Six identical particles each of mass ‘m’ are arranged at the corners of a regular
hexagon of side length ‘L’. If the masses of any two adjacent particles are doubled.
The shift in the centre of mass is???
To find the shift in the center of mass, we need to compare the center of mass of the system before and after the masses are doubled for any two adjacent particles.
Before the masses are doubled:
The center of mass of the system is at the centroid of the regular hexagon. The centroid is the point of intersection of the three medians of the hexagon.
After the masses are doubled:
Let's assume that the original mass of each particle is 'm'. When the masses of any two adjacent particles are doubled, their masses become '2m' each.
To calculate the shift in the center of mass, we can use the principle of moments.
Let's consider a reference point as the center of the hexagon. The distance of each particle from this reference point is half of the side length 'L'.
Before the masses are doubled:
The total mass of the system = 6m
The distance of the center of mass from the reference point = 0 (as the center of mass is at the reference point)
After the masses are doubled:
The total mass of the system = 2m + 2m + 2m + 2m + 2m + 2m = 12m
The distance of the center of mass from the reference point = distance from the centroid to the reference point.
The distance of the centroid from the reference point can be calculated using the formula for the distance from the centroid of a regular hexagon to any vertex:
d = (2/3) * L
Now, let's calculate the shift in the center of mass by comparing the moments of the system before and after the masses are doubled:
Before:
Moment = mass * distance = 6m * 0 = 0
After:
Moment = mass * distance = 12m * [(2/3) * L]
The shift in the center of mass = Moments after - Moments before
= (12m * [(2/3) * L]) - 0
= (24/3) * mL
= 8mL
Therefore, the shift in the center of mass is 8mL.
To determine the shift in the center of mass, we need to calculate the coordinates of the original center of mass and the new center of mass after doubling the mass of adjacent particles.
Let's consider the original arrangement of six particles at the corners of a regular hexagon. The center of mass of this system can be found by taking the average of the coordinates of all the particles.
The coordinates of the original center of mass (x_cm, y_cm) can be calculated as follows:
x_cm = (x₁ + x₂ + x₃ + x₄ + x₅ + x₆) / 6
y_cm = (y₁ + y₂ + y₃ + y₄ + y₅ + y₆) / 6
Since the particles are arranged in a regular hexagon, we can simplify these expressions by using the properties of a regular hexagon:
The x-coordinates of the particles are: L, L/2, -L/2, -L, -L/2, L/2
The y-coordinates of the particles are: 0, √3L/2, √3L/2, 0, -√3L/2, -√3L/2
Substituting these values into the formulas for x_cm and y_cm:
x_cm = (L + L/2 - L/2 - L - L/2 + L/2) / 6 = 0
y_cm = (0 + √3L/2 + √3L/2 + 0 - √3L/2 - √3L/2) / 6 = 0
Hence, the original center of mass is located at (0, 0).
Now let's consider the new arrangement where the masses of any two adjacent particles are doubled. This means that the mass of particles at opposite corners of the hexagon remains the same, while the mass of adjacent particles doubles. In this case, the particles at the corners of the hexagon have mass m, and the particles at the midpoints of the sides have mass 2m.
We can repeat the same calculation to find the new coordinates of the center of mass.
The x-coordinates of the particles in the new arrangement are still: L, L/2, -L/2, -L, -L/2, L/2
The y-coordinates of the particles in the new arrangement are still: 0, √3L/2, √3L/2, 0, -√3L/2, -√3L/2
Again, substituting these values into the formulas for x_cm and y_cm:
x_cm = (L + L/2 - L/2 - L - L/2 + L/2) / 6 = 0
y_cm = (0 + √3L/2 + √3L/2 + 0 - √3L/2 - √3L/2) / 6 = 0
We can observe that the coordinates of the new center of mass are the same as the original center of mass.
Therefore, the shift in the center of mass is zero.
first arrangement. Consider the entire 6m at the center, then add the two masses. since the one triangle now has two masses on each base corner, and 6m at the vertix...the triangle is all sides the same (think about that), and each angle is 60 deg.
so the cm will shift exactly 2m*cos30/6m