How Do You Calculate the Center of Mass for a Club-Ax?

[kathyt.25]

Homework Statement

"Alley club-ax consists of a symmetrical 8 kg
stone that is 18cm long, and is attached to the end of a uniform 2.5 kg
stick that is 98 cm long."

Homework Equations

cm = centre of mass

x(cm) = x1m1 + m2x2 / (m1+m2)

The Attempt at a Solution

m1=8
m2=2.5

x1=9
x2=40

This seems like a very straightforward question, but I can't seem to get it right... the answer is 77.3cm from the handle's end, and I keep on getting 18.52cm which is WAY off!

 

[method_man]

Formula is good, but the data you have put in it isn't. You say that the solution is 77.3 cm from the handle's end. Handle length is 80 cm and length of a stone is 18.
First, you have to define the coordinate system. In this case, we'll say that coordinate system starts from the handle's end. That means that the handle's CM is located at 40 cm (x_h=40cm). CM of the stone is at 89 cm because coordinate system starts from the handle's end, so you have to add handle's length which is 80 cm plus 9 cm from the stone (x_s=89cm).

x_m=(m_hx_h+m_sx_s)/(m_h+m_s)=(2.5*40+8*89)/(2.5+8)=77,3333cm

 

[nlightner]

Finding the center of mass in this situation requires using the equation x(cm) = (x1m1 + x2m2) / (m1+m2). In this case, we have two masses, the 8 kg stone and the 2.5 kg stick, with their respective distances from the handle being 18 cm and 98 cm. Plugging in the values, we get x(cm) = (18 cm * 8 kg + 98 cm * 2.5 kg) / (8 kg + 2.5 kg) = 77.3 cm. This is the correct answer and is located 77.3 cm from the handle's end. It is important to carefully check your calculations and units to ensure accuracy in solving for the center of mass.