A wheel of diameter 400 cm starts from rest and rotates with

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.50 rad/s^2. 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 = omega^2r. From the relationship a_rad = u^2/r.

Solution

A) As the wheel turn by 2pi radian in each revolution ,the no of revolution n = angular displacement / 2 pi

Therefore angular displacement = n x2pi = 2x2pi = 4pi rad

Also angular displacement = w₀ t + ( t² ) /2 Where \'w₀ \' is angular velocity and \' \' is angular acceleration

Since the wheel start from rest w₀t is zero

Angular displacement = ( t² ) /2 = 3.50t² /2

4pi = 1.75 t²

t = 2.679 s

Angular velocity = angular displacement / time = 4pi / 2.679 = 4.6883rad/ sec

Radial acceleration a = w² r = 4.6883²x 20x10^-2 = 4.3960

B) radial acceleration = v²/r

V= rw = 20x10^-2x4.6883 = .9377

Radial acceleration = 4.3960 m/s²