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 0.27-m-long cylinder with an outer radius of 1.12 x 10-2 m and a hollow inner core with a radius of 3.9 x 10-3 m. Excluding the arm, the mass of the gymnast is 60 kg. Bone has a compressional Young's modulus of 9.4 x 109 N/m2. (a) What is the compressional strain of the humerus? (b) By how much is the humerus compressed?

Answer 1

Answer:

                  $= 5.241 \times 10^{-5} m$

Explanation:

Given:

 Length of cylinder is, L = 0.27 m

Outer radius of cylinder is, r_out = 1.12×10^{-2} m

Inner radius of cylinder is, r_in = 3.9×10^{-3} m

Mass of person, m = 60 kg

 Young's modulus , Y = 9.4×10^9 N/m2

(a)

     Compressional strain of humerous is,

$Strain = \frac{Stress}{Young's\ modulus}$

     $\frac{\Delta L}{L_0} = \frac{\frac{F}{A}}{Y}$

                  $= \frac{(mg)}{\pi(r_out^2 - r_in^2 )Y}$

                  $= \frac{(60 kg)(9.8 m/s2 )}{(\pi)[( 1.12\times 10^{-2})^2 - (3.9\times 10^{-3} m)^2] (9.4\times 10^9 N/m2 )}[tex] [tex]= 1.80\times 10^{-4} m$

(b)  Let assume that humerous is compressed by ΔL

       Since,   strain = ΔL/L0

      $(1.80 \times 10^{-4} m) = ΔL / 0.29 m$

     Thus,

           $ΔL = (4.56 \times 10^{-4} m)(0.29 m)$

                  $= 5.241 \times 10^{-5} m$