Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².
Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.
Given,
Radius of flywheel (r) = 1.96 cm = 0.0196m
Angular acceleration (α)= 13.0 rad/s²
The tangential acceleration formula is at=rα
where, α is the angular acceleration, and r is the radius of the circle.
using the formula; at=rα = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².
The tangential acceleration is 0.2548 m/s².
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We have the equation of motion
, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.
In this case initial velocity = 2.20 m/s, final velocity = 0 m/s, displacement = 14 m
On substitution we will get 0 = 
On solving we can find the acceleration value as -0.173
So free fall acceleration = 0.173
