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A small merry-go round is spinning about its center in a clockwise direction at 1.6 r a d s and with rotational kinetic energy of 580 J. Calculate the moment of inertia of the merry-go round.

Respuesta :

onyebuchinnaji

Answer:

The moment of inertia of the merry-go round is 453.125 kg.m²

Explanation:

Given;'

angular velocity of the merry-go round, ω = 1.6 rad/s

rotational kinetic energy, K =  580 J

Rotational kinetic energy is given as;

K = ¹/₂Iω²

Where;

I is the moment of inertia of the merry-go round

[tex]I = \frac{2K}{\omega^2} \\\\I = \frac{2*580}{1.6^2} \\\\I = 453.125 \ kg.m^2[/tex]

Therefore, the moment of inertia of the merry-go round is 453.125 kg.m²

Lanuel

Since the small merry-go round is spinning about its center in a clockwise direction, its moment of inertia is equal to 453.13 [tex]Kgm^2[/tex]

Given the following data:

  • Angular velocity = 1.6 rad/s
  • Rotational kinetic energy = 580 Joules.

To calculate the moment of inertia of the small merry-go round:

Mathematically, the rotational kinetic energy of an object is giving by the formula:

[tex]E_{rotational} = \frac{1}{2} Iw^2[/tex]

Where:

  • I is the moment of inertia.
  • w is the angular speed.

Making moment of inertia (I) the subject of formula, we have:

[tex]I = \frac{2E_{rotational}}{w^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]I = \frac{2(580)}{1.6^2}\\\\I = \frac{1160}{2.56}[/tex]

Moment of inertia (I) = 453.13 [tex]Kgm^2[/tex]

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