Two small balls are attached to a rigid rod of mass 2.2 kg that is 6.0 m in length. The first ball has a mass of 2.25 kg and the second ball has a mass of 4.95 kg.
(c) What is the moment of inertia of the rod when it spins around its center of mass?
To find the moment of inertia of the rod when it spins around its center of mass, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia of an object about an axis parallel to and a distance 'd' away from its center of mass is given by the sum of the moment of inertia about its center of mass and the product of the mass and the square of the distance 'd'.
In this case, since the rod is spinning around its center of mass, we only need to find the moment of inertia of the rod itself. The moment of inertia of a uniform rod rotating about its center of mass is given by the formula:
I = (1/12) * m * L^2
where I is the moment of inertia, m is the mass of the rod, and L is the length of the rod.
Given:
Mass of the rod (m) = 2.2 kg
Length of the rod (L) = 6.0 m
Substituting the given values into the formula, we get:
I = (1/12) * 2.2 kg * (6.0 m)^2
Calculating this expression, we get:
I = (1/12) * 2.2 kg * 36.0 m^2
Simplifying, we get:
I = 1.1 kg * 3.0 m^2
Finally, the moment of inertia of the rod when it spins around its center of mass is:
I = 3.3 kg * m^2
To find the moment of inertia of the rod when it spins around its center of mass, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia of a body about an axis parallel to an axis passing through its center of mass is equal to the moment of inertia about the center of mass plus the product of the mass of the body and the square of the distance between the two axes.
Here, we have a rod with two small balls attached to it. The center of mass of the rod is at its midpoint.
The moment of inertia of the rod about its center of mass can be calculated using the equation for the moment of inertia of a slender rod rotated about one end:
I_rod = (1/3) * m_rod * L^2
Where:
- I_rod is the moment of inertia of the rod about its center of mass.
- m_rod is the mass of the rod.
- L is the length of the rod.
Substituting the values in, we get:
I_rod = (1/3) * 2.2 kg * (6.0 m)^2
I_rod ≈ 26.4 kg•m^2.
Now, let's calculate the moment of inertia of the two balls rotating around the center of mass of the rod.
The moment of inertia of a point mass rotating about an axis passing through its center of mass is given by the equation:
I_ball = m_ball * r^2
Where:
- I_ball is the moment of inertia of the ball.
- m_ball is the mass of the ball.
- r is the distance from the axis of rotation to the center of mass of the ball.
Since the balls are attached to the rod at its ends, the distance from the axis of rotation (center of mass of the rod) to the center of mass of each ball is half the length of the rod.
So, for the first ball:
I_1 = m_1 * (L/2)^2
I_1 = 2.25 kg * (6.0 m/2)^2
I_1 = 2.25 kg * 9.0 m^2
I_1 ≈ 20.25 kg•m^2
Similarly, for the second ball:
I_2 = m_2 * (L/2)^2
I_2 = 4.95 kg * (6.0 m/2)^2
I_2 = 4.95 kg * 9.0 m^2
I_2 ≈ 44.55 kg•m^2
Now, we can use the parallel axis theorem to find the total moment of inertia of the system:
I_total = I_rod + I_1 + I_2
I_total = 26.4 kg•m^2 + 20.25 kg•m^2 + 44.55 kg•m^2
I_total ≈ 91.2 kg•m^2
Therefore, the moment of inertia of the rod when it spins around its center of mass is approximately 91.2 kg•m^2.