is so hard :cry:
 
  • #11
freshbox said:
Then for other questions how am I going to know whether to lnclude the GPE or not??
It just depends on the system you choose to analyze. Which is often up to you. (Many problems can be solved in multiple ways.)

In this particular problem it seems to me that they gave you too much information. They didn't need to give you the acceleration. But perhaps they wanted you to use a certain method of solving it.
 
  • #12
Can I say that if the question give me info on parcel B and nothing on A, there will be GPE?

And also my lecturer says that everything is same for work energy for linear and rotational. But the only difference is that kinetics for rotational has an additional "1/2Iω2

I am confused on when to include linear K and rotational K. Can you tell me how do i spot?


Thanks alot..
 
  • #13
freshbox said:
Can I say that if the question give me info on parcel B and nothing on A, there will be GPE?
If you choose to analyze mass B there will certainly be a change in GPE since it moves vertically.
And also my lecturer says that everything is same for work energy for linear and rotational. But the only difference is that kinetics for rotational has an additional "1/2Iω2

I am confused on when to include linear K and rotational K. Can you tell me how do i spot?
For rotational K, something with mass has to be rotating. For example: A spinning wheel or a rolling cylinder.

For linear K, something needs to translate--its center of mass has to move.
 
  • #14
Doc Al said:
It just depends on the system you choose to analyze. Which is often up to you. (Many problems can be solved in multiple ways.)

In this particular problem it seems to me that they gave you too much information. They didn't need to give you the acceleration. But perhaps they wanted you to use a certain method of solving it.

We need angular acceleration to find angular velocity at the end of 8rev. I'm pretty sure they have assumed α is constant, so we shall use ω202=2*α*Δθ. But the info on mass B is totally irrelevant.
 
  • #15
For the above mentioned question, how come linear K is not included? The parcel is attached to the axle and the parcel moves, there should be linear K right?
 
  • #16
Find final omega at the end of 8 rev using:

ω202=2αΔθ. (ω0=0).Make sure to convert revolution to radians.

Find Io=k2mA, k: radius of gyration

T=0.5mAvA2+0.5Ioω2

Since A doesn't have any translational motion, vA=0. Now you can find T(kinetic energy) easily.

For part B, ƩMo=Ioα, Find α from here(which is negative if we take clockwise positive).

Use ω202=2αΔθ again to find Δθ and finally convert it to revolutions. Note also that your final ω now is zero and your initial ω is what you found earlier(at the end of 8rev).
 
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  • #17
supernova1387 said:
We need angular acceleration to find angular velocity at the end of 8rev. I'm pretty sure they have assumed α is constant, so we shall use ω202=2*α*Δθ. But the info on mass B is totally irrelevant.
You are given the angular acceleration, but you don't need it. You should be able to figure it out from the other information given.
 
  • #18
You mentioned that "It just depends on the system you choose to analyze. Which is often up to you"

For the 1st diagram, at first i tried to include A & B in my E equation but you said that I can choose a system (1) which is A to solve since they gave me a lot of info on A.


For the diagram below, I solved this question with my E equation having A and B (2 System)

My initial understanding is that you need to include whatever thing you see in the picture and solve since it's connected together.

But you said that i can choose the system and solve it, so can this question be solve by just using 1 system eg. A only?
 

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  • #19
freshbox said:
For the above mentioned question, how come linear K is not included? The parcel is attached to the axle and the parcel moves, there should be linear K right?
Since you are given the angular acceleration, you don't have to worry about the kinetic energy of the falling mass. All you need to analyze is mass A, which has no linear kinetic energy. (If the angular acceleration wasn't given, and you wanted to use energy methods, then you'd have to include the kinetic energy of the falling mass in your analysis.)
 
  • #20
freshbox said:
For the diagram below, I solved this question with my E equation having A and B (2 System)
Good.
My initial understanding is that you need to include whatever thing you see in the picture and solve since it's connected together.
Not always. Sometimes they give you extra information that makes it easier. (Like they did in your original problem.)
But you said that i can choose the system and solve it, so can this question be solve by just using 1 system eg. A only?
No. (I said that you can often choose the system. But sometimes you have no choice.)

In your original problem they gave you all the information needed so you just had to analyze mass A. But that's not the case here.
 
  • #21
I see. Ok thanks for your explanation and time. I will practise on other question, see you.
 
  • #22
Doc Al said:
You are given the angular acceleration, but you don't need it. You should be able to figure it out from the other information given.

OK and how exactly are we supposed to get alpha? I can think of a solution but my answer doesn't match what the problem says. Here is what I think:

The axle rotates 8rev. I assume the height h that mass B falls down is also 8rev. That is:

h=8(2πrB) , by B I mean diameter of axle

The final velocity of block B would then be : v2-v02=-2gh where v0 is zero. This way we get v for string after 8rev. We can now say:
v=rBω

And we can find ω=161.05 rad/s. Using ω202=2αΔθ , we don't get the same value for α as the problem says. What do you reckon?

_____________________________________________________________________________
Edit to the above solution:

I realized where is the problem:smile:. I can't neglect the string force on block B and assume free fall. To get alpha I did the following:

1- Take moment about O (of wheel) and get: TrB=Ioα
2- write down Newtons 2nd law for block B. That is: mg-T=mBa=mB(rBα)
3- Using 2 equations and 2 unknowns, we can get the correct answer for alpha. That is α=1.056 rad/s2:wink:

My coordinate system is +x→right, +y→down and +z→into the page(clockwise positive) which obeys right hand rule.
 
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  • #23
Good. At least we know the data is consistent. :wink:
 

1. What is the work-energy method for rotational motion?

The work-energy method for rotational motion is a technique used to analyze the motion of objects that are rotating. It involves calculating the work done by the forces acting on the rotating object, and using the work-energy theorem to determine the object's final rotational kinetic energy.

2. How is work defined in rotational motion?

In rotational motion, work is defined as the product of the tangential force applied to an object and the distance over which the force acts. This is similar to the definition of work in linear motion, where it is the product of force and displacement.

3. What is the work-energy theorem for rotational motion?

The work-energy theorem for rotational motion states that the net work done on an object is equal to the change in its rotational kinetic energy. This means that if the net work done on an object is positive, its rotational kinetic energy increases, and if the net work done is negative, its rotational kinetic energy decreases.

4. How is the work-energy method used to solve problems in rotational motion?

The work-energy method for rotational motion is used by first calculating the work done by all the forces acting on the rotating object. This is then equated to the change in the object's rotational kinetic energy, which can be solved for the final rotational velocity or angular acceleration of the object.

5. What are the advantages of using the work-energy method for rotational motion?

The work-energy method for rotational motion is advantageous because it allows for the analysis of complex rotational motion problems without having to use calculus. It also provides a direct relationship between work and energy, making it easier to understand and apply in problem-solving.

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