Question

К шару радиусом 10 см и массой 2 кг присоединен стержень длиной 40 см и массой 6 кг. Ось стержня проходит через центр шара. Найти момент инерции системы относительно оси, проходящей через шар перпендикулярно отрезку, соединяющему центр шара и стержень.

Answer 1

Момент инерции самого шара $I_{1}= \frac{2}{5}M R^{2}$.
Центр стержня длиной L удалён от оси на расст. d=R+0.5L.
Отсюда, учитывая теорему Штейнера и складывая оба момента инерции, получим:
$I= \frac{2}{5}M R^{2}$+m$d^{2}$+$\frac{1}{12}m L^{2}$

Answer 2

Calculation of Moment of Inertia

To find the moment of inertia of the system, we need to consider the moment of inertia of the sphere and the moment of inertia of the rod separately.

The moment of inertia of a solid sphere about an axis passing through its center is given by the formula:

I_sphere = (2/5) * m * R^2 Where: - m is the mass of the sphere - R is the radius of the sphere

Given that the sphere has a radius of 10 cm and a mass of 2 kg, we can substitute these values into the formula to calculate the moment of inertia of the sphere.

I_sphere = (2/5) * 2 kg * (0.1 m)^2

Simplifying the equation, we get:

I_sphere = 0.08 kg * m^2

Next, let's calculate the moment of inertia of the rod. The moment of inertia of a rod about an axis perpendicular to the rod 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 - l is the length of the rod

Given that the rod has a length of 40 cm and a mass of 6 kg, we can substitute these values into the formula to calculate the moment of inertia of the rod.

I_rod = (1/12) * 6 kg * (0.4 m)^2

Simplifying the equation, we get:

I_rod = 0.08 kg * m^2

Now, to find the total moment of inertia of the system about an axis passing through the sphere and perpendicular to the rod, we can add the moment of inertia of the sphere and the moment of inertia of the rod.

Total moment of inertia = I_sphere + I_rod

Substituting the values we calculated earlier, we get:

Total moment of inertia = 0.08 kg * m^2 + 0.08 kg * m^2

Simplifying the equation, we get:

Total moment of inertia = 0.16 kg * m^2

Therefore, the moment of inertia of the system about an axis passing through the sphere and perpendicular to the rod is 0.16 kg * m^2.

Conclusion

The moment of inertia of the system, consisting of a sphere with a radius of 10 cm and a mass of 2 kg, and a rod with a length of 40 cm and a mass of 6 kg, about an axis passing through the sphere and perpendicular to the rod, is calculated to be 0.16 kg * m^2.