Question

Please derive these equations and break them down into steps . thank you very muvj

Pumping from a Standing Position The pumped swing is modeled as a pendulum with variable length L. The rider is modeled as a point mass m, and L is the distance from the rider's center of mass to the fixed swing support point O. Conservation of angular momentum for a point mass undergoing plane motion is where H is the angular momentum of the body about O, and N is the net torque about O due to all forces acting on the point mass [7]. In our case H is mL2φ and the torque about O due to the gravitational force is the product of the transverse component of this force,-mg sin φ, and the lever arm L. The torque about O due to the tension T in the ropes is 0 (Figure ). Thus, after dividing out the mass, we have the equation of motion As the rider stands or crouches, the effective length L of the pendulum varies with time; however, as the rider does not decide when to change his position by looking at his watch, the length of the pendulum is not well modeled as an explicit function of time L L(t). Instead, the rider modifies his stance in accordance with the position and velocity of the pendulum. Therefore, we model the pendulum length as an autonomous function of the state of the pendulum; that is, L-: L(d,d) Following Tea and Falk [11, we assume that the rider squats for the first half of each swing of the pendulum, then suddenly stands up as the swing passes through φ 0 and remains standing for the upward part of the motion. When the pendu- lum reaches its maximum height and comes to rest instantaneously, the rider again squats and repeats the forcing cycle, this time with the swing moving in the opposite direction (Figure 2) Suddenly standing up causes a decrease in L, and squatting causes an equal increase in L. We mathematically model the decrease in L when ф ~ 0 as beginning with the rider in a squatting position when t to and ending at a time Δt later with the rider standing upright, where lol st for to t to Δ. Integrating equation Figure 2. Pumping strategy for a standing rider (2) over to,to At] gives an tand auat- to Since isin ф1S €, the integral on the right-hand side is O(e), so as €-+0 we have To simplify the notation for the angular velocities just before and just after the de- crease in L caused by the rider's suddenly standing up, we write ф_ for aquat-d(to) and φ+ for atand d(to + Δt). Thus the rider's standing up as φ passes through 0 produces the following boost in the angular velocity -(논), What is the effect of the increase in L caused by returning to a squat when φ ~ 0 at the high points in the motion? Assume the rider is standing at some time fı, and a bit later at time t-+ Δt the rider is fully crouched, with Ιφ1S e for t1S t S t1 + Δ Then the values of ф before and after the rider returns to a crouch differ by no more than 2e. In the limit, therefore, ase → 0, returning to the squatting position produces no increase or decrease in the angular velocity, and hence in the kinetic energy, of the pendulum. To see whether the swing angle is affected, we integrate equation (2) from ti to some time tti At and obtain Dividing by L(t) and integrating again, this time from ti to ti + At, gives at- Lstand 268 THE COLLEGE MATHEMATICS JOURNAL This content dounloaded from 132.23627.111 on Tue, 07 Mar 2017 08:36:52 UTC All use subjoct to bpWabout jstoronglerms

Answer 1

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