Answer:
The right answer is c (48º)
Explanation:
If you model the leg as a uniform thin rod of length L=0.81m ([tex]I_{cm} =\frac{1}{12} ML^{2}[/tex]), 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:
([tex]I_{icr} =\frac{1}{3} ML^{2}[/tex])
The relation between tangential acceleration and angular acceleration in this case is:
[tex]a_{f}=g =\gamma L\rightarrow \gamma=\frac{g}{L}[/tex]
Considering torque from the instantaneous center of rotation:
[tex]T_{ICR} =W\frac{L}{2} Cos(\theta)=\gamma I_{ICR}[/tex]
[tex]Mg\frac{L}{2} Cos(\theta)=\frac{g}{L} \frac{1}{3} ML^{2} \Rightarrow Cos(\theta)=\frac{2}{3}[/tex]
Therefore:
[tex]\theta= Cos^{-1}(\frac{2}{3})\simeq 48º[/tex]
