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rewona [7]
3 years ago
7

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent ho

op about that axis, if the hoop has the same mass M and a radius k given by The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.
Physics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

Explanation:

Let mass of cylinder be M

Moment of inertia of cylinder

= 1/2 M R² r is radius of cylinder

If radius of equivalent  hoop be k

Mk² = 1/2 x MR²

k = R / √2

1.2 / 1.414

Radius of gyration = 0.848 m

b )

moment of inertia of spherical shell

= 2 / 3 M R²

Moment of inertia of equivalent hoop

Mk²

So

Mk² = 2 / 3 M R²

k = √2/3 x R

= .816 X 1.2

Radius of gyration = .98 m

c )

Moment of inertia of solid sphere

= 2/5 M R²

Moment of inertia of equivalent hoop

= Mk²

Mk² = 2/5 M R²

k √ 2/5 R

Radius of gyration = .63 R

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Answer:

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Explanation:

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Answer:

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Explanation:

Step one:

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Si la aceleración de gravedad en la superficie del planeta Mercurio es de 3,7 m/s2, entonces ¿cuál sería el peso de una persona
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Answer:

W = 222 N.

Explanation:

The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface? "

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