A Flywheel's Energy

solved RotationalDynamics

Q. A cord of negligible mass is wound round the rim of a flywheel of mass 20 kg $M = 20$ kg and radius 20 cm $R = 20$ cm, at rest $ω_{0} = 0$ rad/s on frictionless bearings. A steady pull of 25 N $F_{c} = 25$ N is applied on the cord until 2 m $l = 2$ m of cord is unwound. What is the kinetic energy of the wheel at this time, in joules?

Inputs

mass

M

20 kg

radius

R

20 cm

pull

F_{c}

25 N

unwound

l

2.00 m

Outputs

energy

K_{rot,1}

How it's solved

Solved with RotationalDynamics

Given

$M = 20$ kg
$R = 20$ cm
$F_{c} = 25$ N
$l = 2$ m
$ω_{0} = 0$ rad/s from “at rest”

Find $K_{rot,1}$ .

Working

$R = 20$ cm $= 0.2$ m
torque of a pull at the rim(flywheel, cord, turning) $τ = F_{c} \cdot R$
$τ = 25 \cdot 0.2$
$τ = 5$ J
a disc s moment of inertia(turning) $I = \frac{M \cdot R^{2}}{2}$
$I = \frac{20 \cdot {0.2}^{2}}{2}$
$I = 0.4$ kg m²
second law of rotation(turning) $τ = I \cdot α_{r}$
$5 = 0.4 \cdot α_{r}$
$α_{r} = 12.5$ rad/s²
cord unwound by the turn(turning) $l = R \cdot θ$
$2 = 0.2 \cdot θ$
$θ = 10$
angular velocity after an angle(turning) ${ω_{1}}^{2} = {ω_{0}}^{2} + 2 \cdot α_{r} \cdot θ$
${ω_{1}}^{2} = 0^{2} + 2 \cdot 12.5 \cdot 10$
$ω_{1} = 15.8113883$ rad/s
kinetic energy of turning(flywheel, turning, 2) $K_{rot,1} = \frac{I \cdot {ω_{1}}^{2}}{2}$
$K_{rot,1} = \frac{0.4 \cdot {15.8113883}^{2}}{2}$
$K_{rot,1} = 50$ J

Conclusion

The pull does work equal to the force times the cord it unwinds, and with no friction all of it becomes the wheel's energy of turning.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_flywheel_s_energy do4  kind :numerical5  describes "A cord of negligible mass is wound round the rim of a flywheel of mass %{mass} and radius %{radius}, " \6            "at rest on frictionless bearings. A steady pull of %{pull} is applied on the cord until %{unwound} of " \7            "cord is unwound. %{energy}, in joules?"8  given mass: "20 kg"9  given radius: "20 cm"10  given pull: "25 N"11  given unwound: "2 m"12  asks energy: "What is the kinetic energy of the wheel at this time"13end

solution.rb

1require_relative "../questions/a_flywheel_s_energy"2require_relative "../science/systems_of_particles_and_rotational_motion/rotational_dynamics"34solve :a_flywheel_s_energy, using: RotationalDynamics do5  objects :flywheel, :cord6  flywheel.goes(:turning)78  take mass: flywheel.mass9  take radius: flywheel.radius10  take pull: flywheel.pull(by: cord)11  take unwound: flywheel.cord_unwound1213  deduce_from "at rest", flywheel.initial_angular_velocity => "0 rad/s"1415  energy = flywheel.rotational_kinetic_energy(at: turning.end)16  answer energy:, in: "J"1718  conclusion "The pull does work equal to the force times the cord it unwinds, and with no friction all " \19             "of it becomes the wheel's energy of turning."20end

