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. The dimensions of the
club-ax are that the stcik is 80 cm across and the stone is tied to the stick is 18 cm long. How far is
the center of mass from the handle end of the club-ax?
and what is your thinking?
To determine the distance of the center of mass from the handle end of the club-ax, we can use the principle of the center of mass. The center of mass of an object is calculated by finding the weighted average of the positions of all its individual parts.
In this case, we have a stone attached to a stick. To simplify the problem, let's consider the stone and the stick as two separate objects.
First, let's find the center of mass of the stone. The center of mass of a symmetrical object is located at the geometric center. Since the stone is symmetrical, its center of mass will be at its exact midpoint. Therefore, the center of mass of the stone is located 9 cm from both ends (18 cm long stone / 2).
Next, let's find the center of mass of the stick. The center of mass of a uniform object is located at its geometric center, which is also the midpoint. Since the stick is 98 cm long, its center of mass is located 49 cm from both ends (98 cm long stick / 2).
Now, we need to consider the weighted average of the two centers of mass based on their masses. The stone has a mass of 8 kg, and the stick has a mass of 2.5 kg.
To calculate the distance of the center of mass from the handle end, we can use the following formula:
center of mass = (m1 * x1 + m2 * x2) / (m1 + m2)
where m1 and x1 are the mass and position of the stone, and m2 and x2 are the mass and position of the stick.
Plugging in the values:
center of mass = (8 kg * 9 cm + 2.5 kg * 49 cm) / (8 kg + 2.5 kg)
Simplifying the equation:
center of mass = (72 kg.cm + 122.5 kg.cm) / 10.5 kg
center of mass = 194.5 kg.cm / 10.5 kg
center of mass ≈ 18.52 cm
Therefore, the center of mass of the club-ax is located approximately 18.52 cm from the handle end.