Swing and Levitate
The animation shows a football player, Josh Imatorbhebhe, jumping high in the air.
He swings his arms in the air and appears to levitate in the air for a brief moment of time.
How does swinging his arms help him accomplish this feat?
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9 solutions
During the interval of apparent levitation, his center of mass is moving downwards and his arms are also moving downwards. How is this considered the opposite direction?
While I don't think there's a good choice as an answer to this question, saying that the downward acceleration of his arms causes an opposing force that pulls him up (A) seems like a better answer.
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His center of mass is moving upward during the first half, downward during the second half.
You are right that the CM and his arms move in the same direction.
I believe either of the first two answers might be acceptable. The jumper swings his arms up gaining upward momentum, with the force being supplied be the interaction of his feet with the ground. Then, having established that vertical momentum, the jumper uses his legs to give his body additional vertical momentum. Actually, the jumper should coordinate his arms and legs so that he leaves the ground just as his arms a vertical.
CAB
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The first answer is incorrect. If the jumper would keep his arms downward instead of upward in the middle of the jump, his body would end up going higher. By swinging his arms upward, the jumper pushes the rest of his body downward so it stops going upward.
Tonight, I will post graphs of the heights of the various centers of mass to illustrate this.
Sorry, I don't understand that xD
The provided answers are all partial choices, possibly intended to mislead readers into a simplistic interpretation.
CAB's suggestion is almost right about vertical arms, and they were. Bear in mind that the CG of his arms was accelerated in a vertical circle (which is already transferring vertical momentum to his shoulders), so that had to be a compromise with the precise timing of his subsequent arm swings.
His timing was perfect for each of his 3 arm swings: a) to maximize his vertical momentum, and add more back (not just traps) & shoulder muscle effort to the jump by swinging his arms forward & (head & shoulders) upward during acceleration. b) to compensate (avoid rot. decel. around the X axis) for the consequent backward rotation to keep his feet under him for the landing, c) to allow his feet to hold ('magically') still at their apex by accelerating his arms downward, allowing his CG to follow the start of the downward trajectory while his hips & feet paused during that 2nd down sweep of his arms. It is difficult to notice the difference in the trajectory of his CG and his feet, except at the apex, where it is dramatic. Also it is easy to see the extreme effort he puts into that down stroke. And does beg the question of how much aerodynamic lift that powerful arm movement does provide (and his hands are not vertical during the greatest -Vy!)
Magnificent demonstration of a skilled athlete blending physics, bio-mechanics, and showmanship. I would have liked to ask him if what I observed him do was intentional or intuitive(unconsciously trained).
I would also ask to video an identical jump with NO arm movement, to run concurrently with the original on this page for timing, position, & energy comparison.
PS: Realize that if he had not gone to the effort of 'pausing' his feet, I think they could have gone a little higher (He could even have bent his legs & torso at apogee to achieve more ground clearance, or after apogee to achieve a less 'magical' looking foot pause.)
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A rough estimate of aerodynamic lift comes from the Raleigh drag equation, F D = 2 1 ρ v 2 C D A . Order of magnitudes are
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ρ (density of air) = 1 . 2 × 1 0 − 3 k g / m 3 ;
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v (speed of arm relative to air) may be estimated as v ≈ π ℓ / T ≈ 1 5 m / s ;
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C D (drag coefficient) ≈ 1 ;
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A (cross-sectional area of both arms) ≈ 2 × 0 . 9 m × 0 . 2 m ≈ 0 . 4 m 2 .
This gives F D ≈ 0 . 0 5 N , which is only a fraction of a percent of the gravitational force. It is safe to conclude that aerodynamics plays no role here. (And, incidentally, that people really can't fly.)
Sir could u pls explain why u did not multiply m with u and h in the equations!!
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Here is the full derivation: let h s be the shoulder height and h t be the height of the center of mass of the rest of the body above the feet. Then y C M = M m y arms + ( M − m ) y rest = M m ( h s + u ± ℓ ) + ( M − m ) ( h t + u ) = M m h s + ( M − m ) h t + M m u + ( M − m ) u ± M m ℓ = h + u ± ℓ m / M , where h is the weighted average of h s and h t .
