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katen-ka-za [31]

3 years ago

15

A horizontal disk with a radius of 23 m rotates about a vertical axis through its center. The disk starts from rest and has a co

nstant angular acceleration of 5.7 rad/s 2 . At what time will the radial and tangential components of the linear acceleration of a point on the rim of the disk be equal in magnitude? Answer in units of s

1 answer:

lys-0071 [83]3 years ago

8 0

Answer:

time is 0.42 sec

Explanation:

Given data

radius = 23 m

angular acceleration = 5.7 rad/s²

to find out

time

solution

we know that radius is constant so that

tangential acceleration At = angular acceleration × radius ............. 1

tangential acceleration = 5.7 × 23 = 131.1 m/s²

and

radial acceleration Ar = (angular velocity)² × radius ........................2

we consider angular velocity = ω

this is acting toward center

so

compare 1 and 2

At = Ar

5.7 r =ω³ r

ω = √5.7 = 2.38746 rad/s

so

ω = 5.7 t

2.387 = 5.7 t

t = 2.387 / 5.7

t = 0.4187

time is 0.42 sec

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then we calculate t

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$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

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