aa r X i v : . [ phy s i c s . pop - ph ] A ug Simple Model of a Standing Vertical Jump
Chris Lin
Department of Physics, University of Houston, Houston, TX77204-5005, United States of America.
August 27, 2020
Free fall occupies a large chunk of a standard course in introductory mechan-ics. For completeness, once Newton’s laws are learned, the physics of the takeoffshould be discussed. Although humans are in a privileged position of not havingto jump to navigate or to escape predators, jumping still holds enchantment,an expression of joy, or a climactic flourish to a sequence of moves in sportsand dance [1]. One can leverage the popularity of sports [2] and dance to getstudents excited at applying the physics they learned to investigate the jumpthrough construction of a simple model.Most discussions on jumping take as a starting point a force that developsbetween the object and the ground. Although a lot can be extracted fromthis model, particularly if a force plate is used to measure the ground force[3, 4, 5, 6], one drawback is that we know that jumping is initiated internallythrough muscle contractions before the force is communicated to the ground, sofrom a pedagogical perspective it would be nice to have a simple solveable modelthat attempts to show how internal forces causally lead to the development ofexternal forces. Moreover, since this model considers the internal structure ofthe jumper, we will be able to predict the ground force generated by a humanjumper, finding good agreement with force plate measurements.In this paper we use Newton’s 3rd law to deduce the simplest model of anobject that can perform a standing vertical jump – a two-segmented objectwith an initial constant repulsive force between the segments, followed by anabrupt attractive force. Such an object, when placed on a sturdy ground, willjump, and the motion can be calculated using only the constant accelerationequations, making the example suitable for algebra-based physics. We thenproceed to solve for the motion of an n -segmented object, and determine theoptimal number of segments for jumping. We then discuss a few similarities and1 Figure 1: Left – an internal repulsion causes separation of the two segments.Right – after a max separation distance d , the opposite force-pair become at-tractive.differences of this simple model from jumping robots and jumping humans, andthen conclude by arguing the model’s pedagogical merits. Consider an organism made of two parts, a top segment of mass m , and abottom segment of mass m , resting on a hard surface. Given that internalforces between the top and bottom segments must be equal but opposite, one canask students whether the internal force pair that initiates the jump is attractiveor repulsive. Students can reason that the top segment must move upwardsbefore the bottom segment can, so there must be an upward force producedinternally on the top segment, and from Newton’s 3rd law, an equal downwardsforce on the bottom segment, hence initially there is repulsion: see Fig. 1. Thetop segment accelerates upward, but the bottom segment is prevented fromaccelerating downward due to the ground, which therefore supplies an externalupwards force. As the top segment displaces upward, its separation from thebottom segment (which remains static, pressed into the ground) increases. Thisseparation cannot continue indefinitely if the object is to not break apart, sothe internal forces must change their direction and become attractive as theupper segment pulls the bottom segment upwards, and from Newton’s 3rd law,is itself pulled downwards by the bottom segment. One can then ask studentsto make an analogy of this model with jumping by humans, and would likely getanswers discussing how, from a crouched position, muscles exert internal forcescausing humans to unfold, but once the human has completely straightened to2each maximum separation, muscle forces switch to joint forces to prevent thetwo halves from separating. To calculate the speed at which the object leaves the ground, we model theinitial phase of the jump as comprising a constant repulsive force F exertedover a maximum separation distance of d between the two segments. Once thedistance d is