A light, rigid rod l = 5.55 m in length rotates
in the xy plane about a pivot through the
rod’s center. Two particles of masses m1 =
9.8 kg and m2 = 2.4 kg are connected to its
ends
find the moment of inertia
The moment of inertia of a system depends on the masses and their positions relative to the axis of rotation. In this case, we are given the masses of the two particles, m1 = 9.8 kg and m2 = 2.4 kg, and the length of the rod, l = 5.55 m.
To find the moment of inertia of this system, we can use the parallel axis theorem, which states that the moment of inertia of a system about an axis parallel to and a distance "d" away from an axis through the center of mass is equal to the moment of inertia about the center of mass plus the product of the total mass of the system and the square of the distance "d".
Let's find the moment of inertia of each particle about the center of mass of the system first. Since the center of mass of the system is at the center of the rod, the distance from each particle to the center of mass is half the length of the rod, l/2.
The moment of inertia of each particle about the center of mass is given by the formula:
I = m * r^2
where m is the mass of the particle and r is the perpendicular distance from the particle to the axis of rotation.
For particle 1:
I1 = m1 * (l/2)^2
For particle 2:
I2 = m2 * (l/2)^2
Now, using the parallel axis theorem, we can find the moment of inertia of each particle about the pivot axis (through the center of the rod) by adding the moment of inertia about the center of mass to the product of the total mass of the system and the square of the distance between the center of mass and the pivot axis.
Let's denote the distance between the center of mass and the pivot axis as "x". Since the pivot axis passes through the center of the rod, the distance "x" is zero.
For particle 1:
I1_total = I1 + m1 * x^2
= m1 * (l/2)^2 + m1 * 0^2
= m1 * (l/2)^2
For particle 2:
I2_total = I2 + m2 * x^2
= m2 * (l/2)^2 + m2 * 0^2
= m2 * (l/2)^2
The total moment of inertia of the system is the sum of the individual moment of inertia of the particles:
I_total = I1_total + I2_total
= m1 * (l/2)^2 + m2 * (l/2)^2
Now we can substitute in the given values:
I_total = (9.8 kg) * (5.55 m/2)^2 + (2.4 kg) * (5.55 m/2)^2
Calculating this expression will give you the moment of inertia of the system.
To find the moment of inertia of the system, we'll need to consider the individual moment of inertias of the rod and the two particles, and then sum them up.
1. Moment of Inertia of the Rod:
The moment of inertia of a thin rod rotating about an axis perpendicular to its length and passing through its center is given by the formula: I_rod = (1/12) * m * L^2, where m is the mass of the rod and L is its length. In this case, the rod's mass is assumed to be concentrated at its center.
Given that the length of the rod is L = 5.55 m, we need the mass of the rod to calculate its moment of inertia. However, the mass of the rod is not given in the question. We'll assume the rod is of negligible mass compared to the particles and will not contribute to the moment of inertia.
2. Moment of Inertia of Mass m1:
The moment of inertia of a particle rotating about an axis passing through its center is given by the formula: I_particle = m * r^2, where m is the mass of the particle and r is the perpendicular distance between the particle and the axis of rotation.
Given that m1 = 9.8 kg, we need the distance between the particle and the axis of rotation. The distance is not mentioned in the question, so we'll assume that the particles are located at the ends of the rod, which is L = 5.55 m. For a particle located at the edge of the rod, the perpendicular distance from the axis of rotation is half the length of the rod.
So, for the particle with mass m1, the moment of inertia is: I_m1 = m1 * (L/2)^2 = (9.8 kg) * (2.775 m)^2.
3. Moment of Inertia of Mass m2:
Same as above, the moment of inertia of the other particle will be: I_m2 = m2 * (L/2)^2 = (2.4 kg) * (2.775 m)^2.
Total Moment of Inertia:
Finally, we sum up the moment of inertias of the rod (negligible in this case), mass m1, and mass m2 to get the total moment of inertia of the system:
I_total = I_rod + I_m1 + I_m2 = I_m1 + I_m2.
Substituting the values we found earlier:
I_total = (9.8 kg) * (2.775 m)^2 + (2.4 kg) * (2.775 m)^2 = (9.8 kg + 2.4 kg) * (2.775 m)^2.
Simplifying the equation should give you the moment of inertia of the system.
Easy.
MI= rigid rod I + m1r2^2 + m2r2^2
as I read the problem r1=r2=5.55/2