Question

A flywheel with a radius of .340 m starts from rest and accelerates with a constant angular acceleration of .730 rad/ s2.

A) compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim at the start.

B) compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 60.0.

C) compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 120

Answer 1

a_r=r*ω^2

a_t =r*α

r= 0.34 m ,ω is the angular velocity, α = .730 rad/s²

A)

a_t =r*α = 0.34*0.730 = 0.2482 m/s²

a_r=r*ω^2 = 0.34*0 = 0 m/s²

resustant = sqrt(a_t ^2 +a_r ^2) =   0.2482 m/s²

B)​

a_t =r*α = 0.34*0.730 = 0.2482 m/s²

θ = 60 deg = 60 *π/ 180 = π/3  

ωf^2 = ωi^2 + 2*α*θ​

ωf = sqrt (2*α*θ​) = sqrt( 2*.730*π/3) = 1.23649037 rad/s^2

a_r = 0.34*1.23649037^2 =0.519828868 m/s²

resultant acceleration = sqrt( 0.2482^2 +0.519828868^2) =0.576042787 m/s²

C)

same ways  

a_t =r*α = 0.34*0.730 = 0.2482 m/s²

θ = 120 deg =120 *π/ 180 = 2π/3  

ωf^2 = ωi^2 + 2*α*θ​

ωf = sqrt (2*α*θ​) = sqrt( 2*.730*2π/3) = 1.74866145 rad/s^2

a_r = 0.34*1.74866145^2 =1.03965773 m/s²

resultant acceleration = sqrt( 0.2482^2 +1.03965773^2) =1.06887391 m/s²