reached, we model the attractive force between the two segmentsas an abrupt completely inelastic collision, after which the segments are lockedtogether.The velocity of the upper segment under constant acceleration a = F − m gm overa distance d is v f = s (cid:18) F − m gm (cid:19) d + v i . (1)We model the second phase as a completely inelastic collision between the lowerand upper segments, where the final velocity of the combined system is: v ′ f = m v f m + m (2)= m m + m s (cid:18) F − m gm (cid:19) d + v i . Once in the air, the object is in free-fall and the height of the jump is given by h = v ′ f g (3)= m ( m + m ) (cid:18) F − m gg (cid:19) d, where v i was set to zero because the upper segment is initially at rest.Traditionally, collisions are treated long after Newton’s 3rd law and the con-stant acceleration equations [7]. We therefore in Appendix B offer an alternatederivation of Eq.(2) using only the constant acceleration equations, so that themodel can be introduced earlier. We simplify the problem by having all segments be the same, m = m = ... = m N = m , and that the n th segment does not turn on its repulsion until the3 Figure 2: The multi-segment system treated like a two-segment system, whereone of the segments is the dotted box comprising n-1 segments, and the n th segment lies just below.( n − th segment has collided with the ( n − th segment: see Fig. 2. Atthat moment denote the velocity of the entire upper mass of ( n − m as v n − .After completing the interaction with the n th mass, the new velocity v n of thecombined mass nm is, using Eq. (2), v n = ( n − m ( n − m + m s (cid:18) F − ( n − mg ( n − m (cid:19) d + v n − . (4)Plugging in a few numbers v = 12 s (cid:18) F − mgm (cid:19) d + 0 = r r dm p F − mg (5) v = 23 s (cid:18) F − mg m (cid:19) d + v = r r dm r F − mgv = 34 s (cid:18) F − mg m (cid:19) d + v = r r dm r F − mg. From the pattern we guess v n = q n − n q dm q F − n − mg which we can verifyby indeed showing that it satisfies Eq. (4). Therefore for N identical segmentsthe velocity upon takeoff and maximum height attained are:4 f = r N − N r dm r F − N − mg (6) h = v f g = N − N (cid:18) m (cid:19) F − N − mgg ! d. Eq. (6) is also derived in Appendix A using the work-kinetic energy theorem.
If we express the force F between segments as a multiple α of the weight mg of each segment ( F = α mg , so that α is the strength-to-weight ratio of eachsegment), then the height in Eq. (6) becomes: h = 12 (cid:18) N − N (cid:19) (cid:18) α + 13 − N (cid:19) d. (7)The height of the jump is proportional to the uncoiling length d between seg-ments, but for a given α , different values of N maximize the height h . Studentsin algebra-based physics courses can numerically plug in different α values anduse graphing tools to find the N value which maximizes hd (see Fig. 3). α = α = α = N - / d Jump Height vs Number of Segments
Figure 3: Jump height ratio h/d vs number of segments N , for different valuesof the strength-to-weight ratio α of each segment. The curves have a decreasingtail that goes as h/d = α +12 − N + O [ N ], which is the large- N limit of Eq. (7).Calculus yields the exact result of N max ( α ) = q α +12 and h max ( α ) = h ( N max ( α )) =5igure 4: Two segments that repel via the hydraulic cylinder. ( √ α +1 −√ d , which implies that the greater the force each segment can exertrelative to its weight, the more segments you can add to maximize your jumpbefore adding becomes counterproductive. When α = 2 . α = 5 . N = 2and N = 3 (e.g. one can uncoil at both the knees and the hips using thequadriceps and glutes, respectively), suggesting 2 . < α < .
7. In Eq. (8) weshow that the ground force (as would be measured by a force plate) throughoutthe jump varies from a minimum value of F ming = ( α + 1) mg to a maximumvalue of F maxg = ( α + N − mg , which when written in terms of the totalmass M = N m corresponds to F ming = ( α +1) N M g and F maxg = ( α + N − N M g .Plugging in α = 2 . N = 2 and α = 5 . N = 3 predicts that a forceplate would measure 1 . M g for 2 segments and between 2 . M g and 2 . M g for3 segments. This is not too far from force plate measurements which put thethe force measured as between 2
M g and 2 . M g [5].
