Classical-Mechanics · Unit 25 · Video 7 · Interactive Practice

Same Spin About Any Point: A Puck Strikes a Stick on Frictionless Ice

IKey Formulas

Formula	Name	What you need
$d_{\text{cm}} = \dfrac{m(l/2)}{m + m} = \dfrac{l}{4}$	Centre of mass after	Equal masses; measured from the stick's midpoint toward the struck end
$m v_i = (2m)\,v_{\text{cm},f} \;\Rightarrow\; v_{\text{cm},f} = \dfrac{v_i}{2}$	Linear momentum	No net external force in the plane of the ice
$\dfrac{l}{4}\,m v_i = \Big(I_{\text{cm}} + \dfrac{ml^2}{8}\Big)\,\omega_{\text{cm},f} \;\Rightarrow\; \omega_{\text{cm},f} = \dfrac{6}{5}\,\dfrac{v_i}{l}$	Angular momentum about $S$	$S$ fixed at the system cm; $I_{\text{cm}} = \tfrac{1}{12}ml^2$ for a uniform stick
$x_{\text{cm}} = 2\pi\,\dfrac{v_{\text{cm},f}}{\omega_{\text{cm},f}} = 2\pi\,\dfrac{I_{\text{cm}} + ml^2/8}{m(l/2)} = \dfrac{5}{6}\pi l$	Distance per turn	Period $T = 2\pi/\omega_{\text{cm},f}$ ; independent of $v_i$

Key Insight: Momentum fixes the slide and angular momentum fixes the spin, and the spin comes out the same about every fixed point. Kinetic energy is not conserved: $K_f = \tfrac{1}{4}mv_i^2 + \tfrac{3}{20}mv_i^2 = \tfrac{4}{5}K_i$ , so one fifth goes into the putty.

IIVisualization 1 — Same Spin About Any Point

Angular momentum is conserved about every fixed point $P$ on the ice, and every choice of $P$ returns $\omega_{\text{cm},f} = 6.00\ \text{rad/s}$ .

💡 Only the height of $P$ matters: both momenta are horizontal, so sliding $P$ sideways changes no moment arm. At $S$ the orbital term is zero; at $C$ the initial angular momentum is zero.

IIIVisualization 2 — Slide and Spin

After the collision the cm slides at $v_i/2$ while the pair spins at $\tfrac{6}{5}v_i/l$ , covering $\tfrac{5}{6}\pi l$ per turn.

Turns after impact −0.20

IVVisualization 3 — Distance per Turn

The distance per turn depends on how the stick's mass is spread, through $I_{\text{cm}}$ , and never on $v_i$ .

Puck speed v_i (m/s) 5.0

💡 A faster puck makes the pair slide faster and spin faster in the same proportion, so $v_i$ cancels from $x_{\text{cm}} = 2\pi v_{\text{cm},f}/\omega_{\text{cm},f}$ .

VQuiz Questions

Problem 1 · Unequal Masses

Given: A uniform stick of mass $M = 0.60\ \text{kg}$ and length $l = 1.20\ \text{m}$ rests on frictionless ice. A puck of mass $m = 0.20\ \text{kg}$ slides at $v_i = 8.0\ \text{m/s}$ perpendicular to the stick, hits one end and sticks. Find the system's centre of mass and its velocity after the collision.

Distance of the system cm from the stick's midpoint

Speed of the system cm after the collision

✅ Correct!

d_{\text{cm}} = 0.15\ \text{m}

toward the struck end and

v_{\text{cm},f} = 2.0\ \text{m/s}

; both weight the puck by

m/(M+m) = \tfrac{1}{4}

❌ Not quite.

l/4

holds only for equal masses. Weight the puck's distance

l/2

m/(M+m)

❌ Not quite. You weighted the distance

l/2

by the stick's mass

M

. The stick's own centre is at the midpoint (distance

0

); only the puck, of mass

m

, sits at

l/2

❌ Not quite. Use

d_{\text{cm}} = \dfrac{M\cdot 0 + m(l/2)}{M + m}

, with the total mass in the denominator.