It's late here and I haven't thought this through for any length of time, but I don't like the wording of the answer either. The simplest 'handwaving' argument that occurs to me is:- From the topmost position, gravity requires that his centre of mass moves downwards a certain amount in a certain time. The athlete satisfies this requirement by moving his arms downwards, so his arms are surely moving in the same direction as his centre of mass? Edit:- And before he reaches the topmost position, he starts to move his arms up, slowing the rest of the body down, but his arms are still moving in the same direction as his (total) centre of mass.
We have to take into account the fact that when Josh is throwing his arms upward, he is infact throwing his entire body downward with the same force (Newton's 3rd law). In the beginning of his hang time he throws his arms upward by throwing his body downward. His center of mass doesn't go anywhere, it remains at the same place relative to ground or i'd rather say he moves with equal velocity both up and down for his hang time. Hang time = the interval of time in which he's upward velocity greater or equal to his downward velocity ≈ amount of time his arms go a rull rotation.
Your solution is equally as correct as mine, you're saying that he hangs up in the air because his c.m shifts its position up. I'm saying that he hangs in the air because his c.m goes up and down at the same time.
P.S air also adds up to the equation a little. Summary: There is no wrong answer really. Everyone has his own explaination and everyone might be equally correct
you did a mistake in solving the equation: ycm(t) = ycm(±T) h + u + lm/M - 0.5gt^2 = h + u -lm/M 2lm/M = .5gT^2 (t = T) T = 2sqrt(lm/Mg) {you missed to multiply 2}
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Going up and down at the same time is not possible.
"His center of mass doesn't go anywhere, it remains at the same place relative to ground or i'd rather say he moves with equal velocity both up and down for his hang time. "
That would be true if the force between his arms and there rest of his body were the only force. However, the force of gravity is an external force and will accelerate the center of mass downward at a rate of g , no matter what Josh does with his arms.
As for the influence of air: the drag force on his arms is about 0.05 newtons, which is negligible in comparison to his weight of nearly 1000 N.
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That's what i'm saying. His center of mass moves up with constant velocity (newton's third law) and gravity is pulling on him with the same velocity well.. we can consider constant velocity because it's for very very small interval (about 0.06 seconds) . So even with gravity included his center of mass will still stay put relative to ground ( what i meant by moving up and down at the same time). P.S you forgot to multiply 2 in that equation.
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@Rougeami Cypher – Are you talking about the center of mass of his trunk or about the center of mass of his entire body ?
The former stays in place because, as you imply, the upward force from his arms balances the downward force of gravity.
You are right that T is half of the "hang time". Note that I did multiply by two in the last line when I wrote "0.2 seconds".
Let M be the mass of Josh's entire body ----> includes mass of his arms and snickers ... ----> center (height) of Josh's entire body mass in a change (varies up/down) until his arms either upward or downward. At the instant time t when velocity is near ----> 0 and Josh's moving his arms at a max possible acceleration upward, his whole body receives additional impulse upward when v approaches -----> 0 and Josh's body yet not in a free fall. So, for a small fraction of the time near 0.1s in average human eye recognizes changes. Example. Alternate current and a light from a bulb. Note. If Josh would have some weight in both arms and at the highest point of the jump (the same presentation) pushes objects upward , then what will be expected free fall delay? ( T = 0.09, or longer)
Isn't there a missing 2 in the T equation or I am wrong?
After the athlete has reached the maximum height, his lower portion of the body will have a downward velocity which tries to move the center of mass downwards.
Because his arms have upward velocity at that instant, it will result in zero velocity of the center of mass.
The athlete is rising and spinning his arms in a circular path. As he approaches his apex, he is swinging his arms up, halting the ascent of his feet, legs and torso, while his arms are going up, so his center of mass continues to rise, while his feet have stopped rising - his ascent appears to have been stopped. Then as his arms swing down, his center of mass shifts down, without his feet or the rest of his body descending, yet - his descent appears to have been delayed, while in fact it has been confined to his arms. After his arms have fallen, he is obviously fallen.
