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3 years ago

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If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th

ree rotor blades about the axis of rotation.

1 answer:

devlian [24]3 years ago

7 0

Answer:

1470kgm²

Explanation:

The formula for expressing the moment of inertial is expressed as;

I = 1/3mr²

m is the mass of the body

r is the radius

Since there are three rotor blades, the moment of inertia will be;

I = 3(1/3mr²)

I = mr²

Given

m = 120kg

r = 3.50m

Required

Moment of inertia

Substitute the given values and get I

I = 120(3.50)²

I = 120(12.25)

I = 1470kgm²

Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²

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During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient

ElenaW [278]

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

$g=9.8 m/s^2$

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

b)

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

$W=F_f d$

where

$F_f$ is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

c)

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

$K_f = U_i -E_t$

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

d)

The kinetic energy of the rock at the bottom of the hill can be rewritten as

$K_f=\frac{1}{2}mv^2$

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

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Drupady [299]

Answer:

initial velocity is v = 4.95 m / s

Explanation:

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y = y₀ + v_{oy} t - ½ g t²

y-y₀ = 0 -1/2 g t²

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t = 2.02 s

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x = v₀ₓ t

v₀ₓ = x / t

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v₀ₓ = 10 / 2.02

v₀ₓ = 4.95 m / s

initial velocity is v = 4.95 m / s

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