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Unit 6: Energy and Momentum of Rotating Systems

Definition Of Angular Momentum For A Particle

$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}$

Where:

$\vec{r}$ = position from reference point
$\vec{p}$ = linear momentum

Units: kg - m2/s

$L = \mathbf{r} \times \mathbf{p}$ (Particle)

Vector cross product form.

L = Iω (rigid body about fixed axis):

Simplified angular momentum.

$L = I\omega$

For extended rigid body rotating about fixed axis.

$L = I\omega$ (Rigid Body About Fixed Axis)

Simplified angular momentum.

$L = I\omega$

For extended rigid body rotating about fixed axis.

Total Angular Momentum: $L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)$ Or $L_{total} = I_{total} \omega$ (For Rigid Bodies)

Sum of angular momenta.

Discrete system: $\vec{L}_{total} = \sum_{i} \vec{r}_i \times \vec{p}_i$

Rigid body: $L = I\omega$

Net External Torque: $\Sigma\tau_{ext} = 0$ (Or Negligible).

Conservation condition.

When net external torque = 0, angular momentum conserved.

$L_{initial} = L_{final}$ (If $\Sigma\tau_{ext} = 0$ )

Applications:

Ice skater spinning (arms in/out -> changes I, ω changes to conserve L)
Planetary orbits (Kepler's second law)
Collisions with rotation

Orbital Motion: Kepler's Second Law (Equal Area In Equal Time) Is A Consequence Of Angular Momentum Conservation.

Radius vector sweeps equal areas in equal times.

$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$

Direct consequence of angular momentum conservation (central forces exert no torque).

Integral Form: $W = \int \tau d\theta$

Work by torque through angular displacement.

$W = \int_{\theta_i}^{\theta_f} \tau(\theta) \, d\theta$

Constant Torque: $W = \tau \Delta\theta$ (If Torque Is Constant)

$W = \tau \Delta \theta$

Work-energy Theorem For Rotation: $W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2$

Net work equals change in rotational kinetic energy.

$W_{net} = \Delta K_{rot} = K_{rot,f} - K_{rot,i}$

$W_{net} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$

Instantaneous Power: $P = \frac{dW}{dt} = \tau \omega$

$P = \tau \omega$

Relationship To Linear Power: Analogous To $P = \mathbf{F} \cdot \mathbf{v}$ In Linear Motion.

Rotational power analogous to translational.

$\tau \leftrightarrow F$
$\omega \leftrightarrow v$

Rolling Without Slipping Condition: $v_{cm} = R\omega$ (Where $R$ Is Radius)

$v_{cm} = R\omega$

Constraint for pure rolling.

Total Kinetic Energy: $\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^2$

$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$

Using rolling constraint: $K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\left(\frac{v_{cm}}{R}\right)^2$

Conservation Of Energy ( $K_{trans} + K_{rot} + U$ )

Complete energy conservation.

$E_{total} = K_{trans} + K_{rot} + U = \text{constant}$

$\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2 + U = \text{constant}$

Conservation Of Linear And Angular Momentum

Both can be conserved (different conditions).

Linear momentum: $\vec{F}_{net,ext} = 0$
Angular momentum: $\vec{\tau}_{net,ext} = 0$

Example: collision with rotation conserves both if no external forces/torques.

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Unit 6: Energy and Momentum of Rotating Systems

Definition Of Angular Momentum For A Particle

$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}$

Where:

$\vec{r}$ = position from reference point
$\vec{p}$ = linear momentum

Units: kg - m2/s

$L = \mathbf{r} \times \mathbf{p}$ (Particle)

Vector cross product form.

L = Iω (rigid body about fixed axis):

Simplified angular momentum.

$L = I\omega$

For extended rigid body rotating about fixed axis.

$L = I\omega$ (Rigid Body About Fixed Axis)

Simplified angular momentum.

$L = I\omega$

For extended rigid body rotating about fixed axis.