❌ Not quite.

v_i/2

holds only for equal masses. The total mass after is

M + m = 0.80\ \text{kg}

❌ Not quite. Conserve momentum:

m v_i = (M + m)\,v_{\text{cm},f}

, with both bodies moving together after.

Show solution

Measure from the stick's midpoint. The stick's mass sits at $0$ , the puck at $l/2 = 0.60\ \text{m}$ :

d_{\text{cm}} = \frac{M\cdot 0 + m(l/2)}{M + m} = \frac{(0.20)(0.60)}{0.80} = 0.15\ \text{m}

No net external force acts in the plane, so total momentum is conserved:

m v_i = (M + m)\,v_{\text{cm},f} \;\Rightarrow\; v_{\text{cm},f} = \frac{(0.20)(8.0)}{0.80} = 2.0\ \text{m/s}

Dividing by $M$ alone gives $2.7\ \text{m/s}$ and leaves the puck out of the moving mass.

Problem 2 · The Stick's Own Centre

Given: The video's collision: equal masses $m$ , a uniform stick of length $l = 1.00\ \text{m}$ , puck speed $v_i = 5.0\ \text{m/s}$ , so $v_{\text{cm},f} = 2.5\ \text{m/s}$ and $\omega_{\text{cm},f} = 6.0\ \text{rad/s}$ . Find the speed of the stick's own midpoint just after the collision.

✅ Correct!

v_{\text{stick}} = v_i/2 - \omega_{\text{cm},f}(l/4) = 2.5 - 1.5 = 1.0\ \text{m/s} = v_i/5

❌ Close, but…

2.5\ \text{m/s}

is the velocity of the system's centre of mass. The stick's midpoint sits

l/4

away from it and also turns about it.

❌ Check the direction. The midpoint is on the far side from the puck, so the counterclockwise spin carries it backward: subtract

\omega_{\text{cm},f}(l/4)

. Adding gives the puck's speed.

❌ Not quite. The midpoint shares the forward

v_{\text{cm},f}

and gets a backward

\omega_{\text{cm},f}(l/4)

from the rotation.

Show solution

The system cm is $l/4$ below the stick's midpoint, so the midpoint is $l/4 = 0.25\ \text{m}$ above it, on the side away from the puck. Its velocity is the cm's velocity plus its turning velocity about the cm:

v_{\text{stick}} = v_{\text{cm},f} - \omega_{\text{cm},f}\,\frac{l}{4} = 2.5 - (6.0)(0.25) = 1.0\ \text{m/s}

In general, $\omega_{\text{cm},f}\tfrac{l}{4} = \tfrac{6}{5}\cdot\tfrac{1}{4}v_i = \tfrac{3}{10}v_i$ , so $v_{\text{stick}} = \tfrac{v_i}{2} - \tfrac{3}{10}v_i = \tfrac{v_i}{5}$ . The puck end, below the cm, moves at $2.5 + 1.5 = 4.0\ \text{m/s}$ .

Problem 3 · Spin and Energy Lost

Given: A uniform stick and a puck, each of mass $m = 0.30\ \text{kg}$ ; stick length $l = 0.80\ \text{m}$ ; the puck hits one end at $v_i = 6.0\ \text{m/s}$ , perpendicular to the stick, and sticks. Find $\omega_{\text{cm},f}$ and the kinetic energy lost to the putty.

Angular velocity after the collision

Kinetic energy lost

✅ Correct!

\omega_{\text{cm},f} = \tfrac{6}{5}(6.0/0.80) = 9.0\ \text{rad/s}

, and one fifth of

K_i = 5.40\ \text{J}

is lost:

1.08\ \text{J}

❌ Not quite. You kept the stick's shifted axis

m(l/4)^2

but dropped the puck's own

m(l/4)^2

about

S

I_{\text{cm},f}

has both.

❌ Not quite.

