Fred Flintstone’s club–ax is made up of two parts: an uniform 3.0 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.
Wood handle:80cm
Stone club:18cm
To find the center of mass of the club-ax, we need to consider the masses and the distances of the two parts (the stick and the stone).
The center of mass of an object is the point where the mass is evenly distributed around that point. For a two-part object like this club-ax, we can find the center of mass by using the principle of moments.
The formula for the center of mass is given by:
Center of mass = (m1 * d1 + m2 * d2) / (m1 + m2)
Where:
m1 and m2 are the masses of the two parts, and
d1 and d2 are the distances of the center of mass of each part from a reference point (in this case, the handle end of the club-ax).
Given:
m1 = 3.0 kg (mass of the stick)
m2 = 10 kg (mass of the stone)
d1 = 80 cm (distance of the center of mass of the stick from the handle end)
d2 = 18 cm (distance of the center of mass of the stone from the handle end)
Plugging these values into the formula, we have:
Center of mass = (3.0 kg * 80 cm + 10 kg * 18 cm) / (3.0 kg + 10 kg)
Calculating using these values, we get:
Center of mass = (240 cm * kg + 180 cm * kg) / 13 kg
Center of mass = (420 cm * kg) / 13 kg
Center of mass ≈ 32.31 cm
Therefore, the center of mass of the club-ax is approximately 32.31 cm from the handle end.