Fred Flintstone’s club–ax is made up of two parts: an uniform 2.5 kg stick and a symmetrical 10 kg stone attached to the end of the stick. The dimensions of the club-ax are shown in the figure. How far is the center of mass from the handle end of the club-ax? Express your answer in cm.
Handle: 80cm
Stone: 18cm
Hint: Assume stone is drilled and the stick passes through it
the c-o-m is between the c-o-m of the stick and the c-o-m of the stone
the centers of mass for the stick and the stone are in their physical centers
using some distance (d) from the center of the stick toward the center of the stone
... 2.5 * d = 10 * (40 - 18/2 - d)
find d ... add to 40 cm
To find the distance of the center of mass from the handle end of the club-ax, we need to consider the masses and positions of the two parts (the stick and the stone).
Let's begin by finding the center of mass of the stick. Since the stick is uniform, its center of mass will be located at its midpoint. Given that the stick has a length of 80 cm, the center of mass of the stick will be:
Center of mass of stick = Length of stick / 2 = 80 cm / 2 = 40 cm
Next, let's find the center of mass of the stone. Since the stone is symmetrical, its center of mass will be located at its geometric center. Given that the stone has a length of 18 cm, the center of mass of the stone will be:
Center of mass of stone = Length of stone / 2 = 18 cm / 2 = 9 cm
Now, we need to consider the masses of the stick and the stone. The stick has a mass of 2.5 kg, and the stone has a mass of 10 kg.
To determine the distance of the center of mass from the handle end of the club-ax, we can use the formula:
Center of mass = (center of mass of stick * mass of stick + center of mass of stone * mass of stone) / total mass
Total mass = mass of stick + mass of stone = 2.5 kg + 10 kg = 12.5 kg
Plugging in the values we have:
Center of mass = (40 cm * 2.5 kg + 9 cm * 10 kg) / 12.5 kg
Calculating this expression gives us the position of the center of mass:
Center of mass = (100 cm kg + 90 cm kg) / 12.5 kg = 190 cm kg / 12.5 kg
Simplifying further:
Center of mass = 15.2 cm
Therefore, the center of mass of the club-ax is 15.2 cm from the handle end of the club-ax.