Question

3.

A flywheel comprises a uniform circular disk with a mass of 112.0
kg and a radius of 1.3 m. It rotates with an angular velocity of
1213 rev/min. A constant tangential force is applied at a radial
distance of 0.8 m. What is the initial kinetic energy of the
wheel?

4.

If the wheel is brought to rest in 111.0 s, what is the tangential
force?

5.

How many revolutions does the flywheel make while it is stopping in
time 111.0 s?

Answer #1

3.

kinetic energy:

K = (1/2)I ω²

= (1/2)(0.5 m r²) (2πf)²

=(1/2)(0.5* 112* 1.3²) (2π * 1213/60)²

= **7.6275 * 10^5 J**

4.

Angular acceleration: α = -F d/(0.5*m r²)

the force is calculated as follows:

ω = ωo + α t

0 = 2πf -F t d/(0.5 m r²)

0 = 2π * 1213/60 -F (111)(0.8)/(0.5 * 112* 1.3²)

F = **135.3 N**

5.

rotational kinematic equation: ω² = ωo² + 2α θ

the angular displacement is,

0 = (2πf)² + 2α θ

0 = (2πf)² + 2(-F d/(0.5 m r²)) θ

0 = (2π * 1213/60)² +2(-135.3 * 0.8/(0.5 * 112* 1.3²)) θ

θ =**7046.855 rad**

or

θ = **1122.11 rev**

**note**: 1 rev = 2π rad

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