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Sagot :
The linear acceleration of the disk is [tex]8.17 m/s^{2}[/tex].
We know that, a rotating disk has two acceleration:
Radial acceleration ([tex]a_{r}[/tex]) = ω^2 * r and
Tangential acceleration[tex](a_{t} )[/tex] = [tex]\alpha r[/tex]
where ω = angular velocity
[tex]\alpha[/tex] = angular velocity
r = radius
Given, ω = 4 m/s , [tex]\alpha[/tex] = 3.4 [tex]m/s^{2}[/tex] , r = 0.5m
Putting these values in above equation we get
[tex]a_{r} = 8 m/s^{2} \\a_{t} = 1.7 m/s^{2}[/tex]
The linear acceleration is given by:
[tex]a = \sqrt{a_{r} ^{2} + a_{t} ^{2} }[/tex]
Putting respected values in above equation we get a = 8.17 [tex]m/s^{2}[/tex].
So the linear acceleration of the disk is [tex]8.17 m/s^{2}[/tex].
To know more about rotational mechanics visit:
https://brainly.com/question/13262346?
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