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angular motion

Discussion in 'Mathematics and Physics' started by PG1995, Jun 6, 2015.

  1. PG1995

    PG1995 Active Member

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    Hi

    Please have a look here.

    The text on left doesn't really make sense to me. When the arms are closed the moment of inertia decreases and to preserve angular momentum and rotational kinetic energy the angular velocity should increase. Naturally, to bring them arms inward one has to do biological work but in my view this doesn't mean it counts toward increasing rotational kinetic energy.

    Let me say it differently. The skater would still have to do 'biological' work to bring her arms closer in to her body (assuming her arms were extended outward) even when she is not rotating. Suppose that she has to 10 J of work to bring her extended arms inward while being stationary. I believe that she would need to expend the same amount work to bring her arms inward even when she is rotating.

    Thank you.

    Regards
    PG

    Helpful links:
    Types of motion:
    http://www.bbc.co.uk/schools/gcsebitesize/design/systemscontrol/mechanismsrev8.shtml

    Three important assumptions about friction between surfaces:
    http://hyperphysics.phy-astr.gsu.edu/hbase/frict3.html

    Difference between circular and rotational motion:
    http://www.differencebetween.com/difference-between-circular-motion-and-vs-rotational-motion/
    https://in.answers.yahoo.com/question/index?qid=20111017095951AANRUEm
    https://answers.yahoo.com/question/index?qid=20080627221013AAE9zDJ

    Difference between rotational and angular motion:
    They are basically different names for the same thing.

    Angular motion:
    http://francesa.phy.cmich.edu/people/andy/physics110/book/chapters/chapter6.htm
     

    Attached Files:

  2. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Just a quick note...

    If you have a mass on the end of a string with a spring scale tied in the middle of the string so you can measure the pull force the mass exerts as you hold the other end of the string, when the mass is still you will register the force from the weight of the mass only. But when you twirl it around in a circle you will measure more force because it is trying to fly away from the center.
    Likewise, if the mass on the string was on a frictionless horizontal plate you could move it easily from side to side, but if you rotate it in a circle with the other end of the string at the center, it will be harder to pull in toward the center and easier to let back out to the original rotation diameter.
    Thus the girls arms are harder to pull in toward her body when rotating and easier to let back out to full length to her sides, which would be much different if she were standing still. In fact, there may be some speed where she doesnt even have the strength to pull her arms in, although this might require a rotational speed faster than humanly possible.
     
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  3. jpanhalt

    jpanhalt Well-Known Member Most Helpful Member

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    Hi PG, it's been awhile.

    Try your own experiment. Stand still with your arms extended. Do that for a few hours, and your arms will probably drop from their own weigh. If you are spinning; however, it requires work to more your arms in. You can test that with socks and a hardwood floor. I am not sure of your exact question, but the section in green that you have highlighted is correct.

    upload_2015-6-6_8-7-25.png

    John
     
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  4. dave

    Dave New Member

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  5. PG1995

    PG1995 Active Member

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    Thank you, MrAl, John.

    So, according to both of you the text in green is correct and the skater has to expend a certain amount of work to bring her arms closer to her body and this expended work get transferred to her rotational kinetic energy. Right?

    That certain amount of work is not fixed and would vary depending upon her rotational speed before she starts bringing her arms inward. For example, she will need to do more work to bring her arms closer to her body when she is rotating at 160 RPM than when she was rotating at 80 RPM. Right?

    Suppose that before the skater decides to bring her arms closer to body, her rotational kinetic energy is 120 J and that energy should be preserved. Now she starts bringing her arms inward and hence she expends certain amount of 'biological' work/energy for this purpose. Let's suppose that the expended 'biological' energy/work is 30 J. Once she has completely brought her arms closer to body, her total rotational kinetic energy should be now 120+30=150J. Most probably what I'm saying is not correct but I believe that now you can see better where I'm having confusion. Thanks.

    Regards
    PG
     
  6. Ratchit

    Ratchit Well-Known Member

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    I don't see where the confusion is. The rotational kinetic energy is proportional to the moment of inertia and the angular velocity squared. If the skater retracts the arms so as to reduce the moment of inertia by one-half, then the angular velocity will double, and the square of the angular velocity will quadruple. The one-half moment of inertia and the quadruple angular velocity will double the rotational kinetic energy when the skater retracts. The skater's muscles will have to provide the needed energy to do the maneuver.

    Ratch
     
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  7. PG1995

    PG1995 Active Member

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    Thank you, Ratch.


