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

0.00018784

Step-by-step explanation:

`L_0` = Initial length of cylinder = 0.26 m

`\Delta L` = Change in length

`r_o` = Outer radius =
`1.07* 10^(-2)\ m`

`r_i` = Inner radius =
`3.7* 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* 10^9\ N/m^2`

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

`(\Delta L)/(L_0)=(F)/(A)* (1)/(Y)\\\Rightarrow (\Delta L)/(L_0)=(mg)/(\pi(r_o^2-r_i^2))* (1)/(Y)\\\Rightarrow (\Delta L)/(L_0)=(57* 9.81)/(\pi((1.07* 10^(-2))^2-(3.7* 10^(-3))^2))* (1)/(9.4* 10^9)\\\Rightarrow (\Delta L)/(L_0)=0.00018784`

Compressional strain of the humerus is 0.00018784