Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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?
#SPJ4
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
