Finding the Center of Mass of a Meter Stick with Multiple Masses

[sallychan]

Homework Statement

A meter stick is 200 g. A mass of 1 kg is placed in 20 cm, and another mass of 5 kg is placed in 100 cm.

So the diagram will be like:

Homework Equations

Why do we have to include M2? And why is the mass of M2 is 200g? Isn't the whole meter stick weight 200g, and ithe mass of M2 should be lighter because it is just a point on the meter stick.

The Attempt at a Solution

Center of Mass = [(1)(0.2) + (0.2)(0.5) + (5)(1)] / (1+0.2+5) = 0.8548

 

[Chestermiller]

Homework Statement

A meter stick is 200 g. A mass of 1 kg is placed in 20 cm, and another mass of 5 kg is placed in 100 cm.

So the diagram will be like:
View attachment 80612

Homework Equations

Why do we have to include M2? And why is the mass of M2 is 200g? Isn't the whole meter stick weight 200g, and ithe mass of M2 should be lighter because it is just a point on the meter stick.

The distributed mass of the meter stick can be taken as being concentrated at the center of mass of the stick itself.

The Attempt at a Solution

Center of Mass = (1)(0.2) + (0.2)(0.5) + (5)(1) = 0.8548

This is not the equation for the center of mass. There is supposed to be a denominator on the right hand side. Please write the correct equation for the location of the center of mass.

Chet