Classical-Mechanics · Unit 23 · Video 6 · Interactive Practice

Zero Torque About the Pivot: Conservation of Angular Momentum and the Struck Rod

IKey Formulas

Formula	Name	What you need
$\vec{\tau}_S^{\,\text{ext}} = \vec{0} \;\Rightarrow\; (\vec{L}_S^{\,\text{sys}})_f = (\vec{L}_S^{\,\text{sys}})_i$	Conservation of angular momentum	A point $S$ about which every external torque vanishes
$l m_1 v_i = l m_1 \tfrac{v_i}{2} + I_S\omega_a \;\Rightarrow\; \omega_a = \dfrac{3m_1 v_i}{2m_2 l}$	The collision, $i \to a$ , about the pivot	$I_S = \tfrac{1}{3}m_2 l^2$ for a rod about its end
$\tfrac{1}{2}I_S\omega_a^2 = m_2 g\,\tfrac{l}{2} \;\Rightarrow\; v_i = \dfrac{m_2}{m_1}\sqrt{\dfrac{4gl}{3}}$	The swing, $a \to f$ , to the ceiling	The centre of mass rises $l/2$ ; the object's kinetic energy cancels
$\tfrac{1}{2}m_1 v_i^2 = \dfrac{3m_1^2 v_i^2}{8m_2} + \tfrac{1}{8}m_1 v_i^2 \;\Rightarrow\; \dfrac{m_2}{m_1} = 1$	Elastic condition	$E_i^{mech} = E_a^{mech}$ through the collision

Key Insight: One constant quantity per stage. Through the blow the pivot changes the momentum, but it has no moment arm about itself, so $\vec{L}_S$ is constant; through the swing gravity changes $\vec{L}_S$ , but the mechanical energy is constant.

IIVisualization 1 — Choosing the Point S

Through the blow, angular momentum is unchanged about the pivot, where the unknown pivot impulse has no moment arm.

💡 Here the pivot impulse turns out to be horizontal, so any point level with $S$ also works — but only $S$ is guaranteed before you know the pivot force.

IIIVisualization 2 — What Survives the Collision

About $S$ the angular momentum is unchanged for every mass ratio; momentum never is, and kinetic energy only at $m_2 = m_1$ .

Mass ratio m₂/m₁ 2.00

💡 Below $m_2/m_1 = 1$ the kinetic energy would have to grow, which no collision can do without a source of energy — a rod that light cannot leave the object moving at $v_i/2$ .

IVVisualization 3 — Strike, Then Swing

Angular momentum fixes $\omega_a$ ; energy then decides how high the rod climbs, and the ceiling needs $v_i = \tfrac{m_2}{m_1}\sqrt{4gl/3}$ .

Speed v_i (m/s) 6.00 Rod mass m₂ Time i

💡 Challenge: for each rod mass, find the speed that brings the rod to rest exactly against the ceiling — doubling $m_2$ should double it.

VQuiz Questions

Problem 1 · The Rod's Angular Speed

Given: An object of mass $m_1 = 0.5\ \text{kg}$ moving at $v_i = 6\ \text{m/s}$ strikes the lower end of a uniform rod ( $m_2 = 1.5\ \text{kg}$ , $l = 0.9\ \text{m}$ ) hanging from a frictionless pivot, and leaves at $v_i/2$ . Find the rod's angular speed $\omega_a$ immediately after the collision.

✅ Correct! The object keeps half its angular momentum about

S

, so the rod carries the other half:

I_S\omega_a = \tfrac{1}{2}l m_1 v_i

❌ Not quite. That gives the rod all of

l m_1 v_i

— but the object leaves at

v_i/2

and keeps

l m_1 v_i/2

of the angular momentum about

S

❌ Not quite. That comes from conserving momentum, which the pivot impulse spoils. Conserve the angular momentum about

S

instead.

❌ Not quite. The rod rotates about the fixed pivot, so use

I_S = \tfrac{1}{3}m_2 l^2

, not

I_{cm} = \tfrac{1}{12}m_2 l^2

❌ Not quite. Set

l m_1 v_i = l m_1 v_i/2 + I_S\omega_a

with

I_S = \tfrac{1}{3}m_2 l^2

Show solution

No external torque acts about the pivot $S$ during the blow, so $\vec{L}_S$ is the same immediately before and immediately after:

l m_1 v_i = l m_1\frac{v_i}{2} + I_S\omega_a \quad\Rightarrow\quad \omega_a = \frac{l m_1 v_i}{2I_S} = \frac{3m_1 v_i}{2m_2 l}

\omega_a = \frac{3(0.5)(6)}{2(1.5)(0.9)} = \frac{9}{2.7} \approx 3.33\ \text{rad/s}

Problem 2 · Where the Momentum Went

Given: The struck rod of the video: the object arrives with momentum $m_1 v_i$ , leaves at $v_i/2$ , and the rod starts turning at $\omega_a = \dfrac{3m_1 v_i}{2m_2 l}$ . Find the total momentum of object and rod immediately after the collision.

✅ Correct! The system gains

\tfrac{1}{4}m_1 v_i

— the pivot's forward impulse on the top of the rod.