playground.rb

1require_relative "../solutions/a_flywheel_s_energy"23play_with :a_flywheel_s_energy do4  title "A Flywheel's Energy"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	RotationalKinematics
position	$x$	m	—	Velocity
displacement	$Δ x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average velocity	$v_{av}$	m/s	—	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
angular velocity	$ω$	rad/s	—	RotationalKinematics
angular acceleration	$α_{r}$	rad/s²	—	RotationalKinematics
angle turned	$θ$	—	—	RotationalKinematics
turns	$n_{rev}$	—	—	RotationalKinematics
starting angular velocity	$ω_{0}$	rad/s	—	RotationalKinematics
final angular velocity	$ω$	rad/s	—	RotationalKinematics
mass	$M$	kg	0 … ∞	RotationalDynamics
radius	$R$	m	0 … ∞	RotationalDynamics
moment of inertia	$I$	kg m²	—	RotationalDynamics
pull	$F$	N	—	RotationalDynamics
torque	$τ$	J	—	RotationalDynamics
cord unwound	$l$	m	—	RotationalDynamics
rotational kinetic energy	$K_{rot}$	J	—	RotationalDynamics
angular momentum	$L$	J s	—	RotationalDynamics
displacement of flywheel during turning	$Δ x$	m	—
position of flywheel at end of turning	$x_{1}$	m	—
position of flywheel at start of turning	$x_{0}$	m	—
average velocity of flywheel during turning	$v_{av}$	m/s	—
time taken of turning	$Δ t$	s	0 … 1e+05
average speed of flywheel during turning	$c_{av}$	m/s	—
path length of flywheel during turning	$d$	m	—
angular velocity of flywheel at end of turning	$ω_{1}$	rad/s	—
angular velocity of flywheel at start of turning	$ω_{0}$	rad/s	—
angular acceleration of flywheel during turning	$α_{r}$	rad/s²	—
angle turned of flywheel during turning	$θ$	—	—
turns of flywheel during turning	$n_{rev}$	—	—
moment of inertia of flywheel	$I$	kg m²	—
mass of flywheel	$M$	kg	0 … ∞
radius of flywheel	$R$	m	0 … ∞
torque on flywheel during turning	$τ$	J	—
cord unwound of flywheel during turning	$l$	m	—
pull on flywheel by ground during turning	$F_{g}$	N	—
pull on flywheel by cord during turning	$F_{c}$	N	—
rotational kinetic energy of flywheel at start of turning	$K_{rot,0}$	J	—
rotational kinetic energy of flywheel at end of turning	$K_{rot,1}$	J	—
angular momentum of flywheel at start of turning	$L_{0}$	J s	—
angular momentum of flywheel at end of turning	$L_{1}$	J s	—
mass of cord	$M_{c}$	kg	0 … ∞
radius of cord	$R_{c}$	m	0 … ∞
moment of inertia of cord	$I_{c}$	kg m²	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

ω_{0} \equiv ω_{i} (start of motion)

starting angular velocity

ω \equiv ω_{i} (end of motion)

final angular velocity

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

Δ x = x_{1} - x_{0}

change of position(turning)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(turning)

c_{av} = \frac{d}{Δ t}

path a time(turning)

ω_{1} = ω_{0} + α_{r} \cdot Δ t

angular velocity after a time(turning)

θ = ω_{0} \cdot Δ t + \frac{α_{r} \cdot {Δ t}^{2}}{2}

angle after a time(turning)

{ω_{1}}^{2} = {ω_{0}}^{2} + 2 \cdot α_{r} \cdot θ

angular velocity after an angle(turning)

n_{rev} = \frac{θ}{2 \cdot π}

turns in an angle(turning)

I = \frac{M \cdot R^{2}}{2}

a disc s moment of inertia(turning)

τ = I \cdot α_{r}

second law of rotation(turning)

l = R \cdot θ

cord unwound by the turn(turning)

τ = F_{g} \cdot R

torque of a pull at the rim(flywheel, ground, turning)

τ = F_{c} \cdot R

torque of a pull at the rim(flywheel, cord, turning)

K_{rot,0} = \frac{I \cdot {ω_{0}}^{2}}{2}

kinetic energy of turning(flywheel, turning, 1)

K_{rot,1} = \frac{I \cdot {ω_{1}}^{2}}{2}

kinetic energy of turning(flywheel, turning, 2)

L_{0} = I \cdot ω_{0}

angular momentum of turning(flywheel, turning, 1)

L_{1} = I \cdot ω_{1}

angular momentum of turning(flywheel, turning, 2)