Two particles (m1 = 0.20 kg, m2 = 0.30 kg) are positioned at the ends of a 2.0-m long rod of negligible mass. What is the moment of inertia of this rigid body about an axis perpendicular to the rod and through the center of mass?
.2*1^2 + .3 * 1^2
Why did the particle go to the gym? Because it wanted to be an "axis" of power!
To find the moment of inertia of the rigid body, we need to consider the moment of inertia of each particle individually and then combine them together.
The moment of inertia of a particle rotating about an axis is given by the formula:
I = m * r^2
Where:
I is the moment of inertia,
m is the mass of the particle, and
r is the distance of the particle from the axis of rotation.
Since the two particles are located at the ends of the rod, their distances from the axis of rotation will be half the length of the rod, which is 2.0 m / 2 = 1.0 m.
Now, we can calculate the moment of inertia of each particle:
For particle 1:
m1 = 0.20 kg
r1 = 1.0 m
I1 = m1 * r1^2
For particle 2:
m2 = 0.30 kg
r2 = 1.0 m
I2 = m2 * r2^2
Finally, we can combine the moment of inertia of the two particles by using the parallel axis theorem. According to this theorem, the moment of inertia of a rigid body about an axis parallel to an axis through its center of mass is given by:
I_total = I1 + I2
Substituting the values, we get:
I_total = I1 + I2
= m1 * r1^2 + m2 * r2^2
Calculating the values:
I1 = 0.20 kg * (1.0 m)^2
I2 = 0.30 kg * (1.0 m)^2
I_total = 0.20 kg * (1.0 m)^2 + 0.30 kg * (1.0 m)^2
Simplifying:
I_total = 0.20 kg * 1.0 m^2 + 0.30 kg * 1.0 m^2
= 0.20 kg + 0.30 kg
= 0.50 kg * m^2
Therefore, the moment of inertia of the rigid body about an axis perpendicular to the rod and through the center of mass is 0.50 kg * m^2.
To find the moment of inertia of this rigid body, we need to consider the individual moments of inertia of the two particles and the parallel axis theorem.
The moment of inertia of a point mass rotating about an axis that passes through its center of mass is given by the formula:
I = m*r^2
Where I is the moment of inertia, m is the mass of the object, and r is the perpendicular distance from the axis of rotation to the point mass.
For the particle at one end of the rod:
m1 = 0.20 kg
r1 = 1.0 m (half the length of the rod)
I1 = m1*r1^2
For the particle at the other end of the rod:
m2 = 0.30 kg
r2 = 1.0 m (half the length of the rod)
I2 = m2*r2^2
The total moment of inertia is the sum of the individual moments of inertia:
I_total = I1 + I2
Now, we need to consider the parallel axis theorem. The moment of inertia of a system about an axis parallel to and a distance 'd' away from an axis through its center of mass is given by:
I_parallel_axis = I_center_of_mass + m*d^2
In this case, the axis we want to find the moment of inertia about is through the center of mass, and it is perpendicular to the rod. The distance between this axis and the center of mass of the system is the length of the rod divided by 2:
d = 2.0 m / 2 = 1.0 m
Now we can apply the parallel axis theorem:
I_total_parallel_axis = I_total + m_total*d^2
Where m_total is the sum of the masses:
m_total = m1 + m2
Finally, we can substitute the values and calculate the moment of inertia:
I_total_parallel_axis = (I1 + I2) + (m1 + m2)*d^2
Substituting the values:
I_total_parallel_axis = (m1*r1^2 + m2*r2^2) + (m1 + m2)*(1.0 m)^2
Calculate I_total_parallel_axis to find the moment of inertia of the rigid body about the desired axis.