On a weekend archeological dig, you discover an old club-ax that consists of a symmetrical 8.3 kg stone attached to the end of a uniform 1.8 kg stick. You measure the dimensions of the club-ax as shown in the figure: the stick is 98 cm long, and the 18 cm long stone is drilled through its center and mounted on the stick. How far is the center of mass of the club-ax from the handle end of the club-ax?
CM of the stone is 9 cm from the edge, for the stick, 40 cm from the end, both by symmetry.
putting the origin at the center of the stone
r = 49/(1 + 8.3/1.8) = 8.73 cm
that is from the center of the stone, which is 89 cm from the left end
d = 89-8.73 = 80.27 cm
as a check, make the stick the larger
r = 49/(1 + 1.8/8.3) = 40.27 cm
that + 40
= 80.27 cm
To find the center of mass of the club-ax, you need to consider the mass and the distribution of mass along its length. The center of mass is the point at which the entire mass of the object can be considered to be concentrated.
First, let's calculate the center of mass of the stone and the stick separately, and then combine them to find the overall center of mass.
1. Center of mass of the stone:
The stone is symmetrical, so its center of mass will be at its geometric center. Since the stone is 18 cm long, the center of mass will be located at a distance of 9 cm from each end.
2. Center of mass of the stick:
The stick is a uniform rod, so the center of mass will be at its midpoint. The stick is 98 cm long, so the center of mass will be located at a distance of 49 cm from each end.
3. Combining the center of mass of the stone and stick:
To find the overall center of mass of the club-ax, we need to take into account the masses of the stone and the stick. The distance from the center of mass of each component to the handle end of the club-ax should be weighted by their respective masses.
Let's denote the distance from the handle end of the club-ax to the center of mass of the stone as x1 and the distance to the center of mass of the stick as x2.
Using the principle of moments, we can set up the following equation:
(m1 * x1) + (m2 * x2) = 0
Where:
m1 = mass of the stone = 8.3 kg
m2 = mass of the stick = 1.8 kg
Substituting the values, we get:
(8.3 kg * x1) + (1.8 kg * x2) = 0
Since we know that the center of mass of the stone is at a distance of 9 cm from each end, we can substitute x1 = 9 cm.
(8.3 kg * 9 cm) + (1.8 kg * x2) = 0
Simplifying the equation further, we can solve for x2:
74.7 kg * cm + (1.8 kg * x2) = 0
1.8 kg * x2 = -74.7 kg * cm
x2 = -74.7 kg * cm / 1.8 kg
x2 ≈ -41.5 cm
Therefore, the center of mass of the club-ax is approximately 41.5 cm from the handle end of the club-ax. Note that the negative sign indicates that the center of mass is on the opposite side of the handle end.
To find the center of mass of the club-ax, we need to calculate the distances from the center of mass to both ends of the club-ax.
Let's assume that the center of mass is located at a distance "x" from the handle end of the club-ax.
To find the center of mass, we can use the formula:
(m1 * x1) + (m2 * x2) = m * x
Where:
m1 = mass of the stick
x1 = distance from the center of mass of the stick to the handle end of the club-ax
m2 = mass of the stone
x2 = distance from the center of mass of the stone to the handle end of the club-ax
m = total mass of the club-ax
x = distance from the center of mass of the club-ax to the handle end of the club-ax
Given:
m1 = 1.8 kg
m2 = 8.3 kg
x1 = 98 cm = 0.98 m
x2 = 9 cm = 0.09 m
We can substitute these values into the formula:
(1.8 kg * 0.98 m) + (8.3 kg * 0.09 m) = (1.8 kg + 8.3 kg) * x
Simplifying the equation gives us:
(1.764 kg * m) + (0.747 kg * m) = 10.1 kg * x
Combining the terms on the left side of the equation gives us:
2.511 kg * m = 10.1 kg * x
Now we can solve for x:
x = (2.511 kg * m) / (10.1 kg)
x ≈ 0.248 m
Therefore, the center of mass of the club-ax is approximately 0.248 meters from the handle end of the club-ax.