Awheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s^2. at the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship arad=w^2r and (b) from the relationship arad=v^2/r.

ac= 15.07 m/s²

Explanation:

The Wheel rotates with a constant angular acceleration.:

Centripetal acceleration is calculated as follows:

ac =ω² *R   Formula (1)

ac =v² / R    Formula (2)

Kinematics of the wheel

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2αθ  Formula (3)

v = ω* R Formula (4)

Where:

θ : angle that the wheel has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed   ( rad/s)

v: tangential velocity of a point on the rim ( m/s)

R : radius of  wheel (m)

ac: centripetal acceleration, (m/s²)

Data:

D = 40.0 cm : diameter of the wheel

R = D/2= 40.0 cm/ 2 = 20 cm = 0.2m

α = 3.00 rad/s^2

ω₀ = 0

n = 2 revolutions : number of revolutions

θ =2πn (rad) = 2π*2 (rad) = 4π rad

Calculate of the ωf

We replace data in the formula (3)

ωf²= ω₀² + 2αθ

ωf²= 0 + 2(3)(4π)

ωf²= 24π

ωf = 8.68 rad/s

Calculate of the v

We replace data in the formula (4)

v = ω*R

v = (8.68)*(0.2)

v = 1.736 m/s

Calculate of the ac

We replace data in the formula (1)

ac = ( ω)²*(R)

ac = (8.68)²*(0.2)

ac = 15.07  m/s²

We replace data in the formula (2)

ac = v²/ R

ac = ( 1.736  )²/(0.2)

ac = 15.07  m/s²

answer; ///(c); ///bronchi i do believe this is the correct answer,; plz check first..good luck

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right