Classical-Mechanics · Unit 23 · Video 7 · Interactive Practice

Angular Impulse and the Dropped Washer: Only the Bearing Friction Removes Angular Momentum

IKey Formulas

Formula	Name	What you need
$\vec{J}_S^{\,\text{ext}} \equiv \displaystyle\int_{t_i}^{t_f} \vec{\tau}_S^{\,\text{ext}}\,dt = \vec{L}_{S,f}^{\,\text{sys}} - \vec{L}_{S,i}^{\,\text{sys}}$	Angular impulse for a system	External torques only — internal torques cancel in pairs
$J_z = -\tau_f\,\Delta t, \quad \tau_f = -I_0\,\alpha_1 = I_0\,\dfrac{\omega_0 - \omega_a}{\Delta t_1}$	Angular impulse of constant bearing friction	The coasting deceleration and the time it acts
$\omega_b = \dfrac{I_0}{I_0 + I_w}\,\omega_a$	Common speed, instantaneous collision	$J_z \to 0$ as $\Delta t_{\text{int}} \to 0$
$\omega_b = \dfrac{I_0}{I_0 + I_w}\left[(\omega_a - \omega_0)\dfrac{\Delta t_{\text{int}}}{\Delta t_1} + \omega_a\right]$	Common speed, finite collision	$(I_0 + I_w)\,\omega_b - I_0\,\omega_a = J_z$

Key Insight: Take the system to be the rotor and both washers: the sliding friction between the washers is internal and drops out, so only the bearing friction changes $L_z$ — at the same steady rate $\tau_f$ before, during and after the drop.

IIVisualization 1 — Angular Impulse Is an Area

The bearing friction's angular impulse during the collision is the rectangle $-\tau_f\,\Delta t_{\text{int}}$ , which vanishes as $\Delta t_{\text{int}} \to 0$ .

Collision time Δt_int (s) 1.00

IIIVisualization 2 — One Straight Line for $L_z$

The angular speeds turn a corner at the drop, but the system's $L_z$ falls along one straight line to zero at $t_f = I_0\,\omega_0/\tau_f$ .

Dropped washer I_w (×10⁻³ kg·m²) 1.0 Collision time Δt_int (s) 1.0 Time t (s) 2.4

IVVisualization 3 — Where the Energy Equation Could Not Go

An instantaneous drop keeps $L_z$ but turns the fraction $I_w/(I_0 + I_w)$ of the kinetic energy into heat.

Dropped washer I_w (×10⁻³ kg·m²) 1.0

💡 With a finite collision the bearing friction also does negative work while the washers slide, so the heat between the washers is no longer the whole energy drop — but $L_z$ , through the angular impulse, still fixes $\omega_b$ first.

VQuiz Questions

Problem 1 · Instantaneous Drop

Given: A motor rotor plus washer, $I_0 = 6.0\times10^{-4}\ \text{kg}\cdot\text{m}^2$ , coasts at $\omega_a = 50\ \text{rad/s}$ when a non-spinning washer with $I_w = 2.0\times10^{-4}\ \text{kg}\cdot\text{m}^2$ is dropped onto it. Treat the collision as instantaneous — find the common angular speed $\omega_b$ .

✅ Correct!

L_z = I_0\,\omega_a

is shared by the larger moment of inertia

I_0 + I_w

, so

\omega_b = \tfrac{6}{8}\cdot 50 = 37.5\ \text{rad/s}

❌ The prefactor is $I_0/(I_0 + I_w)$ , not

I_w/(I_0 + I_w)

: the angular momentum was carried in by

I_0

, and the dropped washer brought none.

❌ That conserves kinetic energy. Sliding friction between the washers turns an unknown amount of kinetic energy into heat; only

L_z

is unchanged through an instantaneous collision.

❌ The ratio is upside down. The same

L_z

with a larger moment of inertia must give a smaller angular speed.

