# For continuum bodies r cm r dm dm r cm i r i m i m i

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For continuum bodies r CM = r dm dm r CM = i r i m i m i becomes, in the limit r CM = 1 M r dm or Usually we know the density: r CM = 1 M V r ρ ( r ) dV

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Phys 2A - Mechanics Example: determine the position of the CM of the example we saw previously: “the density of a long thin metal rod of length L is engineered to increase linearly: it is given as a function of distance x along the length of the rod starting from one end of the rod, by λ ( x ) = λ 0 (1 + cx ) where c is a constant with units of 1/meter. What is the mass of the rod?” ANS: M = λ 0 L (1 + 1 2 cL ) So we need Recall we had x dm = x λ dx = L 0 x λ ( x ) dx = L 0 x λ 0 (1 + cx ) dx = λ 0 ( 1 2 L 2 + 1 3 cL 3 ) and finally x CM = λ 0 L 2 ( 1 2 + 1 3 cL ) λ 0 L (1 + 1 2 cL ) x CM = 1 + 2 3 cL 1 + 1 2 cL L 2 or which is not in the middle (unless c=0) x=L x x+dx dm = λ ( x ) dx
Phys 2A - Mechanics CM of two (or more) bodies: If the CM of body 1 of mass m1 is at and the CM of body 2 of mass m2 is at the CM of the system is r 1 r 2 r CM = m 1 r 1 + m 2 r 2 m 1 + m 2 That is, the CM is as if the two bodies are replaced by point particles of their masses located at their CMs. r 1 r 2 r CM m 1 m 2

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Phys 2A - Mechanics Example: Our previous examples with point masses could also be: Where is the CM of two solid spheres of uniform density of masses 1.0kg and 2.0kg when their centers are 3.0 m apart? ANS: The CM of each sphere is at their respective centers (WHY?). The computation of the CM is then the same as before! Now (in meters) so 0 1m 3m x 2m x CM = m 1 x 1 + m 2 x 2 m 1 + m 2 = (1 . 0)(0) + (2 . 0)(3 . 0) 1 . 0 + 2 . 0 m = 2 . 0 m (and y CM =0) r 1 = (0 , 0) , r 2 = (3 . 0 , 0) 1kg 2kg Copying from previous slide:
Phys 2A - Mechanics Example: Our previous examples with point masses could also be: Where is the CM of two solid spheres of uniform density of masses 1.0kg and 2.0kg when their centers are 3.0 m apart? ANS: The CM of each sphere is at their respective centers (WHY?). The computation of the CM is then the same as before! Now (in meters) so 0 1m 3m x 2m x CM = m 1 x 1 + m 2 x 2 m 1 + m 2 = (1 . 0)(0) + (2 . 0)(3 . 0) 1 . 0 + 2 . 0 m = 2 . 0 m (and y CM =0) r 1 = (0 , 0) , r 2 = (3 . 0 , 0) 1kg 2kg 0 1m 3m x 2m CM 1kg 2kg Copying from previous slide:

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Phys 2A - Mechanics Example: Our previous examples with point masses could also be: Where is the CM of two solid spheres of uniform density of masses 1.0kg and 2.0kg when their centers are 3.0 m apart?
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