ω = 10.44rad/s ,
ε = 51.57rad/s^2
R =D/2.
Normal (centripetal) acceleration
a(n) = ω^2•R.
Tangential acceleration
a(τ) = ε•R.
Resultant linear acceleration
a =sqrt(a(n)^2 + a(τ)^2).
A flywheel accelerates from rest to an angular velocity of 10.44rad/s at an angular acceleration of 51.57rad/s^2. The diameter of the flywheel is 1.16m.
At this instant, calculate the resultant linear acceleration, in m/s^2, of a point on the circumference of the flywheel.
1 answer
1 answer