The linear velocity is 2.2 m/s
Given:
r = 0.1 m
ω₀ = 0 rad/s
α = 4.4 rad/s²
t = 5 s
So from the equation for angular acceleration we will get,
ω = αt + ω₀
ω = (4.4 rad/s²) (5 s) + 0 rad/s
ω = 22 rad/s
Angular velocity ω is analogous to linear velocity v
The relation between velocity and angular acceleration is given which is:
v = ωr
v = (22 rad/s) (0.1 m)
v = 2.2 m/s
Therefore, linear velocity of the point on its rim after 5s is 2.2 m/s.
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