Classical-Mechanics · Unit 25 · Video 2 · Interactive Practice

One Push, Two Motions: Rigid-Body Equations, Work-Energy, and Angular Impulse

IKey Formulas

Formula	Name	What it says
$\vec{F}^{\,\text{ext}} = m\vec{A}_{\text{cm}}$	Translation	The same $\vec{A}_{\text{cm}}$ wherever on the body the force acts
$\tau_{\text{cm},z}^{\,\text{ext}} = I_{\text{cm}}\,\alpha_{\text{cm},z}$	Rotation about the cm	Fixed axis direction through the cm; holds even while the cm accelerates
$W = \Delta K_{\text{trans}} + \Delta K_{\text{rot}}, \quad K = \tfrac{1}{2}mV_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{cm}}\,\omega_{\text{cm}}^2$	Work-energy	Relates speeds at two times; a force's work uses the displacement of its point of application
$F\Delta t = m\,\Delta V_{\text{cm}}, \quad \int \tau_{\text{cm}}\,dt = I_{\text{cm}}\,\Delta\omega$	Linear and angular impulse	One brief push moves the cm and spins the body about it

Key Insight: For the struck rod ( $2M$ at A, $M$ at B, length $D$ , impulse $F\Delta t$ at B): $V_{\text{cm}} = \dfrac{F\Delta t}{3M}$ , $I_{\text{cm}} = \tfrac{2}{3}MD^2$ , $\omega = \dfrac{F\Delta t}{MD}$ , and two-thirds of the kinetic energy $\dfrac{(F\Delta t)^2}{2M}$ is rotational.

IIWhere the Push Lands

The same impulse always gives $V_{\text{cm}} = F\Delta t/3M$ ; only its lever arm $h$ about the cm sets the spin.

Strike point, height above B 0

IIIWork Is Paid at the Point of Application

During the push the force's work is $F$ times the distance B moves, which pays for the spin too.

Time t / Δt 0.60

💡 The push is brief, so the rod barely turns while it acts: the force stays perpendicular to the rod with lever arm $\tfrac{2}{3}D$ , and B's speed grows in step with $t$ .

IVOrbital Plus Spin Equals Zero

About B's starting point O the angular momentum stays zero, so the orbital and spin parts cancel.

Mass ratio m_A / m_B 2

VQuiz Questions

Problem 1 · A Push at the Corner

Given: A $3.0\ \text{kg}$ plank floats in space. A $6.0\ \text{N}$ force, perpendicular to the plank, pushes at one corner. Find the acceleration of the plank's center of mass at that instant.

✅ Correct!

\vec{F}^{\,\text{ext}} = m\vec{A}_{\text{cm}}

does not care where the force acts:

A_{\text{cm}} = 6.0/3.0 = 2.0\ \text{m/s}^2

❌ Not quite. A net external force always accelerates the center of mass; the point of application decides only whether the body also turns.

❌ Close, but no force is "used up". The translational equation holds in full for any point of application; the rotation comes from the torque about the cm, which is a separate equation.

❌ Not quite.

I_{\text{cm}}

and the lever arm enter only the rotational equation

\tau_{\text{cm},z} = I_{\text{cm}}\alpha_{\text{cm},z}

; the cm's acceleration needs only the force and the mass.

Show solution

The momentum of a rigid body is $m\vec{V}_{\text{cm}}$ with constant $m$ , so

\vec{F}^{\,\text{ext}} = m\vec{A}_{\text{cm}} \quad\Rightarrow\quad A_{\text{cm}} = \frac{6.0\ \text{N}}{3.0\ \text{kg}} = 2.0\ \text{m/s}^2.

The same push through the center of mass would give the same $A_{\text{cm}}$ . Pushing at the corner adds an angular acceleration about the cm, but it does not reduce $A_{\text{cm}}$ .

