A Tumbling Oxygen Molecule

Q. The oxygen molecule has a mass of 5.30 × 10⁻²⁶ kg and a moment of inertia of 1.94 × 10⁻⁴⁶ kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule, in rad/s? expects 6.75 × 10¹² rad/s

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Q. The oxygen molecule has a mass of 5.30 × 10⁻²⁶ kg $M = 5.30 \times 10^{−26}$ kg and a moment of inertia of 1.94 × 10⁻⁴⁶ kg m² $I = 1.94 \times 10^{−46}$ kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s $c_{av} = 500$ m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation $K_{rot,0} = 0.666666667 \cdot \frac{M \cdot {c_{av}}^{2}}{2}$ . Find the average angular velocity of the molecule, in rad/s?

The playground

Inputs

mass

M

5.3e-26 kg

inertia

I

1.94e-46 kg m²

speed

c_{av}

500.0 m/s

Outputs

angular velocity

ω_{0}

The solution

Solved with RotationalDynamics

Given

$M = 5.30 \times 10^{−26}$ kg
$I = 1.94 \times 10^{−46}$ kg m²
$c_{av} = 500$ m/s
$K_{rot,0} = 0.666666667 \cdot \frac{M \cdot {c_{av}}^{2}}{2}$ from “two thirds of its kinetic energy of translation”

Find $ω_{0}$ .

Working

two thirds of its kinetic energy of translation $K_{rot,0} = 0.666666667 \cdot \frac{M \cdot {c_{av}}^{2}}{2}$
$K_{rot,0} = 0.666666667 \cdot \frac{5.3 \times 10^{−26} \cdot 500^{2}}{2}$
$K_{rot,0} = 4.41666667 \times 10^{−21}$ J
kinetic energy of turning(molecule, tumbling, 1) $K_{rot,0} = \frac{I \cdot {ω_{0}}^{2}}{2}$
$4.41666667 \times 10^{−21} = \frac{1.94 \times 10^{−46} \cdot {ω_{0}}^{2}}{2}$
$ω_{0} = −6.74778823 \times 10^{12}, 6.74778823 \times 10^{12}$ rad/sthe positive one is taken
$ω_{0} = 6.748 \times 10^{12}$ rad/s
$ω_{0} \approx 6.75 \times 10^{12}$ rad/sto 3 significant figures

Conclusion

The molecule is light but its atoms sit very close to the axis, so the share of energy that goes into turning makes it spin enormously fast.

The Ruby

1require_relative "../physics"23question :a_tumbling_oxygen_molecule do4  kind :numerical5  describes "The oxygen molecule has a mass of %{mass} and a moment of inertia of %{inertia} about an axis through " \6            "its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule " \7            "in a gas is %{speed} and that its kinetic energy of rotation is two thirds of its kinetic energy of " \8            "translation. %{angular_velocity}, in rad/s?"9  given mass: "5.30e-26 kg"10  given inertia: "1.94e-46 kg m2"11  given speed: "500 m/s"12  asks angular_velocity: "Find the average angular velocity of the molecule"13end

1require_relative "../questions/a_tumbling_oxygen_molecule"2require_relative "../science/systems_of_particles_and_rotational_motion/rotational_dynamics"34solve :a_tumbling_oxygen_molecule, using: RotationalDynamics do5  objects :molecule6  molecule.goes(:tumbling)78  take mass: molecule.mass9  take inertia: molecule.moment_of_inertia10  take speed: molecule.average_speed1112  deduce_from "two thirds of its kinetic energy of translation",13              molecule.rotational_kinetic_energy(at: tumbling.start) =>14                2.0 / 3 * (molecule.mass * molecule.average_speed**2 / 2)1516  angular_velocity = molecule.initial_angular_velocity17  answer angular_velocity:, in: "rad/s", figures: 3, root: :positive1819  conclusion "The molecule is light but its atoms sit very close to the axis, so the share of energy that goes " \20             "into turning makes it spin enormously fast."21end

1require_relative "../solutions/a_tumbling_oxygen_molecule"23play_with :a_tumbling_oxygen_molecule do4  title "A Tumbling Oxygen Molecule"5end

The laws

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 molecule during tumbling	$Δ x$	m	—
position of molecule at end of tumbling	$x_{1}$	m	—
position of molecule at start of tumbling	$x_{0}$	m	—
average velocity of molecule during tumbling	$v_{av}$	m/s	—
time taken of tumbling	$Δ t$	s	0 … 1e+05
average speed of molecule during tumbling	$c_{av}$	m/s	—
path length of molecule during tumbling	$d$	m	—
angular velocity of molecule at end of tumbling	$ω_{1}$	rad/s	—
angular velocity of molecule at start of tumbling	$ω_{0}$	rad/s	—
angular acceleration of molecule during tumbling	$α_{r}$	rad/s²	—
angle turned of molecule during tumbling	$θ$	—	—
turns of molecule during tumbling	$n_{rev}$	—	—
moment of inertia of molecule	$I$	kg m²	—
mass of molecule	$M$	kg	0 … ∞
radius of molecule	$R$	m	0 … ∞
torque on molecule during tumbling	$τ$	J	—
cord unwound of molecule during tumbling	$l$	m	—
pull on molecule by ground during tumbling	$F_{g}$	N	—
rotational kinetic energy of molecule at start of tumbling	$K_{rot,0}$	J	—
rotational kinetic energy of molecule at end of tumbling	$K_{rot,1}$	J	—
angular momentum of molecule at start of tumbling	$L_{0}$	J s	—
angular momentum of molecule at end of tumbling	$L_{1}$	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(tumbling)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(tumbling)

c_{av} = \frac{d}{Δ t}

path a time(tumbling)

ω_{1} = ω_{0} + α_{r} \cdot Δ t

angular velocity after a time(tumbling)

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

angle after a time(tumbling)

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

angular velocity after an angle(tumbling)

n_{rev} = \frac{θ}{2 \cdot π}

turns in an angle(tumbling)

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

a disc s moment of inertia(tumbling)

τ = I \cdot α_{r}

second law of rotation(tumbling)

l = R \cdot θ

cord unwound by the turn(tumbling)

τ = F_{g} \cdot R

torque of a pull at the rim(molecule, ground, tumbling)

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

kinetic energy of turning(molecule, tumbling, 1)

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

kinetic energy of turning(molecule, tumbling, 2)

L_{0} = I \cdot ω_{0}

angular momentum of turning(molecule, tumbling, 1)

L_{1} = I \cdot ω_{1}

angular momentum of turning(molecule, tumbling, 2)

K_{rot,0} = 0.666666667 \cdot \frac{M \cdot {c_{av}}^{2}}{2}

two thirds of its kinetic energy of translation