Question

A low-mass rod of length 0.40 m has a metal ball of mass 1.5 kg at each end. The center of the rod is located at the origin, and the rod rotates in the xz plane about its center. The rod rotates clockwise around its axis when viewed from a point on the +y axis, looking toward the origin. The rod makes one complete rotation every 0.8 s. What is the moment of inertia of the object (rod plus two balls)?

Answer 1

Answer:

$I = 0.12 kg.m^2$

Explanation:

To solve the problem it is necessary to take into account the concepts related to moments of inertia. For the particular case in rods

According our values we have that,

Length of rod (L) is 0.4 m

And the mass (m) in each part of the rod is 1.5kg.

If the center of the bar is at the origin of coordinates, the distance between the balls and the center is L / 2, thus the moment of inertia for the two is,

$I = m * \frac{L}{2}^2 + m *\frac{L}{2}^2$

$I = 2 * 1.5 * \frac{0.4}{2}^2$

$I = 0.12 kg.m^2$

Therefore the moment of inertia of the object is $0.12kg.m^2$

Answer 2

The moment of inertia of the object is mathematically given as

I = 0.12 kg.m^2

What is the moment of inertia of the object (rod plus two balls)?

Question Parameter(s):

A low-mass rod of length 0.40 m has a metal ball of mass 1.5 kg at each end.

The rod makes one complete rotation every 0.8 s.

Generally, the equation for the moment of Inertia   is mathematically given as

$I = m * \frac{L}{2}^2 + m *\frac{L}{2}^2$

Therefore

I = 2 * 1.5 *( 0.4/2^2)

I = 0.12 kg.m^2

In conclusion, the moment of inertia

I = 0.12 kg.m^2

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