I_{\text{cm}}

is about the stick's midpoint, but the pair turns about

S

. Add

m(l/4)^2

for the stick (parallel axis) and

m(l/4)^2

for the puck.

❌ Not quite. Solve

(l/4)\,m v_i = \big(I_{\text{cm}} + ml^2/8\big)\,\omega_{\text{cm},f}

with

I_{\text{cm}} = ml^2/12

❌ Close, but… That counts only the translational

\tfrac{1}{4}mv_i^2

after. The spin keeps

\tfrac{1}{2}I_{\text{cm},f}\,\omega_{\text{cm},f}^2

as well.

❌ Not quite. The collision is totally inelastic, so some energy is lost; find

K_f

from translation plus spin and subtract it from

K_i

Show solution

Spin. About $S$ , $I_{\text{cm},f} = \tfrac{ml^2}{12} + \tfrac{ml^2}{8} = \tfrac{5}{24}ml^2$ , so

\omega_{\text{cm},f} = \frac{m(l/4)\,v_i}{\tfrac{5}{24}ml^2} = \frac{6}{5}\,\frac{v_i}{l} = \frac{6}{5}\cdot\frac{6.0}{0.80} = 9.0\ \text{rad/s}

Energy. $K_i = \tfrac{1}{2}(0.30)(6.0)^2 = 5.40\ \text{J}$ . After, $v_{\text{cm},f} = 3.0\ \text{m/s}$ and $I_{\text{cm},f} = \tfrac{5}{24}(0.30)(0.80)^2 = 0.040\ \text{kg}\cdot\text{m}^2$ :

K_f = \tfrac{1}{2}(0.60)(3.0)^2 + \tfrac{1}{2}(0.040)(9.0)^2 = 2.70 + 1.62 = 4.32\ \text{J} = \tfrac{4}{5}K_i

Lost to the putty: $5.40 - 4.32 = 1.08\ \text{J} = \tfrac{1}{5}K_i$ .

Problem 4 · A Different Stick

Given: The stick is replaced by a light rod of length $l$ carrying a small mass $m/2$ at each end, so its total mass is still $m$ and $I_{\text{cm}} = \tfrac{1}{4}ml^2$ . A puck of mass $m$ hits one end at $v_i$ , perpendicular to the rod, and sticks. Find how far the system's centre of mass travels while the rod makes one full turn.

✅ Correct!

x_{\text{cm}} = 2\pi\big(\tfrac{1}{4} + \tfrac{1}{8}\big)ml^2/(ml/2) = \tfrac{3}{2}\pi l \approx 4.7\,l

: mass farther from the centre spins slower, so the pair slides farther per turn.

❌ Not quite.

\tfrac{5}{6}\pi l

uses the uniform stick's

I_{\text{cm}} = ml^2/12

. Put

ml^2/4

into the general formula.

❌ Not quite. You kept only the

ml^2/8

term; the rod's own

I_{\text{cm}} = ml^2/4

belongs in the numerator too.

❌ Not quite. Both

v_{\text{cm},f}

and

\omega_{\text{cm},f}

are proportional to

v_i

, so

v_i

cancels from

x_{\text{cm}} = 2\pi v_{\text{cm},f}/\omega_{\text{cm},f}

Show solution

The rod's cm is still its midpoint, so the system cm is still $l/4$ from it and $v_{\text{cm},f} = v_i/2$ . About $S$ :

I_{\text{cm},f} = I_{\text{cm}} + \frac{ml^2}{8} = \frac{ml^2}{4} + \frac{ml^2}{8} = \frac{3}{8}ml^2, \qquad \omega_{\text{cm},f} = \frac{m(l/4)\,v_i}{\tfrac{3}{8}ml^2} = \frac{2}{3}\,\frac{v_i}{l}

One turn takes $T = 2\pi/\omega_{\text{cm},f}$ , so

x_{\text{cm}} = v_{\text{cm},f}\,T = 2\pi\,\frac{v_i/2}{\tfrac{2}{3}v_i/l} = \frac{3}{2}\pi l \approx 4.71\,l

Solved: 0 / 4