8.45 To keep the calculations fairly simple, but still reasonable, we shall mode

Question

8.45

To keep the calculations fairly simple, but still reasonable, we shall model a human leg that is 92.0 cm long (measured from the hip joint) by assuming that the upper leg and the lower leg (which includes the foot) have equal lengths and that each of them is uniform. For a 70.0 kg person, the mass of the upper leg would be 8.35 kg , while that of the lower leg (including the foot) is 5.50 kg . Take that x-axis is directed horizontally and the y-axis is directed vertically downward.

A. Find the x-coordinate of the center of mass of this leg, relative to the hip joint, if it is stretched out horizontally.

ANSWER: Xc= _________ cm

B. Find the y-coordinate of the center of mass of this leg, relative to the hip joint, if it is stretched out horizontally.

ANSWER: Yc= _________cm

C. Find the x-coordinate of the center of mass of this leg, relative to the hip joint, if it is bent at the knee to form a right angle with the upper leg remaining horizontal.

  ANSWER: Xc= _________ cm

D. Find the y-coordinate of the center of mass of this leg, relative to the hip joint, if it is bent at the knee to form a right angle with the upper leg remaining horizontal.

ANSWER: Yc= _________cm   

Explanation / Answer

a) set x=0 as the position of the hip joint

xcm = (x1 m1 + x2 m2)/(m1+m2)

where x1=location of mass 1; m1=mag of mass1

x2=location of mass 2; m2=mag of mass2

x1=23 cm; m1=8.35kg (x1=23 cm since the upper part is 46 cm, and a uniform object has its center of mass at the

midpoint

x2=69cm; m2=5.25kg

xcm=(23*8.35+69*5.25)/(8.35+5.25)

xcm=40.75 cm

b)

y1 = 0 and y2 = 0

ycm = 0

c) x1 is the same in this case, but now x2=46 cm since the lower part lies along the line x=46 cm

xcm=(23 * 8.35 + 46 * 5.25)/13.6= 31.88 cm

d) here, y1=0 since the upper part lies along the line y=0

y2=-23 cm

ycm=(y1 m1 +y2 m2)/(13.6)

ycm=(0+(-23)(5.25))/13.6 = -8.88cm

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