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freshbox
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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.freshbox said:You mean there is GPE? o_0
Not sure what you mean.freshbox said:Because i see the working for E2 and E3 equation the GPE is 0 and they got 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.)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.
The GPE doesn't change because its center of mass doesn't change height.freshbox said:GPE of A doesn't change because it's a wheel.
Right. They are equating the work done on A to its change in energy.So the "E" equation is all about wheel A?
It just depends on the system you choose to analyze. Which is often up to you. (Many problems can be solved in multiple ways.)freshbox said:Then for other questions how am I going to know whether to lnclude the GPE or not??
If you choose to analyze mass B there will certainly be a change in GPE since it moves vertically.freshbox said:Can I say that if the question give me info on parcel B and nothing on A, there will be GPE?
For rotational K, something with mass has to be rotating. For example: A spinning wheel or a rolling cylinder.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?
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.
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.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 ω2-ω02=2*α*Δθ. But the info on mass B is totally irrelevant.
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.)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?
Good.freshbox said:For the diagram below, I solved this question with my E equation having A and B (2 System)
Not always. Sometimes they give you extra information that makes it easier. (Like they did in your original problem.)My initial understanding is that you need to include whatever thing you see in the picture and solve since it's connected together.
No. (I said that you can often choose the system. But sometimes you have no choice.)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?
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.
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.
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.
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.
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.
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.