Question

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°

Answer 1

The right answer is c (48º)

Step-by-step explanation:

If you model the leg as a uniform thin rod of length L=0.81m (
`I_(cm) =(1)/(12) ML^(2)`), 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:

(
`I_(icr) =(1)/(3) ML^(2)`)

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

`a_(f)=g =\gamma L\rightarrow \gamma=(g)/(L)`

Considering torque from the instantaneous center of rotation:

`T_(ICR) =W(L)/(2) Cos(\theta)=\gamma I_(ICR)`

`Mg(L)/(2) Cos(\theta)=(g)/(L) (1)/(3) ML^(2) \Rightarrow Cos(\theta)=(2)/(3)`

Therefore:

`\theta= Cos^(-1)((2)/(3))\simeq 48º`