Problem 2 · Measure From the Center of Mass

Given: The struck rod with $M = 0.50\ \text{kg}$ (so A has $1.0\ \text{kg}$ , B has $0.50\ \text{kg}$ ), $D = 0.60\ \text{m}$ , and an impulse $F\Delta t = 0.90\ \text{N}\cdot\text{s}$ at B, perpendicular to the rod. Find the angular speed after the push.

✅ Correct! Angular impulse

\tfrac{2}{3}D\,F\Delta t = 0.36\ \text{N}\cdot\text{m}\cdot\text{s}

over

I_{\text{cm}} = 0.12\ \text{kg}\cdot\text{m}^2

gives

3.0\ \text{rad/s} = F\Delta t/MD

❌ Close, but use the center of mass, not the middle of the rod. Both the lever arm and the moment of inertia must be taken about the cm, which sits

D/6

above the middle, nearer the heavier mass.

❌ Check the lever arm. B is

\tfrac{2}{3}D = 0.40\ \text{m}

below the cm, not

D/2 = 0.30\ \text{m}

; the torque is about the cm.

❌ Check the lever arm. The rotational equation uses torque about the cm, so the arm is the distance from the cm to B,

\tfrac{2}{3}D

, not the whole length

D

Show solution

Center of mass: $\tfrac{2}{3}D = 0.40\ \text{m}$ above B and $\tfrac{1}{3}D = 0.20\ \text{m}$ below A.

I_{\text{cm}} = (1.0)(0.20)^2 + (0.50)(0.40)^2 = 0.04 + 0.08 = 0.12\ \text{kg}\cdot\text{m}^2 = \tfrac{2}{3}MD^2.

Angular impulse about the cm equals the change in spin angular momentum:

\tfrac{2}{3}D\,F\Delta t = (0.40)(0.90) = 0.36 = I_{\text{cm}}\,\omega \quad\Rightarrow\quad \omega = \frac{0.36}{0.12} = 3.0\ \text{rad/s}.

Check with the formula: $\omega = \dfrac{F\Delta t}{MD} = \dfrac{0.90}{(0.50)(0.60)} = 3.0\ \text{rad/s}$ .

Problem 3 · Where the Energy Comes From

Given: The same rod and push as Problem 2 ( $M = 0.50\ \text{kg}$ , $D = 0.60\ \text{m}$ , $F\Delta t = 0.90\ \text{N}\cdot\text{s}$ , $\omega = 3.0\ \text{rad/s}$ ). Find the translational kinetic energy after the push and the work done by the force.

Translational kinetic energy?

Work done by the force?

✅ Correct!

K_{\text{trans}} = 0.27\ \text{J}

K_{\text{rot}} = 0.54\ \text{J}

, and the force's work at its point of application,

0.81\ \text{J}

, pays for both.

❌ Check $K_{\text{trans}}$ . Use the total mass

3M = 1.5\ \text{kg}

and

V_{\text{cm}} = F\Delta t/3M = 0.60\ \text{m/s}

❌ Not quite.

F\Delta t/M = 1.8\ \text{m/s}

is B's speed after the push, not the cm's; the impulse changes the momentum of the whole

3M

❌ Check the work. By the work-energy theorem

W = K_{\text{trans}} + K_{\text{rot}}

, starting from rest.

❌ Close, but that is $F$ times the cm's displacement. It gives only

K_{\text{trans}}

; a force's work uses the displacement of its point of application, B.

❌ Not quite. B's speed grows steadily from

0

1.8\ \text{m/s}

during the push, so its distance is

\tfrac{1}{2}(1.8)\Delta t

, not

(1.8)\Delta t

Show solution

Translation: $V_{\text{cm}} = \dfrac{F\Delta t}{3M} = \dfrac{0.90}{1.5} = 0.60\ \text{m/s}$ , so

K_{\text{trans}} = \tfrac{1}{2}(1.5)(0.60)^2 = 0.27\ \text{J} = \frac{(F\Delta t)^2}{6M}.

