B The Simple Pendulum F = a -mg sing=m-2 s=LO mg sin 8 sindo 0\mg cos 6 Small angle approximation sine × d26 6=0 6=ocos(a+中)
Small angle approximation sin 0 2 2 + = L g dt d B. The Simple Pendulum 2 2 2 2 sin sin dt d g L s L dt d s mg m Ft mat − = = − = = cos( ) = 0 t + l g = 2
C. physical pendulum P409 ∑ t=la d26 mgd sin 8=l dt d6 mgd 6 dt o mod mg sin Ga中测mE
I mgd I mgd dt d dt d mgd I I = = − − = = 2 2 2 2 2 sin C. physical pendulum P409
Example: Tyrannosaurus rex and physical pendulum All walking animals, including humans, have a natural walking pace, a number of steps per minute that is more comfortable than a faster or slower pace. Suppose this natural pace is equal to the period of the leg, viewed as a uniform rod pivoted at the hip joint. A) How does the natural walking pace depend on the length L of the leg, measured from hip to foot? B )Fossil evidence shows that Tyrannosaurus rex, a two-legged dinosaur that lived about 65 million years ago at the end of the cretaceous period, had a leg length L= 3.1 m and a stride length(the distance from one foot-print to the next print of the same foot S=4.0 m Estimate the walking speed of Tyrannosaurus rex. (page 410 EX13-10) Solution:
Example: Tyrannosaurus rex and physical pendulum All walking animals, including humans, have a natural walking pace, a number of steps per minute that is more comfortable than a faster or slower pace. Suppose this natural pace is equal to the period of the leg, viewed as a uniform rod pivoted at the hip joint. A) How does the natural walking pace depend on the length L of the leg, measured from hip to foot? B) Fossil evidence shows that Tyrannosaurus rex, a two-legged dinosaur that lived about 65 million years ago at the end of the Cretaceous period, had a leg length L = 3.1 m and a stride length (the distance from one foot-print to the next print of the same foot ) S = 4.0 m. Estimate the walking speed of Tyrannosaurus rex. (page 410 EX13-10) Solution:
4 Leg Stride length s length Copyright 9 2004 Poarson Education, inc, publishing as Addison Wosley
Solution: 2 T=2丌 L=2038m/s2 2(3.lm) 2.9s 13g S4.0m_14m T2.9
Solution: s m s m g L T 2.9 3(9.8 / ) 2(3.1 ) 2 3 2 2 2 = = = m s s m T S v 1.4 / 2.9 4.0 = = =