    [​IMG]


    Do I make sense? Kindly let me know. Thanks.

    Regards
    PG
     

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  8. Ratchit

    Ratchit Well-Known Member

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    Yes, it does make sense. You are repeating what I said previously.

    Ratch
     
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  9. PG1995

    PG1995 Active Member

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    Thanks.

    Yes, your posting paved the way for understanding and I only had to put everything in mathematical order.


    Now let's imagine the situation when the skater decides to extends her arms outward again. Her angular velocity is supposed to be 40 and moment of inertia to be 5. Once she has extended her arms outward completely, her angular velocity would be 20 and moment of inertia 10. Her kinetic energy would be 2000 J. But when her arms were close to her body, her rotational kinetic energy was 4000 J, so where did the rest 2000 J go? I belive this energy go into pulling her arms outward. In other words, she didn't have to do any work to extend her arms, the arms got extended on their own using system's 2000 J of energy. Is this correct?

    Regards
    PG
     
  10. Ratchit

    Ratchit Well-Known Member

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    Yes.

    Ratch
     
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  11. PG1995

    PG1995 Active Member

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    Hi

    Suppose that a bucket full of water is let to fall from the roof of a 5-storey building using the pulley system like the one shown here. By the time, the bucket is just going to hit the ground, the rope is pulled back so that the bucket doesn't touch the ground, and this is done successfully. Naturally, the bucket had a significant kinetic energy before it was stopped by a human, where did this kinetic go? I understand that human muscles (or, biological energy) were used to stop the bucket movement but I can't clearly see the transfer of bucket's kinetic energy into any other form? Please help me with this. Thanks.

    Regards
    PG
     

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  12. Ratchit

    Ratchit Well-Known Member

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    Force through a distance equals energy or work. The rope was pulled on with perhaps a hundred pounds of force by the human with friction on his hands (ouch) or hand under hand grabbing. It took perhaps 10 feet before it stopped. So the kinetic energy that what was dissipated by the human's muscles and blisters was 10 times 100 or 1000 ft-lbs.

    Ratch
     
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  13. PG1995

    PG1995 Active Member

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    Thank you.

    Forget the blisters and let's focus on only dissipation by the human muscles. How does this dissipation take place? I can imagine that the bucket will try to pull the arm outward and the muscles will counteract this pulling. To counteract this pulling the muscles would basically be investing 'biological' energy... Kindly guide me. Thanks.
     
  14. Ratchit

    Ratchit Well-Known Member

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    Yes, each arm will be exerting a force on the rope while extending the length of its reach. So force through a distance equals work, right?

    Ratch
     
  15. PG1995

    PG1995 Active Member

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    Hi

    Q1:


    I'm still confused. Let me rephrase it.

    Suppose the bucket is let to fall directly on the ground. The kinetic energy will be converted into sound energy, thermal energy, etc. So, we can see that how the kinetic energy is being transferred. But we chose the buck not to fall directly on the ground and halted its movement as it was just going to touch the ground. Some of the kinetic energy converted into heat energy as friction. The bucket's kinetic energy also try to pull bones of your arms out of their sockets but your muscles act like massive spring with very large spring constants. It means that besides the conversion of kinetic energy into frictional energy, the rest of energy goes into your muscles in form of thermal energy (temperature of your muscles will rise), elastic energy because your muscles will be minutely stretched and so on. But the muscles in your arms need a lot of 'biological' energy to function like massive spring.

    Do I make sense? I believe that at least now yoiu can see where I'm having confusion. Please guide me. Thanks.

    Q2:
    According to Newton's Third Law there is an equal and opposite reaction force for every action force.

    In a circular motion such as where a stone or ball is whirled round in a horizontal circle at a constant speed, the centripetal force is used to keep the ball in a circular motion. The centripetal force is directed toward the center of motion. It is also said that the centrifugal force, equal and opposite reaction force to the centripetal force, does not exist. I tend to disagree with this statement.

    Think of earth as a perfect sphere. When you are running in direction of east (basically you are pushing the earth in the other direction using motion of your feet), you are making the earth rotate in direction of west. Because the earth is so huge, we cannot discern this rotation. Along the similar lines, I do think there does exist a kind centrifugal force which tend to pull outward. Think of yourself as a pole on earth. You are whirling a stone around you using your arm. Your arm exerts a force on the stone which makes it whirl around you but at the same time the stone exerts an equal and opposite force on your arm (as your arms, body and earth constitute a single system), therefore your arm tend to follow a circle and forms the perimeter of base of cone and your feet which are resting on earth form the vertex. My imagination might be little off but I hope that you get my overall point. Thank you.

