Asked by MUHAMMAD JUNAID
On a weekend archeological dig, you discover an old club-ax that consists of a symmetrical 8.3 kg stone attached to the end of a uniform 1.8 kg stick. You measure the dimensions of the club-ax as shown in the figure: the stick is 98 cm long, and the 18 cm long stone is drilled through its center and mounted on the stick. How far is the center of mass of the club-ax from the handle end of the club-ax?
Answers
Answered by
Wasif
CM of the stone is 9 cm from the edge, for the stick, 40 cm from the end, both by symmetry.
putting the origin at the center of the stone
r = 49/(1 + 8.3/1.8) = 8.73 cm
that is from the center of the stone, which is 89 cm from the left end
d = 89-8.73 = 80.27 cm
as a check, make the stick the larger
r = 49/(1 + 1.8/8.3) = 40.27 cm
that + 40
= 80.27 cm
putting the origin at the center of the stone
r = 49/(1 + 8.3/1.8) = 8.73 cm
that is from the center of the stone, which is 89 cm from the left end
d = 89-8.73 = 80.27 cm
as a check, make the stick the larger
r = 49/(1 + 1.8/8.3) = 40.27 cm
that + 40
= 80.27 cm
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