The two-segment model kind of resembles a robot leg powered by a hydrauliccylinder, where expansion of the cylinder provides the repulsive force and sepa-ration of the upper and lower segments is achieved by opening at the hinge: see One can avoid calculus by noting that Eq. (7) can be written as h = d ( α +1) − d (cid:16) N + kN (cid:17) ,where k = α +12 . (cid:16) N + kN (cid:17) is minimized by noting (cid:16) N + kN (cid:17) = (cid:16) N − kN (cid:17) + 4 k which hasa minimum when N = kN , or N = √ k = q α +12 . RN Figure 5: Two segments that repel via the muscle-joint-kneecap system, thesum of whose forces has an upward component.Fig. 4. It should be noted that the cylinder, being connected to both the lowerand upper segments, must deform as the two segments separate, which it doesby extending.Muscle and joints are more complicated. In Fig. 5, the contraction of the mus-cle, which we model with a tension T in the string, causes an attractive ratherthan a repulsive force. However, the hinge itself provides a reaction force R thatultimately causes a net upward force on the upper segment. The force N comesfrom the normal force of the string wrapped around the kneecap. Analysis ofsuch a device is most natural in terms of torque, where it becomes obvious thatthe tension T causes the center of mass of the top segment to move upwards asit rotates clockwise, despite the fact that it is pulling downward. Accurate biomechanical models of humans can go as far as having 16 rigidbodies joined together comprising 38 degrees of freedom [8], which resists anyattempts at an analytic solution. Analytic solutions for the jumping motion ofsprings [9, 10] have been made, but as the forces are not constant, they requirethe solving of differential equations. In theory at least, students who know theprinciples of physics but lack knowledge of calculus can model a varying forceby partitioning the force over many small intervals and assuming the force isconstant within each interval. Interesting investigations can be made using suchpartitions [11], and are more in the spirit of physics than integrating a function.Although historically calculus and physics are connected, the very existence ofalgebra-based physics courses indicates the belief that the essential features of7hysics do not require calculus. Why limit constant force equations to constantforces?
A N-segments Using Work Kinetic-Energy Equa-tion
When n segments are in the air, the ground must support the weight of N − n segments, along with the force F pressing on these segments: F ( n )ground = ( N − n ) mg + F. (8)Therefore the net external force on the entire system when n segments are inthe air is: F ( n )net = F ( n )ground − N mg = F − nmg. (9)When n segments move upwards a distance d , the center of mass of the systemmoves upward by: ∆ X ( n )COM = ndN . (10)Therefore using the work kinetic-energy equation: N − X n =1 F ( n )net ∆ X ( n )COM = ∆K (11) N − X n =1 ( F − nmg ) ndN = K f Using the famous sums P N − n =1 n = N ( N − , P N − n =1 n = ( N − N (2 N − , alongwith setting K f = N mgh for free-fall, one gets: (cid:18) N ( N − F − ( N − N (2 N − mg (cid:19) dN = N mgh (12) N − N (cid:18) m (cid:19) F − (2 N − mgg ! d = h, which agrees with Eq. (6). B Constant Force Impulse Approximation
To derive Eq. (2) without collision equations, we assume that after the repulsiveforce F acts for a distance d , the constant attractive force F ′ acts for a distance8 , after which both segments lock together and move at the same velocity v ′ f .The top segment deaccelerates to v ′ f = r v f − F ′ m ǫ (13)in time t = v f − v ′ fF ′ m , while the bottom segment accelerates to v ′ f = F ′ m v f − v ′ fF ′ m ! (14) v ′ f = m v f m + m , which proves Eq. (2). Moreover, we see that for Eqs (13) and (14) to be equal,we must have F ′ = v f ǫ ! m m (2 m + m )( m + m ) . (15)Therefore we have the freedom to make ǫ very small with the corresponding F ′ in Eq. (15) very large, such that their product in Eq. (13) is unchanged. Wecan then neglect ǫ whenever compared to d , so that we may still say that twosegments lock and move together when their separation distance reaches d . References [1] K. Laws and M. Swope.
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