Question

A wheel of diameter 50.0 cm starts from rest and rotates with a constant angular acceleration of 5.00 rad/s². At the instant the wheel has completed its second revolution, compute the radial acceleration of a point on the rim in two ways.  

Using the relationship a_rad = ω²r.  

From the relationship a_rad = v²/r.

Answer 1

at the end of second revolution

angular displacement covered is theta = 2*2*3.142 = 12.568 rad

given that radius of the wheel is r = D/2 = 0.5/2 = 0.25 m

angular accelaration is alpha = 5 rad/s^2

initial angular velocity is Wo = 0 rad/s

then using Wf^2 -Wo^2 = 2*alpha*theta

Wf^2 - 0 ^2 = 2*5*12.568

Wf = 11.21 rad/sec

A) a_rad = w^2*r = 11.21^2*0.25 = 31.4 m/s^2

B) linear velocity v = r*w = 0.25*11.21 = 2.8025 m/sec

a_rad = v^2/r = 2.8025^2/(0.25) = 31.4 m/s^2