Answer to Question #83472 in Mechanics | Relativity for Aya

Question

An athelete weighs 90 kg and has a height of 160 cm as shown , determine the change in elevation of his center of mass when he raisee his arm from the sides of his body 180° over his head

Answer 1

Usually length of hands is

arms≈3·head,

head≈1/8 height ⇨ arms≈3/8 height.

Shoulders are

d≈1/5 height

below top of our head.

Mass of hands is approximately 0.09 mass of body.

m≈0.09M.

Center of mass of body without arms is

y_(body without arms)≈0.45height

Calculate center of mass for a body with arms put down above the ground:

y_down≈(0.45height(M-m)+(height-d-arms/2)m)/M=0.46height.

Center of mass of arms up is above the ground for:

y_(arms up)=height+arms/2-d.

Center of mass of body with arms up:

y_up=(y_(arms up) m+y_(body without arms) (M-m))/M=

=((height+arms/2-d)m+0.46height(M-m))/M=

=((height+3/16 height-1/5 height)·0.09M+0.46height·0.91M)/M=0.50height.

Δy=y_up-y_down=height(0.50-0.46)=0.04height=6.4 cm.

Of course approximately.

Usually length of hands is

arms≈3·head,

head≈1/8 height ⇨ arms≈3/8 height.

Shoulders are

d≈1/5 height

below top of our head.

Mass of hands is approximately 0.09 mass of body.

m≈0.09M.

Center of mass of body without arms is

y_(body without arms)≈0.45height

Calculate center of mass for a body with arms put down above the ground:

y_down≈(0.45height(M-m)+(height-d-arms/2)m)/M=0.46height.

Center of mass of arms up is above the ground for:

y_(arms up)=height+arms/2-d.

Center of mass of body with arms up:

y_up=(y_(arms up) m+y_(body without arms) (M-m))/M=

=((height+arms/2-d)m+0.46height(M-m))/M=

=((height+3/16 height-1/5 height)·0.09M+0.46height·0.91M)/M=0.50height.

Δy=y_up-y_down=height(0.50-0.46)=0.04height=6.4 cm.

Of course approximately.