A 4.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the x-axis?
To calculate the moment of inertia of a system of masses about the x-axis, we need to first find the individual moments of inertia for each mass and then add them up.
The formula for the moment of inertia of a point mass about an axis is given by:
I = m * r^2
where:
I is the moment of inertia,
m is the mass of the object, and
r is the perpendicular distance of the mass from the axis of rotation.
In this case, we have two masses: a 4.0 kg mass at (3.0, 4.0) m and a 6.0 kg mass at (3.0, -4.0) m.
The horizontal distance of both masses from the x-axis is zero since they both have the same x-coordinate of 3.0.
Now, let's calculate the moment of inertia for each mass.
For the 4.0 kg mass:
m1 = 4.0 kg (mass of the first object)
r1 = 4.0 m (since the perpendicular distance from the x-axis is 4.0 m)
I1 = m1 * r1^2
For the 6.0 kg mass:
m2 = 6.0 kg (mass of the second object)
r2 = -4.0 m (since the perpendicular distance from the x-axis is -4.0 m)
I2 = m2 * r2^2
Now, let's substitute the values and calculate the moments of inertia:
I1 = 4.0 kg * (4.0 m)^2
I1 = 4.0 kg * 16.0 m^2
I1 = 64.0 kg * m^2
I2 = 6.0 kg * (-4.0 m)^2
I2 = 6.0 kg * 16.0 m^2
I2 = 96.0 kg * m^2
Finally, we add up the moments of inertia of each mass to obtain the moment of inertia of the system about the x-axis:
I_total = I1 + I2
I_total = 64.0 kg * m^2 + 96.0 kg * m^2
I_total = 160.0 kg * m^2
So, the moment of inertia of this system of masses about the x-axis is 160.0 kg * m^2.