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3 years ago

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Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig

htweight material of density 600 kg/m3. (Assume that acceleration due to gravity is g = 9.8 m/s2. Round your answer to one decimal place.)

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = $\frac{1}{3}\pi r^2 h$

thus,

volume of the cone = $\frac{1}{3}\pi 1.2^2\times 4$ = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the $\frac{1}{4}$ times the total height

thus,

center of mass lies at, h' = $\frac{1}{4}\times4=1m$

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

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Answer:

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Explanation:

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$\omega_{f}$ - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$ (2)

Where $M$ is the mass of the turntable, in kilograms.

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$I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}$

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$

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