Four masses (2kg 3kg 1kg and 2kg) are arranged on a square of side 1.2 m. What is the moment of inertia if the axis of rotation is at the center of the square? What is the moment of inertia if you make the axis of rotation through mass 2kg (this mass cannot contribute to moment of inertia because its the origin of axis of rotation)
Thank you!
a. through the center:
Itotal=2r^2+3r^2 + sum them all. r is distance from center to corner.
b. same way, except each distance is different this time. OR you can use the parallel axis theorem.
To calculate the moment of inertia, we need to consider the individual masses and their respective distances from the axis of rotation.
For the first scenario, where the axis of rotation is at the center of the square, we consider all four masses. The moment of inertia for each mass is calculated as the product of the mass and the square of the distance from the axis of rotation. In this case, since the square is symmetric, each of the four masses is equidistant from the axis of rotation.
The moment of inertia for each mass is given by:
I = m * r^2
Where:
I is the moment of inertia
m is the mass of the object
r is the distance from the axis of rotation
In this case, the masses are 2kg, 3kg, 1kg, and 2kg.
Since all masses are equidistant from the axis of rotation, we can calculate the total moment of inertia by summing the individual moments of inertia.
I_total = I_2kg + I_3kg + I_1kg + I_2kg
We can substitute the values into the equation to find the moment of inertia for each mass:
I_2kg = 2kg * (0.6m)^2
I_3kg = 3kg * (0.6m)^2
I_1kg = 1kg * (0.6m)^2
I_2kg = 2kg * (0.6m)^2
Calculating each of these, we get:
I_2kg = 2 * 0.36 = 0.72 kg*m^2
I_3kg = 3 * 0.36 = 1.08 kg*m^2
I_1kg = 1 * 0.36 = 0.36 kg*m^2
I_2kg = 2 * 0.36 = 0.72 kg*m^2
Now we can sum the individual moments of inertia to get the total moment of inertia:
I_total = 0.72 + 1.08 + 0.36 + 0.72
I_total = 2.88 kg*m^2
For the second scenario, where the axis of rotation is through the 2kg mass, we need to exclude that mass from the calculation of our moment of inertia. Since only three masses are considered, the calculation becomes:
I_total = I_3kg + I_1kg + I_2kg
We substitute the values into the equation as before:
I_3kg = 3 * (0.6m)^2
I_1kg = 1 * (0.6m)^2
I_2kg is excluded from this calculation since it is the axis of rotation
Calculating each of these, we get:
I_3kg = 3 * 0.36 = 1.08 kg*m^2
I_1kg = 1 * 0.36 = 0.36 kg*m^2
Now we sum the individual moments of inertia to get the total moment of inertia:
I_total = 1.08 + 0.36
I_total = 1.44 kg*m^2
So, in summary:
- If the axis of rotation is at the center of the square, the moment of inertia is 2.88 kg*m^2.
- If the axis of rotation is through the 2kg mass, the moment of inertia is 1.44 kg*m^2.