Question

The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur.

Answer 1

Answer:

a) the velocity of the implant immediately after impact is 20 m/s

b) the average resistance of the implant is 40000 N

Explanation:

a) The impulse momentum is:

mv1 + ∑Imp(1---->2) = mv2

According the exercise:

v1=0

∑Imp(1---->2) = F(t2-t1)

m=0.2 kg

Replacing:

$0+F(t_{2} -t_{1} )=0.2v_{2}$

if F=2 kN and t2-t1=2x10^-3 s. Replacing

$0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s$

b) Work and energy in the system is:

T2 - U(2----->3) = T3

where T2 and T3 are the kinetic energy and U(2----->3) is the work.

$T_{2} =\frac{1}{2} mv_{2}^{2} \\T_{3} =0\\U_{2---3} =-F_{res} x$

Replacing:

$\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N$