Total Angular Momentum: $L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)$ Or $L_{total} = I_{total} \omega$ (For Rigid Bodies)

Sum of angular momenta.

Discrete system: $\vec{L}_{total} = \sum_{i} \vec{r}_i \times \vec{p}_i$

Rigid body: $L = I\omega$

Net External Torque: $\Sigma\tau_{ext} = 0$ (Or Negligible).

Conservation condition.

When net external torque = 0, angular momentum conserved.

$L_{initial} = L_{final}$ (If $\Sigma\tau_{ext} = 0$ )

Applications:

Ice skater spinning (arms in/out -> changes I, ω changes to conserve L)
Planetary orbits (Kepler's second law)
Collisions with rotation

Orbital Motion: Kepler's Second Law (Equal Area In Equal Time) Is A Consequence Of Angular Momentum Conservation.

Radius vector sweeps equal areas in equal times.

$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$

Direct consequence of angular momentum conservation (central forces exert no torque).

Integral Form: $W = \int \tau d\theta$

Work by torque through angular displacement.

$W = \int_{\theta_i}^{\theta_f} \tau(\theta) \, d\theta$

Constant Torque: $W = \tau \Delta\theta$ (If Torque Is Constant)

$W = \tau \Delta \theta$

Work-energy Theorem For Rotation: $W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2$

Net work equals change in rotational kinetic energy.

$W_{net} = \Delta K_{rot} = K_{rot,f} - K_{rot,i}$

$W_{net} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$

Instantaneous Power: $P = \frac{dW}{dt} = \tau \omega$

$P = \tau \omega$

Relationship To Linear Power: Analogous To $P = \mathbf{F} \cdot \mathbf{v}$ In Linear Motion.

Rotational power analogous to translational.

$\tau \leftrightarrow F$
$\omega \leftrightarrow v$

Rolling Without Slipping Condition: $v_{cm} = R\omega$ (Where $R$ Is Radius)

$v_{cm} = R\omega$

Constraint for pure rolling.

Total Kinetic Energy: $\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^2$

$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$

Using rolling constraint: $K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\left(\frac{v_{cm}}{R}\right)^2$

Conservation Of Energy ( $K_{trans} + K_{rot} + U$ )

Complete energy conservation.

$E_{total} = K_{trans} + K_{rot} + U = \text{constant}$

$\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2 + U = \text{constant}$

Conservation Of Linear And Angular Momentum

Both can be conserved (different conditions).

Linear momentum: $\vec{F}_{net,ext} = 0$
Angular momentum: $\vec{\tau}_{net,ext} = 0$

Example: collision with rotation conserves both if no external forces/torques.

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Unit 6: Energy and Momentum of Rotating Systems

Definition Of Angular Momentum For A Particle

$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}$

Where:

$\vec{r}$ = position from reference point
$\vec{p}$ = linear momentum

Units: kg - m2/s

$L = \mathbf{r} \times \mathbf{p}$ (Particle)

Vector cross product form.

L = Iω (rigid body about fixed axis):

Simplified angular momentum.

$L = I\omega$

For extended rigid body rotating about fixed axis.

$L = I\omega$ (Rigid Body About Fixed Axis)

Simplified angular momentum.

$L = I\omega$

For extended rigid body rotating about fixed axis.

Total Angular Momentum: $L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)$ Or $L_{total} = I_{total} \omega$ (For Rigid Bodies)

Sum of angular momenta.

Discrete system: $\vec{L}_{total} = \sum_{i} \vec{r}_i \times \vec{p}_i$

Rigid body: $L = I\omega$

Net External Torque: $\Sigma\tau_{ext} = 0$ (Or Negligible).

Conservation condition.

When net external torque = 0, angular momentum conserved.

$L_{initial} = L_{final}$ (If $\Sigma\tau_{ext} = 0$ )

Applications:

Ice skater spinning (arms in/out -> changes I, ω changes to conserve L)
Planetary orbits (Kepler's second law)
Collisions with rotation

Orbital Motion: Kepler's Second Law (Equal Area In Equal Time) Is A Consequence Of Angular Momentum Conservation.

