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Learning object 1

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Mark Ng Shun

A model organizer wants to know how long it will take a supermodel to walk down a 30m runway and back while posing for 5 seconds at the end. Using physics concepts of the simple pendulum, the summary calculates that with the model's height of 180cm, foot mass of 1.5kg, leg length of 80cm, and extending their legs 25 degrees, their velocity would be 4.40 m/s. With this velocity, it would take the model 6.82 seconds to walk to the end of the 30m runway and back, for a total time of 18.64 seconds including the 5 seconds posing at the end.

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Learning Object 1: Simple Pendulum Fabiois a fashionshow/runway model organizer andhe would like to know the approximateamountof time it wouldtake for oneof his supermodelstoreach the end of the walkway and back, Fabioalso happenstobe a physicsmajor andwould like to apply the concept of simple pendulumstodetermine this time. Fabioknowsthat the approximateheight of each model is 180cmandthat the length of their legs are roughly 4/9 of their height, Fabioalso assumesthattheir legs themselvesare massless(skinny, of course) and only takesthe weight of their feet intoconsideration, which on average is 1.5 kg where the center of massfrom the center of the foot to where it meets the joint is 5cm. Fabio alsoknowsthat the average angle which a modelslegs will extend from resting positionis 25 degrees bothways. Knowingthe runaway is 30mlong, how long will it takefor a supermodelto go there andback, giventhe model stands to posefor 5 seconds at the endof the runway? Figure 1: A model strutsher stuff downthe runway
Solution: Given: Massof foot:1.5kgheight:1.80m 慮:25o distance:30m g:9.81m/s2 Applicable Formulae: Ip=mLcm 2 +2/5Mr2 W=sqrt(MgLcm/Ip) L=L+r v= wA Solvingfor leg length: 1.80m*4/9=.80m Solvingfor w: W= sqrt(1.5*9.81*(0.8+0.05)/(1.5*0.82 +(2/5)1.5*0.052 )=13.009 Solvingfor Velocity: v=wA = 13.009*Lsin(慮)=13.009*0.8sin(25o ) = 4.40 m/s Solvingfor time: D=v*t 30/4.40=t t= 6.82 Ttotal= 2t+5s= 18.64s

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Learning object 1

  • 1. Learning Object 1: Simple Pendulum Fabiois a fashionshow/runway model organizer andhe would like to know the approximateamountof time it wouldtake for oneof his supermodelstoreach the end of the walkway and back, Fabioalso happenstobe a physicsmajor andwould like to apply the concept of simple pendulumstodetermine this time. Fabioknowsthat the approximateheight of each model is 180cmandthat the length of their legs are roughly 4/9 of their height, Fabioalso assumesthattheir legs themselvesare massless(skinny, of course) and only takesthe weight of their feet intoconsideration, which on average is 1.5 kg where the center of massfrom the center of the foot to where it meets the joint is 5cm. Fabio alsoknowsthat the average angle which a modelslegs will extend from resting positionis 25 degrees bothways. Knowingthe runaway is 30mlong, how long will it takefor a supermodelto go there andback, giventhe model stands to posefor 5 seconds at the endof the runway? Figure 1: A model strutsher stuff downthe runway
  • 2. Solution: Given: Massof foot:1.5kgheight:1.80m 慮:25o distance:30m g:9.81m/s2 Applicable Formulae: Ip=mLcm 2 +2/5Mr2 W=sqrt(MgLcm/Ip) L=L+r v= wA Solvingfor leg length: 1.80m*4/9=.80m Solvingfor w: W= sqrt(1.5*9.81*(0.8+0.05)/(1.5*0.82 +(2/5)1.5*0.052 )=13.009 Solvingfor Velocity: v=wA = 13.009*Lsin(慮)=13.009*0.8sin(25o ) = 4.40 m/s Solvingfor time: D=v*t 30/4.40=t t= 6.82 Ttotal= 2t+5s= 18.64s