radial acceleration of a point

Question

A wheel of diameter 35.0cm starts from rest and rotates with a constant angular acceleration of 3.50rad/s2 . At the instant the wheel has completed its second revolution, compute the radial acceleration of a point on the rim in two ways. 

a)Using the relationship arad=ω(2r.)

b)From the relationship arad=v^2/r

Answer 1

(b)

Initial Angular Velocity = 0 rad/sec.

Angular accerlation = 3.5 rad/sec.

Time for 1 revolution = 2π

Time for 2 revolutions = 2 * 2π

Diameter of the wheel = 35.0 cm

Radius of the wheel = 17.5 cm = 17.5 * 10^-2

Time at the instant it completes 2 revolutions = 4π.

Final angular Velocity .

Final Angular velocity= 87.96 rad/sec.

Now angular accerlation .

Where v is the linear velocity

v = 87.96 * 17.5*10^-2

v = 15.393 m/sec

Radial Accerlation

α = (15.393)² / (17.5*10^-2)

α = 1353 m/sec².

Therefore Radial accerlation is 1353 m/sec².

Answer 2

(a)

Initial Angular Velocity = 0 rad/sec.

Angular accerlation = 3.5 rad/sec.

Time for 1 revolution = 2π

Time for 2 revolutions = 2 * 2π

Diameter of the wheel = 35.0 cm

Radius of the wheel = 17.5 cm

Time at the instant it completes 2 revolutions = 4π.

Final angular Velocity .

Final Angular Velocity = 87.96 rad/sec.

Now angular accerlation .

Angular accerlation α = 87.6(2*17.5*10^-2)

α = 30.66 rad/sec².

Therefore angular accerlation is 30.66 rad/sec².