Radius vector sweeps equal areas in equal times.

$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$

Direct consequence of angular momentum conservation (central forces exert no torque).

Integral Form: $W = \int \tau d\theta$

Work by torque through angular displacement.

$W = \int_{\theta_i}^{\theta_f} \tau(\theta) \, d\theta$

Constant Torque: $W = \tau \Delta\theta$ (If Torque Is Constant)

$W = \tau \Delta \theta$

Work-energy Theorem For Rotation: $W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2$

Net work equals change in rotational kinetic energy.

$W_{net} = \Delta K_{rot} = K_{rot,f} - K_{rot,i}$

$W_{net} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$

Instantaneous Power: $P = \frac{dW}{dt} = \tau \omega$

$P = \tau \omega$

Relationship To Linear Power: Analogous To $P = \mathbf{F} \cdot \mathbf{v}$ In Linear Motion.

Rotational power analogous to translational.

$\tau \leftrightarrow F$
$\omega \leftrightarrow v$

Rolling Without Slipping Condition: $v_{cm} = R\omega$ (Where $R$ Is Radius)

$v_{cm} = R\omega$

Constraint for pure rolling.

Total Kinetic Energy: $\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^2$

$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$

Using rolling constraint: $K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\left(\frac{v_{cm}}{R}\right)^2$

Conservation Of Energy ( $K_{trans} + K_{rot} + U$ )

Complete energy conservation.

$E_{total} = K_{trans} + K_{rot} + U = \text{constant}$

$\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2 + U = \text{constant}$

Conservation Of Linear And Angular Momentum

Both can be conserved (different conditions).

Linear momentum: $\vec{F}_{net,ext} = 0$
Angular momentum: $\vec{\tau}_{net,ext} = 0$

Example: collision with rotation conserves both if no external forces/torques.

Unit 6: Energy and Momentum of Rotating Systems

Definition Of Angular Momentum For A Particle

L=r×pL = \mathbf{r} \times \mathbf{p}L=r×p (Particle)

L=IωL = I\omegaL=Iω (Rigid Body About Fixed Axis)

Total Angular Momentum: Ltotal=Σ(ri×pi)L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)Ltotal​=Σ(ri​×pi​) Or Ltotal=ItotalωL_{total} = I_{total} \omegaLtotal​=Itotal​ω (For Rigid Bodies)

Net External Torque: Στext=0\Sigma\tau_{ext} = 0Στext​=0 (Or Negligible).

Linitial=LfinalL_{initial} = L_{final}Linitial​=Lfinal​ (If Στext=0\Sigma\tau_{ext} = 0Στext​=0)

Orbital Motion: Kepler's Second Law (Equal Area In Equal Time) Is A Consequence Of Angular Momentum Conservation.

Integral Form: W=∫τdθW = \int \tau d\thetaW=∫τdθ

Constant Torque: W=τΔθW = \tau \Delta\thetaW=τΔθ (If Torque Is Constant)

Work-energy Theorem For Rotation: Wnet=ΔKrot=12Iωf2−12Iωi2W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2Wnet​=ΔKrot​=21​Iωf2​−21​Iωi2​

Instantaneous Power: P=dWdt=τωP = \frac{dW}{dt} = \tau \omegaP=dtdW​=τω

Relationship To Linear Power: Analogous To P=F⋅vP = \mathbf{F} \cdot \mathbf{v}P=F⋅v In Linear Motion.

