Question

A gymnast does a one-arm handstand. The humerus, which is the upper arm bone between the elbow and the shoulder joint, may be approximated as a 0.26-m-long cylinder with an outer radius of 1.07 x 10-2 m and a hollow inner core with a radius of 3.7 x 10-3 m. Excluding the arm, the mass of the gymnast is 57 kg. Bone has a compressional Young's modulus of 9.4 x 109 N/m2. (a) What is the compressional strain of the humerus

Answer 1

Answer:

0.00018784

Explanation:

$L_0$ = Initial length of cylinder = 0.26 m

$\Delta L$ = Change in length

$r_o$ = Outer radius = $1.07\times 10^{-2}\ m$

$r_i$ = Inner radius = $3.7\times 10^{-3}\ m$

m = Mass of arm = 57 kg

g = Acceleration due to gravity = 9.81 m/s²

A = Area

Y = Young's modulus = $9.4\times 10^9\ N/m^2$

When we divide stress by young's modulus we get compressional strain

$\frac{\Delta L}{L_0}=\frac{F}{A}\times \frac{1}{Y}\\\Rightarrow \frac{\Delta L}{L_0}=\frac{mg}{\pi(r_o^2-r_i^2)}\times \frac{1}{Y}\\\Rightarrow \frac{\Delta L}{L_0}=\frac{57\times 9.81}{\pi((1.07\times 10^{-2})^2-(3.7\times 10^{-3})^2)}\times \frac{1}{9.4\times 10^9}\\\Rightarrow \frac{\Delta L}{L_0}=0.00018784$

Compressional strain of the humerus is 0.00018784