contestada

A majorette takes two batons and fastens them together in the middle at right angles to make an x shape. Each baton was 0.80 m long and each ball on the end is 0.20 kg. (Ignore the mass of the rods.) What is the moment of inertia if the arrangement is spun around an axis through the center perpendicular to both rods

Respuesta :

Answer:

Explanation:

The system essentially consists of 4 balls of .2 kg each  at .8 /2 m from center

Total moment of inertia of the 4 balls

4 x m r²

= 4 x .2 x .4²

= .128 kg m².

Lanuel

The moment of inertia of the two batons is equal to [tex]0.064\;Kgm^2[/tex]

Given the following data:

  • Mass = 0.20 kg.
  • Length (diameter) = 0.80 m.

Radius = [tex]\frac{0.80}{2} =0.40\;m[/tex]

How to calculate moment of inertia.

Mathematically, the moment of inertia of a baton is given by the formula:

[tex]I=mr^2[/tex]

Where:

  • I is the moment of inertia.
  • m is the mass.
  • r is the radius.

Since there are two batons, we have:

[tex]I= \sum mr^2\\\\I=2[0.20 \times 0.40^2]\\\\I=2 \times 0.032\\\\I=0.064\;Kgm^2[/tex]

Read more on moment of inertia here: https://brainly.com/question/3406242

Otras preguntas