Rotation: $K_{\text{rot}} = \tfrac{1}{2}I_{\text{cm}}\,\omega^2 = \tfrac{1}{2}(0.12)(3.0)^2 = 0.54\ \text{J}$ .

Work: B ends at $v_B = V_{\text{cm}} + \omega\,\tfrac{2}{3}D = 0.60 + (3.0)(0.40) = 1.8\ \text{m/s}$ , its speed growing steadily, so it travels $\Delta x_B = \tfrac{1}{2}v_B\,\Delta t$ :

W = F\,\Delta x_B = \tfrac{1}{2}(F\Delta t)\,v_B = \tfrac{1}{2}(0.90)(1.8) = 0.81\ \text{J} = 0.27 + 0.54.

Problem 4 · A Uniform Stick Struck at One End

Given: A uniform stick of mass $1.2\ \text{kg}$ and length $0.80\ \text{m}$ floats in space. A brief impulse $J = 0.60\ \text{N}\cdot\text{s}$ , perpendicular to the stick, hits one end. ( $I_{\text{cm}} = \tfrac{1}{12}mL^2$ .) Find the angular speed and the kinetic energy after the push.

Angular speed?

Kinetic energy?

✅ Correct!

\omega = 6J/(mL) = 3.75\ \text{rad/s}

, and

K = 0.15 + 0.45 = 0.60\ \text{J}

, three-quarters of it rotational.

❌ Check $\omega$ . Angular impulse about the cm,

J\cdot\tfrac{L}{2}

, equals

I_{\text{cm}}\,\omega

with

I_{\text{cm}} = \tfrac{1}{12}mL^2

❌ Not quite. Nothing pins the far end: a free body spins about its cm, so use

I_{\text{cm}}

and the lever arm

L/2

, not the far end's

\tfrac{1}{3}mL^2

and

L

❌ Check the lever arm. The end is

L/2 = 0.40\ \text{m}

from the cm, not

L

❌ Not quite.

F\Delta t/(MD)

belonged to the two-mass rod; a uniform stick has a different

I_{\text{cm}}

and lever arm, so redo

J\cdot\tfrac{L}{2} = I_{\text{cm}}\,\omega

❌ Check $K$ . Add

\tfrac{1}{2}mV_{\text{cm}}^2

and

\tfrac{1}{2}I_{\text{cm}}\,\omega^2

❌ Close, but that is only the translational part. The spin carries

\tfrac{1}{2}I_{\text{cm}}\,\omega^2 = 0.45\ \text{J}

more.

❌ Not quite. The struck end's speed grows from

0

2.0\ \text{m/s}

during the push, so the work is

\tfrac{1}{2}J\,v_{\text{end}}

, not

J\,v_{\text{end}}

Show solution

Translation: $V_{\text{cm}} = J/m = 0.60/1.2 = 0.50\ \text{m/s}$ .

Rotation: $I_{\text{cm}} = \tfrac{1}{12}(1.2)(0.80)^2 = 0.064\ \text{kg}\cdot\text{m}^2$ , lever arm $L/2 = 0.40\ \text{m}$ :

J\,\tfrac{L}{2} = I_{\text{cm}}\,\omega \quad\Rightarrow\quad \omega = \frac{(0.60)(0.40)}{0.064} = \frac{0.24}{0.064} = 3.75\ \text{rad/s}.

Energy:

K = \tfrac{1}{2}(1.2)(0.50)^2 + \tfrac{1}{2}(0.064)(3.75)^2 = 0.15 + 0.45 = 0.60\ \text{J}.

Check by work: the struck end leaves at $v_{\text{end}} = V_{\text{cm}} + \omega\tfrac{L}{2} = 0.50 + 1.50 = 2.0\ \text{m/s}$ , so $W = \tfrac{1}{2}J\,v_{\text{end}} = \tfrac{1}{2}(0.60)(2.0) = 0.60\ \text{J}$ .

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