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olya-2409 [2.1K]
3 years ago
8

A wind farm generator uses a two-bladed propeller mounted on a pylon at a height of20 m. The length of each propeller blade is 1

2 m. A small piece from the tip of thepropeller breaks off the top propeller when the propeller is vertical. At that instant,the angular speed of the propeller is 5.2 rad/s.The fragment flies off horizontally, falls with negligible air resistance, and strikes theground at pointP. How far is pointPfrom the base of the pylon
Physics
1 answer:
alekssr [168]3 years ago
7 0

Answer:

159.37 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

r = Radius = 12 m

Time taken to fall

s=ut+\frac{1}{2}at^2\\\Rightarrow 0.809=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{(20+12)\times 2}{9.81}}\\\Rightarrow t=2.554\ s

Speed is given by

v=r\omega\\\Rightarrow v=12\times 5.2\\\Rightarrow v=62.4\ m/s

Distance is given by

s=vt\\\Rightarrow s=62.4\times 2.554\\\Rightarrow s=159.37\ m

The point P is 159.37 m from the base.

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3 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
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Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

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so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

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I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

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