A disc of radius 0.1m starts from rest with an angular acceleration of 4.4 rad s2 . Then linear velocity of

the point on its rim after 5s is

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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2.2 m/s

Explanation:

r = 0.1 m

ω₀ = 0 rad/s

α = 4.4 rad/s²

t = 5 s

ω = αt + ω₀

ω = (4.4 rad/s²) (5 s) + 0 rad/s

ω = 22 rad/s

v = ωr

v = (22 rad/s) (0.1 m)

v = 2.2 m/s

i think its

this could be newton#'s law of cooling. the cooling curve is exponential