❌ Not quite. Set

(L_z)_{\text{after}} = (L_z)_{\text{before}}

(I_0 + I_w)\,\omega_b = I_0\,\omega_a

Show solution

System: rotor and both washers. The washer–washer friction is internal, and the bearing friction has no time to deliver an angular impulse, so $\Delta L_z = 0$ :

(I_0 + I_w)\,\omega_b = I_0\,\omega_a + I_w\cdot 0

\omega_b = \frac{I_0}{I_0 + I_w}\,\omega_a = \frac{6.0\times10^{-4}}{8.0\times10^{-4}}\,(50\ \text{rad/s}) = 37.5\ \text{rad/s}

Problem 2 · Which Torque Counts?

Given: During a drop lasting $\Delta t_{\text{int}} = 0.10\ \text{s}$ , the lower washer exerts a friction torque of magnitude $0.050\ \text{N}\cdot\text{m}$ on the dropped washer, while the bearing exerts a constant friction torque of magnitude $\tau_f = 0.0020\ \text{N}\cdot\text{m}$ on the rotor, opposing the rotation ( $+z$ along $\vec{\omega}$ ). Find the angular impulse $J_z$ on the system of rotor plus both washers during the collision.

✅ Correct! Only the bearing friction is external:

J_z = -\tau_f\,\Delta t_{\text{int}} = -(0.0020)(0.10) = -2.0\times10^{-4}\ \text{N}\cdot\text{m}\cdot\text{s}

❌ That holds only in the instantaneous limit. Over a finite

0.10\ \text{s}

the bearing friction's rectangle has finite width, so it removes some

L_z

❌ The washer–washer torque is internal. The pair of friction forces acts at the same contact point, so their torques on the two washers cancel — however large they are.

❌ Not quite. Keep only the external torque along the axis, with its sign.

Show solution

The friction between the washers is internal to the chosen system: it speeds up the dropped washer by $+(0.050)(0.10) = 5.0\times10^{-3}\ \text{N}\cdot\text{m}\cdot\text{s}$ and slows the lower one by exactly the same amount, so the net is zero.

The only external axial torque is the bearing friction, $-\tau_f\hat{k}$ , constant during the collision:

J_z = \int \tau^{\text{ext}}_{z}\,dt = -\tau_f\,\Delta t_{\text{int}} = -(0.0020\ \text{N}\cdot\text{m})(0.10\ \text{s}) = -2.0\times10^{-4}\ \text{N}\cdot\text{m}\cdot\text{s}

Negative: the angular impulse points along $-z$ , against the rotation.

Problem 3 · Finite Collision (Part D)

Given: $I_0 = 2.0\times10^{-3}\ \text{kg}\cdot\text{m}^2$ . With the power off, the rotor slows from $\omega_0 = 30\ \text{rad/s}$ to $\omega_a = 25\ \text{rad/s}$ in $\Delta t_1 = 5.0\ \text{s}$ . A washer with $I_w = 1.0\times10^{-3}\ \text{kg}\cdot\text{m}^2$ , not spinning, is dropped on; the washers slide for $\Delta t_{\text{int}} = 0.60\ \text{s}$ . The bearing torque is constant throughout — find $\omega_b$ .

✅ Correct! The bearing removes

0.60\ \text{s}

worth of angular momentum during the slide, so

\omega_b

is a little below the instantaneous

16.7\ \text{rad/s}

❌ That is the instantaneous-collision answer. Over

0.60\ \text{s}

the bearing friction delivers

J_z = -\tau_f\,\Delta t_{\text{int}} \neq 0

❌ Check the sign of $J_z$ . Friction opposes the rotation, so

(\omega_a - \omega_0) < 0

and the correction lowers

\omega_b

❌ The prefactor is $I_0/(I_0 + I_w)$ , the share of the new moment of inertia that carried the angular momentum in — not

I_w/(I_0 + I_w)

❌ Not quite. Use

(I_0 + I_w)\,\omega_b - I_0\,\omega_a = J_z

with

J_z = -\tau_f\,\Delta t_{\text{int}}

Show solution

Bearing torque from the coast:

