A Spinning Solid Cylinder's Angular Momentum

solved RotationalDynamics

Q. A solid cylinder of mass 20 kg $M = 20$ kg rotates about its axis with angular speed 100 rad/s $ω_{0} = 100$ rad/s. The radius of the cylinder is 0.25 m $R = 0.25$ m. What is the magnitude of angular momentum of the cylinder about its axis, in J s?

Inputs

mass

M

20 kg

speed

ω_{0}

100 rad/s

radius

R

0.25 m

Outputs

momentum

L_{0}

How it's solved

Solved with RotationalDynamics

Given

$M = 20$ kg
$ω_{0} = 100$ rad/s
$R = 0.25$ m

Find $L_{0}$ .

Working

a disc s moment of inertia(spinning) $I = \frac{M \cdot R^{2}}{2}$
$I = \frac{20 \cdot {0.25}^{2}}{2}$
$I = 0.625$ kg m²
angular momentum of turning(cylinder, spinning, 1) $L_{0} = I \cdot ω_{0}$
$L_{0} = 0.625 \cdot 100$
$L_{0} = 62.5$ J s
$L_{0} \approx 62.5$ J sto 3 significant figures

Conclusion

Angular momentum is how hard the cylinder resists turning times how fast it turns, the turning counterpart of mass times velocity.

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_spinning_solid_cylinder_s_angular_momentum do4  kind :numerical5  describes "A solid cylinder of mass %{mass} rotates about its axis with angular speed %{speed}. The radius of the " \6            "cylinder is %{radius}. %{momentum}, in J s?"7  given mass: "20 kg"8  given speed: "100 rad/s"9  given radius: "0.25 m"10  asks momentum: "What is the magnitude of angular momentum of the cylinder about its axis"11end

solution.rb

1require_relative "../questions/a_spinning_solid_cylinder_s_angular_momentum"2require_relative "../science/systems_of_particles_and_rotational_motion/rotational_dynamics"34solve :a_spinning_solid_cylinder_s_angular_momentum, using: RotationalDynamics do5  objects :cylinder6  cylinder.goes(:spinning)78  take mass: cylinder.mass9  take speed: cylinder.initial_angular_velocity10  take radius: cylinder.radius1112  momentum = cylinder.angular_momentum(at: spinning.start)13  answer momentum:, in: "J s", figures: 31415  conclusion "Angular momentum is how hard the cylinder resists turning times how fast it turns, the turning " \16             "counterpart of mass times velocity."17end

playground.rb

1require_relative "../solutions/a_spinning_solid_cylinder_s_angular_momentum"23play_with :a_spinning_solid_cylinder_s_angular_momentum do4  title "A Spinning Solid Cylinder's Angular Momentum"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 cylinder during spinning	$Δ x$	m	—
position of cylinder at end of spinning	$x_{1}$	m	—
position of cylinder at start of spinning	$x_{0}$	m	—
average velocity of cylinder during spinning	$v_{av}$	m/s	—
time taken of spinning	$Δ t$	s	0 … 1e+05
average speed of cylinder during spinning	$c_{av}$	m/s	—
path length of cylinder during spinning	$d$	m	—
angular velocity of cylinder at end of spinning	$ω_{1}$	rad/s	—
angular velocity of cylinder at start of spinning	$ω_{0}$	rad/s	—
angular acceleration of cylinder during spinning	$α_{r}$	rad/s²	—
angle turned of cylinder during spinning	$θ$	—	—
turns of cylinder during spinning	$n_{rev}$	—	—
moment of inertia of cylinder	$I$	kg m²	—
mass of cylinder	$M$	kg	0 … ∞
radius of cylinder	$R$	m	0 … ∞
torque on cylinder during spinning	$τ$	J	—
cord unwound of cylinder during spinning	$l$	m	—
pull on cylinder by ground during spinning	$F_{g}$	N	—
rotational kinetic energy of cylinder at start of spinning	$K_{rot,0}$	J	—
rotational kinetic energy of cylinder at end of spinning	$K_{rot,1}$	J	—
angular momentum of cylinder at start of spinning	$L_{0}$	J s	—
angular momentum of cylinder at end of spinning	$L_{1}$	J s	—

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(spinning)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(spinning)

c_{av} = \frac{d}{Δ t}

path a time(spinning)

ω_{1} = ω_{0} + α_{r} \cdot Δ t

angular velocity after a time(spinning)

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

angle after a time(spinning)

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

angular velocity after an angle(spinning)

n_{rev} = \frac{θ}{2 \cdot π}

turns in an angle(spinning)

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

a disc s moment of inertia(spinning)

τ = I \cdot α_{r}

second law of rotation(spinning)

l = R \cdot θ

cord unwound by the turn(spinning)

τ = F_{g} \cdot R

torque of a pull at the rim(cylinder, ground, spinning)

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

kinetic energy of turning(cylinder, spinning, 1)

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

kinetic energy of turning(cylinder, spinning, 2)

L_{0} = I \cdot ω_{0}

angular momentum of turning(cylinder, spinning, 1)

L_{1} = I \cdot ω_{1}

angular momentum of turning(cylinder, spinning, 2)