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Mrac [35]
2 years ago
13

What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs

cript of 1, e.i. I1 = ) What is this value of the moment of inertia when the person is on the edge of the merry-go-round?
Physics
1 answer:
Ann [662]2 years ago
7 0

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

  • the quantity expressed by the body resisting angular acceleration.
  • It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

#SPJ4

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Ram jumps onto a cement floor from a height of 1m and comes to rest in 0.1sec.
umka2103 [35]

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3/10 F.

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The space shuttle fleet was designed with two booster stages. If the second stage provides a thrust of 73 ​kilo-newtons and the
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m = 81281.5 pounds.

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16000 mi/h² = 1.98 m/s²

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According to Newton's second law,

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