\alpha_1 = \frac{\omega_a - \omega_0}{\Delta t_1} = \frac{25 - 30}{5.0} = -1.0\ \text{rad/s}^2, \qquad \tau_f = -I_0\,\alpha_1 = 2.0\times10^{-3}\ \text{N}\cdot\text{m}

Angular impulse during the slide:

J_z = -\tau_f\,\Delta t_{\text{int}} = -(2.0\times10^{-3})(0.60) = -1.2\times10^{-3}\ \text{N}\cdot\text{m}\cdot\text{s}

Angular momentum of rotor plus both washers:

(I_0 + I_w)\,\omega_b = I_0\,\omega_a + J_z = (2.0\times10^{-3})(25) - 1.2\times10^{-3} = 4.88\times10^{-2}\ \text{kg}\cdot\text{m}^2/\text{s}

\omega_b = \frac{4.88\times10^{-2}}{3.0\times10^{-3}} = 16.3\ \text{rad/s}

Equivalently, $\omega_b = \tfrac{2}{3}\left[(25 - 30)\tfrac{0.60}{5.0} + 25\right] = \tfrac{2}{3}(24.4) = 16.3\ \text{rad/s}$ .

Problem 4 · When Do the Washers Stop?

Given: The same motor and washers as Problem 3. Measuring time from the moment the power is cut, find when the washers stop, and their angular acceleration $\alpha_2$ after the collision.

When do the washers stop?

What is $\alpha_2$ ?

✅ Correct!

L_z

falls at the steady rate

\tau_f

from

I_0\,\omega_0

, so the washers stop at

t_f = I_0\,\omega_0/\tau_f = 30\ \text{s}

— exactly when the rotor alone would have — while the larger moment of inertia makes

|\alpha_2| = \tau_f/(I_0 + I_w)

smaller than

|\alpha_1|

❌ Check the stopping time. The bearing friction is the only external torque in every stage, so

-\tau_f\,t_f = 0 - I_0\,\omega_0

— the dropped washer carried in no angular momentum.

❌ That is $\Delta t_2$ , measured from the end of the collision at

5.6\ \text{s}

; the question measures from the moment the power is cut.

❌ That uses $(I_0 + I_w)\,\omega_0$ as the starting angular momentum, but the dropped washer carried in none:

t_f = I_0\,\omega_0/\tau_f

❌ That keeps $\alpha_1 = -1.0\ \text{rad/s}^2$ after the collision. The same

\tau_f

now acts on

I_0 + I_w

, so the washers slow more gently and stop later.

❌ Check $\alpha_2$ . The same torque

\tau_f

now acts on

I_0 + I_w

, and the washers are slowing, so

\alpha_2 < 0

Show solution

Stopping time. From Problem 3, $\tau_f = 2.0\times10^{-3}\ \text{N}\cdot\text{m}$ . The total angular impulse from power-off to rest is $-\tau_f\,t_f$ , and it equals $0 - I_0\,\omega_0$ :

t_f = \frac{I_0\,\omega_0}{\tau_f} = \frac{(2.0\times10^{-3})(30)}{2.0\times10^{-3}} = 30\ \text{s}

Deceleration. After the collision, $-\tau_f = (I_0 + I_w)\,\alpha_2$ :

\alpha_2 = -\frac{2.0\times10^{-3}}{3.0\times10^{-3}} = -0.67\ \text{rad/s}^2

Check with part E. The collision ends at $5.0 + 0.60 = 5.6\ \text{s}$ , so $\Delta t_2 = 30 - 5.6 = 24.4\ \text{s}$ and $\alpha_2 = (0 - \omega_b)/\Delta t_2 = -16.27/24.4 = -0.67\ \text{rad/s}^2$ ✓.

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Angular Impulse and the Dropped Washer: Only the Bearing Friction Removes Angular Momentum

IKey Formulas

IIVisualization 1 — Angular Impulse Is an Area

IIIVisualization 2 — One Straight Line for LzL_zLz​

IVVisualization 3 — Where the Energy Equation Could Not Go

VQuiz Questions

IIIVisualization 2 — One Straight Line for $L_z$