    Regards
    PG
     
    Last edited: Jun 18, 2015
  16. Ratchit

    Ratchit Well-Known Member

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    For question #1, I believe you have the correct understanding. I don't see why you are confused, or what the confusion is.

    For question #2, you did described a scenario, but did not articulate what the point of your confusion is.

    Ratch
     
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  17. steveB

    steveB Well-Known Member Most Helpful Member

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    I also tend to disagree with this statement, and agree with you. First, let's clarify what you said. The claim is not that it does not exist, but that it should not be called a centrifugal force.

    This is a subject that can start a holy war among physicists. Physicist have banned the terminology of centrifugal reaction force. It seems they find this term confusing and improper for teaching.

    The confusion arises because normal centrifugal force is a virtual (or fictitious) force that results from a rotating (or non-inertial) reference frame. The concept of "reaction centrifugal force" is different in that it is simply the equal and opposite force to centripetal force. Newton's tells us that for every force, there is an equal and opposite reaction force. Therefore if centripetal force exists, then reaction to that force exists. This reaction force used to be called (long ago) reaction centrifugal force, but the term was often shortened to just centrifugal force. Therein lies the confusion, because the centrifugal force and reaction centrifugal force are two different things, but they came to have the same terminology.
     
    Last edited: Jun 18, 2015
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  18. Ratchit

    Ratchit Well-Known Member

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    For the life of me, I cannot see what the confusion is about. If a satellite rotates around a planet, there are two forces of equal and opposite direction. The first is the gravitational attraction of the planet on the satellite which is centripetal force. The second is the centrifugal force the satellite exerts by its constant acceleration caused by its constant change of direction. The two forces balance out and constitute a stable system. What is confusing about that?

    Ratch
     
  19. steveB

    steveB Well-Known Member Most Helpful Member

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    As far as the original question goes, I'd like to add some further insight. I think that what is answered above is basically correct, but keep in mind that the task of stopping an object in this way is a very complicated thing to analyze and to figure out exactly where and how the energy is dissipated.

    Consider this thought experiment. Instead of a person stopping the bucket, imagine that a weight/mass is used to stop it. A block of steel, for example, could have a eyehook that the rope freely passes through. Then a knot in the rope can be placed at the proper place so that the knot stops at the eyehook and catches the steel block on to the rope and stops the bucket (because it is heavy enough).

    First we have to realize that the rope itself has stretch (spring force) and energy loss (damping force) and the damping in the rope eats up some (probably most) of the energy. If the weight is very heavy, then it will not get lifted off the ground, and the stopping of the bucket will be a very abrupt event. But, the main place the energy goes (assuming the bucket does not shatter due to the stress) is basically in the damping of the rope.

    Also, we could imagine that initially the weight gets lifted off the ground and that potential energy sucks up some of the energy, but then the weight will probably go back down afterwards. This might give a less abrupt event, buy still it is the damping in the rope that mainly dissipates the energy. The kinetic energies and potential energies of the bucket and weight come into play, but in the end, the bucket is closer to the ground and the mass is in the same place and there is no kinetic energy. Hence, the potential energy loss of the bucket must have been dissipated, and the rope is the place where the damping occurs. Consider that if the rope has very low damping, the bucket will oscillate up and down for a long time while the dissipation happens gradually. If the rope has high damping, then the bucket will stop quickly.

    So, in some sense, the person can act like a heavy weight and do the things described above. However, since humans are intelligent, a person can use muscles and energy dissipation in his/her own body to change the way the bucket stops. In other words, a person can take up some of the dissipation that the rope would normally have to provide, and in doing so, can provide a less stressful and less abrupt event in stopping the bucket.
     
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  20. steveB

    steveB Well-Known Member Most Helpful Member

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    Well, all physicists would say that the centrifugal force is a fictitious (or virtual force). That force only exists as a virtual force in a frame of reference that moves with the satellite. It is caused by the non-inertial frame of reference. That's all. I don't find it confusing at all. But, what you described is not "reaction centrifugal force", but ordinary centrifugal force.
     
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  21. Ratchit

    Ratchit Well-Known Member

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    Well, there is nothing fictitious or virtual about the force caused by a centrifuge. And, I don't hear too much discussion about its noninertial frame of reference, either. So, perhaps the physicists should explain themselves better.

    Ratch
     

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