CGCM

Center of MassThe Center of Mass (CM) of an object or objects is the point that moves as if all of the mass were concentrated there and all the external forces were applied there.

For a symmetrical object of uniform density it is the geometric center of the object.

The Center of Gravity (CG) is at the point where the weight above equals the weight below. This is effectively the same place as the CM

If you account for the force of gravity weakening with altitude, the CG would be slightly below the CM.

Stronger Gravity

Weaker Gravity

The velocity of the center of mass of a system of objects is the total momentum divided by the total mass.

Finding CM of a System of 2 Objects

m1 m2

x2

xCM

m2g

m1g

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∑τ =0m1gxCM −m2g(x2 − xCM ) = 0

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m1xCM −m2x2 + m2xCM = 0

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m1xCM + m2xCM = m2x2xCM (m1 + m2) = m2x2

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xCM =m2x2m1 + m2

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m1 = 2 kg,m2 = 4 kg, x2 = 3m

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xCM =4 ⋅32 + 4

= 2 m

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xCM =m1x1 + m2x2 + m3x3m1 + m2 + m3

Finding CM of a System of n Objects

m1 m2

m2g

m1g

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∑τ =0

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m1xCM −m1x1 −m2x2 + m2xCM = 0

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xCM (m1 + m2) = m1x1 + m2x2

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xCM =m1x1 + m2x2m1 + m2

x2

xCM

x1

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m1g(xCM − x1) −m2g(x2 − xCM ) = 0

What if there were 3 masses?

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xCM =1

Mmii=1

n

∑ x i

What if there were n masses?

Where M is the total mass

Find xCM and yCM of 3 Particles

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xCM =1

Mmii=1

n

∑ x i

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yCM =1

Mmii=1

n

∑ y i

If gravity acted sideways we could derive the same result for the y axis:

Find xCM and ycm for the system of 3 particles shown below. M1 = 3kg, M2 = 5kg, M3 = 8kg

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xCM =1

16(3⋅0.5 + 5 ⋅2 + 8 ⋅4)

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xCM =43.5 kg ⋅m16 kg

= 2.72 m

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yCM =44 kg ⋅m16 kg

= 2.75 m

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yCM =1

16(3⋅5 + 5 ⋅1+ 8 ⋅3)

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M = 3+ 5 + 8 =16 kg

Center of Mass and Statics Problems

0.5 kg

m = ?

0.12 m

A meter stick is balanced on a fulcrum placed at the 85 cm mark by a 0.5 kg mass hanging at the 97 cm mark. Find the mass of the meter stick.

mg

N

Mg

0.35 m

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∑τ =0Mg(0.12) −mg(0.35) = 0

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0.5 ⋅0.12 = m ⋅0.35

m = 0.17 kg