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Ivanshal [37]
3 years ago
7

A 10-kg disk-shaped flywheel of radius 9.0 cm rotates with a rotational speed of 320 rad/s. Part A Determine the rotational mome

ntum of the flywheel. Express your answer to two significant figures and include the appropriate units. Part B With what magnitude rotational speed must a 10-kg solid sphere of 9.0 cm radius rotate to have the same rotational momentum as the flywheel? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Leya [2.2K]3 years ago
4 0

Answer:

(A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

Mass of disk = 10 kg

Radius = 9.0 cm

Rotational speed = 320 m/s

(A). We need to calculate the rotational momentum of the flywheel.

Using formula of momentum

L=I\omega

L=\dfrac{1}{2}mr^2\omega

Put the value into the formula

L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320

L=12.96\ kg m^2/s

(B). Rotation momentum of sphere is same rotational momentum of the  flywheel

We need to calculate the magnitude of the rotational speed of sphere

Using formula of rotational momentum

L_{sphere}=L_{flywheel}

I\omega_{sphere}=I\omega_{flywheel}

\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}

\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}

Put the value into the formula

\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}

\omega_{sphere}=400\ rad/s

Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

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Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

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Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

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