## Center of Mass Problem Common Problems

- Category: High School
- Written by fisikastudycenter

Fisikastudycenter.com- High School Physics Problems and Solutions Examples, Center of Mass problems. Length and area.

**Problem 1**

Find the center of mass of this shape!

**Solution**

The data:

l_{1} = 20, X_{1} = 20, Y_{1} = 10

l_{2} = 20, X_{2} = 60, Y_{2} = 10

l_{3} = 80, X_{3} = 40, Y_{3} = 20

l_{4} = 20, X_{4} = 0, Y_{4} = 30

l_{5} = 40, X_{5} = 40, Y_{5} = 40

l_{6} = 20, X_{6} = 80, Y_{6} = 30

The center of the mass (X_{0} , Y_{0})

**Problem 2**

Find the location of Y_{o} of the center of the mass of this shape below!

**Solution**

Get the data first:

Shape 1 (the black one)

A_{1} = (20 x 60) = 1200

Y_{1} = 30

Shape 2 (the blue one)

A_{2} = (20 x 60) = 1200

Y_{2} = (60 + 10) = 70

**Problem 3**

Find the ordinate location of the center of the mass from fig below!

**Solution**

First we have to divide this shape into 2 parts. A square on the bottom, and a triangle over it. Here are the data that we can get:

Data :

Shape 1 (a square)

A_{1} = (90 x 90) = 8100

Y_{1} = 90/2 = 45

Shape 2 (triangle)

A_{2} = 1/2(90 x 90) = 4050

Y_{2} = 1/3(90) + 90 = 120

The location of Y_{o} :

**Problem 4**

Find the location of Y_{o} of the center of the mass of below shape

**Solution**

Divide the shape into to parts. First, full shape (a rectangle without removed triangle area), second a hole forms triangle.

The data needed :

Shape 1

A_{1} = (180 x 90) = 16200

Y_{1} = (180/2) = 90

Shape 2

A_{2} = 1/2(90 x 90) = 4050

Y_{2} = 180 − (90/3) = 150

Location of Y_{o} :

**Problem 5**

A cylinder and a cone are arranged shown below fig.

Find the center of mass location with AB as the reference line!

**Solution**

Data :

Shape 1 (cylinder)

V_{1} = π r^{2} t = 12π r^{2}

X_{1} = 6

Shape 2 (cone)

V_{2} = 1/3 π r^{2} t = 4 π r^{2}

X_{2} = 12 + (1/4 t) = 12 + 3 = 15

X_{o} location: