Four objects are placed on a light (zero mass) six meter rod. The object on the

Question

Four objects are placed on a light (zero mass) six meter rod. The object on the left end has a mass of 3.65 kg, and the object on the right end has a mass of 3.85 kg. A third ball has a mass of 2.55 kg, a distance of 2.15 m from the left end. Finally, a fourth ball has a mass of 2.7 kg, a distance of 1.5 m from the right end.

(a) How far is the center of mass of the system from the center of the rod?

(b) How far from its current position would you have to move the fourth object so that the center of mass coincides with the physical center of the rod? (negative is left, positive is right)

Explanation / Answer

Let the center of the rod is taken as origin.

Length of the rod L = 6 m

m1 = 3.65 kg

r1 = -L/2 = 3 m

m2 =3.85 kg

r2 = L/2 = 3 m

m3 = 2.55 kg

r3 = -3+2.15 = -0.85 m

m4 = 2.7 kg

r4 = 3 -1.5 = 1,5 m

the center of mass of the system from the center of the rod ,

r cm = (m1r1+m2r2+m3r3+m4r4)/(m1+m2+m3+m4)

= [(3.65*-3) +(3.85x3) +(2.55*-0.85)+(2.7x1.5)]/(3.65+3.85+2.55+2.7)

= 0.1947 m

(B).

r cm = (m1r1+m2r2+m3r3+m4r4)/(m1+m2+m3+m4)

0 = [(3.65*-3) +(3.85x3) +(2.55*-0.85)+(2.7xr4)]/(3.65+3.85+2.55+2.7)

= [-1.5675 +2.7r4] /(3.65+3.85+2.55+2.7)

[-1.5675 +2.7r4] = 0

1.5675 =2.7r4

r4 = 0.58055 m

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