Suppose that you are laying on your back with one leg pointing up. you let your leg fall, rotating freely about your hip joint until it hits the floor. model this leg (including the foot) as a uniform thin rod of length 0.81m. define the angle the leg makes with the floor to be θ.at some moment during the leg movement, the tangential acceleration of the foot is g (the acceleration of gravity). what is the angle of the leg, θ at this instant? a. at any angle as the tangential acceleration of the foot is always g. b .71°c. c. 48°d. 80°e. 85°

The right answer is c (48º)

Explanation:

If you model the leg as a uniform thin rod of length L=0.81m (), you are considering:

The foot is in the free end and pivot end is in your hip, therefore the instantaneous center of rotation is in the hip

Using Steiner in the ICR results:

()

The relation between tangential acceleration and angular acceleration in this case is:

Considering torque from the instantaneous center of rotation:

Therefore:

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

not moving. pos accel,