❌ That assumes momentum is conserved. The pivot holds the top of the rod against the blow, and its impulse is comparable to the blow's, so the system's momentum changes.

❌ That is only the object's share. The rod's centre of mass is moving too, at

v_{cm} = \omega_a\,l/2

❌ That is only the rod's share. Add the object's

m_1 v_i/2

Show solution

The rod's centre of mass is $l/2$ below the pivot, so it moves at

v_{cm} = \omega_a\frac{l}{2} = \frac{3m_1 v_i}{2m_2 l}\cdot\frac{l}{2} = \frac{3m_1 v_i}{4m_2}, \qquad m_2 v_{cm} = \tfrac{3}{4}m_1 v_i

p_a = m_1\frac{v_i}{2} + m_2 v_{cm} = \tfrac{1}{2}m_1 v_i + \tfrac{3}{4}m_1 v_i = \tfrac{5}{4}m_1 v_i

The change $\tfrac{5}{4}m_1 v_i - m_1 v_i = \tfrac{1}{4}m_1 v_i$ is the pivot's impulse, pointing forward: without it the top of the rod would kick backward. It acts at $S$ , so its torque about $S$ is zero — which is why $\vec{L}_S$ survives while $\vec{p}$ does not. The result holds for every mass ratio.

Problem 3 · Just Reaching the Ceiling

Given: The rod and object of Problem 1: $m_1 = 0.5\ \text{kg}$ , $m_2 = 1.5\ \text{kg}$ , $l = 0.9\ \text{m}$ , $g = 9.8\ \text{m/s}^2$ , and the object leaves at $v_i/2$ . Find the speed $v_i$ for which the rod just touches the ceiling on its first swing.

✅ Correct! Angular momentum through the collision, then energy through the swing:

v_i = \tfrac{m_2}{m_1}\sqrt{4gl/3} = 3\sqrt{11.76}

❌ Not quite. That gives the rod all of the object's angular momentum; the object leaves at

v_i/2

, so the rod gets only half and

\omega_a = \tfrac{3m_1 v_i}{2m_2 l}

❌ Not quite. When the rod lies along the ceiling its centre of mass has risen

l/2

, not

l

❌ Not quite. That is

\sqrt{4gl/3}

alone — the mass ratio

m_2/m_1 = 3

multiplies it.

Show solution

Collision ( $i \to a$ ), angular momentum about $S$ : $\omega_a = \dfrac{3m_1 v_i}{2m_2 l}$ .

Swing ( $a \to f$ ), mechanical energy: the object's kinetic energy is the same on both sides and cancels; the rod's rotational energy becomes the rise $l/2$ of its centre of mass.

\tfrac{1}{2}\left(\tfrac{1}{3}m_2 l^2\right)\left(\frac{3m_1 v_i}{2m_2 l}\right)^2 = \frac{3m_1^2 v_i^2}{8m_2} = m_2 g\frac{l}{2}

v_i = \frac{m_2}{m_1}\sqrt{\frac{4gl}{3}} = \frac{1.5}{0.5}\sqrt{\frac{4(9.8)(0.9)}{3}} = 3\sqrt{11.76} \approx 10.3\ \text{m/s}

Problem 4 · The Object Stops Dead

Given: The same pivoted rod, but now the object is stopped by the collision ( $v_a = 0$ ) instead of leaving at $v_i/2$ . Find the mass ratio $m_2/m_1$ that makes this collision elastic.

✅ Correct! The rod must be three times as heavy — it rotates about its end, so its effective mass at the struck end is

I_S/l^2 = \tfrac{1}{3}m_2

❌ Not quite. Equal masses stop the object in a head-on elastic collision between two free particles, but this rod turns about a pivot. Balance

\vec{L}_S

and the kinetic energy again.

❌ Check which way up the ratio is. Solving

\tfrac{1}{2}m_1 v_i^2 = \tfrac{3m_1^2 v_i^2}{2m_2}

gives

m_2 = 3m_1

❌ Not quite. The rod turns about its end, so its moment of inertia is

I_S = \tfrac{1}{3}m_2 l^2

, not

\tfrac{1}{12}m_2 l^2

❌ Not quite. Find

\omega_a

from

\vec{L}_S

with

v_a = 0

, then set the kinetic energy before equal to the rod's kinetic energy after.

Show solution

Angular momentum about $S$ (the object now keeps none of it):

l m_1 v_i = I_S\omega_a = \tfrac{1}{3}m_2 l^2\omega_a \quad\Rightarrow\quad \omega_a = \frac{3m_1 v_i}{m_2 l}

Elastic: nothing changes height during the blow, so the kinetic energy before equals the rod's kinetic energy after:

\tfrac{1}{2}m_1 v_i^2 = \tfrac{1}{2}\left(\tfrac{1}{3}m_2 l^2\right)\frac{9m_1^2 v_i^2}{m_2^2 l^2} = \frac{3m_1^2 v_i^2}{2m_2}

\Rightarrow\quad m_2 = 3m_1, \qquad \frac{m_2}{m_1} = 3

As in the video, $v_i^2$ cancels, so the ratio does not depend on the initial speed.

Solved: 0 / 4