Center of Mass for 8 kg Stone & 2.5 kg Stick

[tascja]

Homework Statement

alley club-ax consists of a symmetrical 8 kg stone attached to the end of a uniform 2.5 kg stick that is 98 cm long. How far is the center of mass from the angle end of the club-ax?

Homework Equations

Xcm = (sum) x*m / m

The Attempt at a Solution

I don't know how to approach this.. do i put the ax on a coordinate plane?
and it says that it is it is symmetrical and of uniform weight, does that give extra information that i should be using?

 

[LowlyPion]

Not sure which is the angle end, but say it is the handle. You can reverse it if not.

The handle is .98m and uniform density. You could take the integral of that to find the center of mass of just the handle, but I'm sure you know already it's in the middle for the handle part. To complete the sum of the moments for the whole shillelagh, then add the moment of the stone at the end.

So with m_s the mass of the stone, m_h the mass of the handle, and x_h the center of mass of the handle,

X_cm = (x_h*m_h + L*m_s)/(m_h + m_s)

 

[nlightner]

As a scientist, we can approach this problem by using the formula for calculating the center of mass: Xcm = (sum) x*m / m. In this case, the "x" represents the distance from the reference point (in this case, the angle end of the club-ax) to the center of mass, and "m" represents the mass at that distance.

Since the club-ax is symmetrical, we can assume that the center of mass is located at the midpoint of the stick, which is 49 cm from the angle end. Therefore, we can plug in the values for the mass and distance into the formula:

Xcm = (8 kg * 49 cm + 2.5 kg * 49 cm) / (8 kg + 2.5 kg)

Simplifying, we get Xcm = (392 cm + 122.5 cm) / 10.5 kg = 50.24 cm.

Therefore, the center of mass for this club-ax is located 50.24 cm from the angle end of the club-ax. This information can be useful for understanding the balance and stability of the club-ax and its potential for use in activities such as chopping wood or playing a game of golf.