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/s^{2}

A)

**a _{t} =r*α = 0.34*0.730 = 0.2482
m/s^{2}**

**a _{r}=r*ω^2 = 0.34*0 = 0
m/s^{2}**

**resustant = sqrt(a _{t} ^2 +a_{r} ^2)
= 0.2482 m/s^{2}**

B)

a_{t} =r*α = 0.34*0.730 = 0.2482 m/s^{2}

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/s^{2}

resultant acceleration = sqrt( 0.2482^2 +0.519828868^2)
=0.576042787 m/s^{2}

C)

same ways

a_{t} =r*α = 0.34*0.730 = 0.2482 m/s^{2}

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/s^{2}

resultant acceleration = sqrt( 0.2482^2 +1.03965773^2)
=1.06887391 m/s^{2}

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