While moving downwards from topmost position, his arms provide a slight upward momentum in opposite to downward momentum of centre of mass.which results in a slight moment of levitation.
The athlete is creating an additional momentum upwards as he accelerates his arms (with their significant portion of muscles). You can see in the video that the arms are most raised when he is in top position. As there is no free lunch he creates additional momentum going down again while his arms spin downwards. Therefore I disagree that this is correctly summarized as that the center of gravity of the arms goes opposite of the center of gravity if the body.
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The athlete does not create any additional momentum for his entire body. The total momentum of a system can only be affected by external forces; in this case, only by the force of gravity.
The muscle action does affect the momentum of his trunk and of his arms, in opposite ways (per Newton's Third Law). For the changes in momentum d p during any time interval d t we have d p trunk + d p arms = − m g d t . During the "hang time", d p trunk is approximately zero, while d p arms is directed downward.
Another way of looking at it: The force at the shoulders transports all of the downward momentum gained by the gravitational force to the arms, causing them to rise slower and then to move downward.
His center of mass stays in the same "parabolic" path, but his center of mass changes as he moves his arms.
The center of mass can be imagined as an "average" of the positions of his limbs, when he swings his arms up, the "average" goes a bit up, and to compensate the rest of the body goes a bit lower since the center of mass needs to stay on the path.
At the top of the "parabolic path", the center of mass starts to fall, but he swings his arms back down, changing the center of mass compared to the rest of his body downwards, but since this would make it seem that he starts to fall faster, to compensate the rest of his body rises a little, and with the center of mass going down and the body going up, they cancel each other out making it look like he's levitating.
Best explanation alongside with Mr. Arjen's one .
The moment of Josh levitation results in the mass of movement and the air time pushing upwards causing him to levitate
bold text اخوكم حسن منورين والله من قلب بورسعيد
For influencing gravity he would need to be heavier (more like an astronomically interesting object) compared to the earth.
For influencing air resistance he would need to be lighter (more like a bird/feather).
Discussions concerning the centre of mass at least seem to be in the right order of magnitude for the possibility of brief levitation.
The Newton's Third Law 'effect' is certainly a major one in this application. If you have ever done 'chin-up' exercises to the point of failure of your arms to pull the rest of your body upward, you may have raised your feet and kicked downward to get your chin up to the bar - and done that repeatedly if the first time didn't suffice. By kicking downward (the action), the body trunk and head - and also the legs when they straightened out because that was the culmination of the action (and of course the fact that they are attached to the hips, trunk, etc.) - experienced the reaction of moving upward (except the arms which are in a fixed position because the hands are holding the chinning bar) -- allowing another chin-up to be done. Try it and see.
In the case of the guy in the video clip, the action is the downward thrust of his outstretched arms and the reaction is upward movement of the entire body. All of this happens in a continuous (and movement changing) manner because of the nature of the bodyparts being connected.
Let M be the mass of Josh's entire body, m the mass of his arms, and ℓ between his shoulders and the distance of the center of mass of the arms. Let h be the height of his center of mass above his feet when his arms are held sideways, and u the height of his feet above the ground.
In the middle of the motion, Josh's arms are stretch above his head, bringing his center of mass to a height of y C M ( 0 ) = u + h + ℓ m / M . At the beginning and end of his "hang time", his arms are stretched downward: y C M ( ± T ) = u + h − ℓ m / M . Now the center of mass y C M is in free projectile motion. Thus y C M ( t ) = u + h + ℓ m / M − 2 1 g t 2 . Substituting t = ± T we find T = 2 M g ℓ m . This is half of Josh's hang time. If his arms have their CM at 0.45 cm from the shouder and the mass of his arms is m = 0 . 1 8 M , then T = 2 1 0 0 . 4 5 ⋅ 0 . 1 8 = 0 . 2 s , allowing for a hang time of about 0.4 seconds. This video must therefore be slowed down about 2 times.