1) An American Bullfrog (Rana catesbeiana) can jump as far as 2 m horizontally f

Question

1) An American Bullfrog (Rana catesbeiana) can jump as far as 2 m horizontally from a sitting position. Let us analyze this champion performance with what we have learned about the standing broad jump. (a) Assuming that the takeoff angle has its optimal value of 45o , what is the frog's takeoff speed, v0? (b) Assuming that the accelerating force Fr is constant and acts over a distance s = 20 cm, corresponding to the length of the frog's fully stretched hind legs, what is its magnitude? The frog's mass is m = 600 g. (c) What is the magnitude of the full force, F, exerted by the hind legs onto the ground, given in terms of the frog's weight, mg? What is the magnitude of F in N? (d) How large is the angle that F makes with the ground?

2) A runner runs with speed v = 2.1 m/s around a curve of radius R = 4.0 m on a horizontal surface. In order to remain stable through the curve, must she lean toward or away from the center of the curve? Draw a free-body diagram for the runner and use it to calculate the angle she has to lean away from the vertical.

3) We want to compare the oscillation periods for a naked human leg and the same leg with a heavy boot on the foot. To do this, we model the leg as a uniform, rigid rod of length l = 0.90 m and mass m1 = 11.0 kg, and the boot as a point mass m2 = 2.0 kg, attached to the lower end of the rod. The hip joint is at the upper end of the rod. (a) What are the moments of inertia about the hip joint of (i) the naked leg, and (ii) the leg with the boot on? (b) What are the leg's natural periods of oscillation about the hip joint in the two cases?

Explanation / Answer

a)

consider the motion along the X-direction

Vox = initial velocity = Vo Cos45

X = horizontal displacement = 2 m

using the equation

X = Vox t

2 = Vo Cos45 t

t = 2/(Vo Cos45) eq-1

consider the motion along the Y-direction

Voy = initial velocity = Vo Sin45

a = acceleration = - 9.8

Y = vertical displacement = 0

t = time of travel

using the equation

Y = Voy t + (0.5) a t2

0 = Vo Sin45 t + (0.5) (- 9.8) t2

using eq-1

0 = Vo Sin45 (2/(Vo Cos45) ) + (0.5) (- 9.8) (2/(Vo Cos45) )2

Vo = 4.43 m/s

b)

Vo = initial velocity = 0 m/s

Vf = final velocity = 4.43 m/s

s = stretch = 20 cm = 0.20 m

a = acceleration during stretch = ?

using the equation

Vf2 = Vo2 + 2 a s

4.432 = 02 + 2 a (0.20)

a = 49.1 m/s2

F = ma = 0.6 x 49.1 = 29.5 N

c)

Fn = force applied

using the force equation

Fn - mg = ma

Fn = m (g + a) = (0.6) (9.8 + 49.1) = 35.34 N

d)

theta = 45

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