Walking Speed of a Dinosaur: How Does Leg Length Affect Natural Walking Pace?

  • Thread starter Thread starter AlonsoMcLaren
  • Start date Start date
  • Tags Tags
    Speed
AI Thread Summary
The discussion focuses on how leg length affects the natural walking pace of dinosaurs, specifically Tyrannosaurus rex, with a leg length of 3.1 meters and a stride length of 4.0 meters. The calculations for the walking speed involve determining the period of motion using the correct formula for a uniform rod, which differs from that of a pendulum. The initial attempt at calculating the period used an incorrect angular frequency equation, leading to an inaccurate walking speed estimate of 1.96 m/s instead of the correct 1.4 m/s. The key takeaway is that the correct formula for the period must account for the leg's uniform rod properties to yield accurate results. Understanding these principles is crucial for estimating the walking speed of large dinosaurs accurately.
AlonsoMcLaren
Messages
89
Reaction score
2

Homework Statement



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.


Homework Equations



I=(m*L^2)/3
T=2*pi/w


The Attempt at a Solution



T=2*pi/w=2*pi/sqrt(Lmg/I)

I=(m*L^2)/3

T=pi*sqrt(4L/3g)=2.04s

v=s/T=4m/2.04s=1.96m/s

But the answer says that T=2pi*sqrt(2L/3g)=2.88s
v=s/T=4m/2.88s=1.4m/s
 
Physics news on Phys.org
Nobody?
 
you've said, you've seen the answer is T=2pi*sqrt(2L/3g), But you have used: T=2pi*sqrt(L/3g), so this is where the problem is. I guess the problem stems from the first line, where you seemed to use w=sqrt(Lmg/I), but this is not true (which is why you end up with an incorrect answer for the period of the motion).
 
BruceW said:
you've said, you've seen the answer is T=2pi*sqrt(2L/3g), But you have used: T=2pi*sqrt(L/3g), so this is where the problem is. I guess the problem stems from the first line, where you seemed to use w=sqrt(Lmg/I), but this is not true (which is why you end up with an incorrect answer for the period of the motion).

Why is w=sqrt(Lmg/I) incorrect?
 
That equation for the angular frequency is correct for a pendulum where there is a small bob on the end of a light, inextensible string. But in this question, the pendulum is a uniform rod. So that equation for the angular frequency is not correct.

Are you meant to derive the equation for the angular frequency of a pendulum made of a uniform rod? If not, you can guess what it is, since you've seen the answer. The rest of your working is correct, its just the angular frequency which was wrong.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top