Rolling Without Slipping Condition: vcm=Rωv_{cm} = R\omegavcm​=Rω (Where RRR Is Radius)

Total Kinetic Energy: 12Mvcm2+12Icmω2\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^221​Mvcm2​+21​Icm​ω2

Conservation Of Energy (Ktrans+Krot+UK_{trans} + K_{rot} + UKtrans​+Krot​+U)

Conservation Of Linear And Angular Momentum

Unit 6: Energy and Momentum of Rotating Systems

Definition Of Angular Momentum For A Particle

L=r×pL = \mathbf{r} \times \mathbf{p}L=r×p (Particle)

L=IωL = I\omegaL=Iω (Rigid Body About Fixed Axis)

Total Angular Momentum: Ltotal=Σ(ri×pi)L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)Ltotal​=Σ(ri​×pi​) Or Ltotal=ItotalωL_{total} = I_{total} \omegaLtotal​=Itotal​ω (For Rigid Bodies)

Net External Torque: Στext=0\Sigma\tau_{ext} = 0Στext​=0 (Or Negligible).

Linitial=LfinalL_{initial} = L_{final}Linitial​=Lfinal​ (If Στext=0\Sigma\tau_{ext} = 0Στext​=0)

Orbital Motion: Kepler's Second Law (Equal Area In Equal Time) Is A Consequence Of Angular Momentum Conservation.

Integral Form: W=∫τdθW = \int \tau d\thetaW=∫τdθ

Constant Torque: W=τΔθW = \tau \Delta\thetaW=τΔθ (If Torque Is Constant)

Work-energy Theorem For Rotation: Wnet=ΔKrot=12Iωf2−12Iωi2W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2Wnet​=ΔKrot​=21​Iωf2​−21​Iωi2​

Instantaneous Power: P=dWdt=τωP = \frac{dW}{dt} = \tau \omegaP=dtdW​=τω

Relationship To Linear Power: Analogous To P=F⋅vP = \mathbf{F} \cdot \mathbf{v}P=F⋅v In Linear Motion.

Rolling Without Slipping Condition: vcm=Rωv_{cm} = R\omegavcm​=Rω (Where RRR Is Radius)

Total Kinetic Energy: 12Mvcm2+12Icmω2\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^221​Mvcm2​+21​Icm​ω2

Conservation Of Energy (Ktrans+Krot+UK_{trans} + K_{rot} + UKtrans​+Krot​+U)

Conservation Of Linear And Angular Momentum

Unit 6: Energy and Momentum of Rotating Systems

Definition Of Angular Momentum For A Particle

L=r×pL = \mathbf{r} \times \mathbf{p}L=r×p (Particle)

L=IωL = I\omegaL=Iω (Rigid Body About Fixed Axis)

Total Angular Momentum: Ltotal=Σ(ri×pi)L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)Ltotal​=Σ(ri​×pi​) Or Ltotal=ItotalωL_{total} = I_{total} \omegaLtotal​=Itotal​ω (For Rigid Bodies)

Net External Torque: Στext=0\Sigma\tau_{ext} = 0Στext​=0 (Or Negligible).

Linitial=LfinalL_{initial} = L_{final}Linitial​=Lfinal​ (If Στext=0\Sigma\tau_{ext} = 0Στext​=0)

Orbital Motion: Kepler's Second Law (Equal Area In Equal Time) Is A Consequence Of Angular Momentum Conservation.

Integral Form: W=∫τdθW = \int \tau d\thetaW=∫τdθ

Constant Torque: W=τΔθW = \tau \Delta\thetaW=τΔθ (If Torque Is Constant)

Work-energy Theorem For Rotation: Wnet=ΔKrot=12Iωf2−12Iωi2W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2Wnet​=ΔKrot​=21​Iωf2​−21​Iωi2​

Instantaneous Power: P=dWdt=τωP = \frac{dW}{dt} = \tau \omegaP=dtdW​=τω

Relationship To Linear Power: Analogous To P=F⋅vP = \mathbf{F} \cdot \mathbf{v}P=F⋅v In Linear Motion.

Rolling Without Slipping Condition: vcm=Rωv_{cm} = R\omegavcm​=Rω (Where RRR Is Radius)

Total Kinetic Energy: 12Mvcm2+12Icmω2\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^221​Mvcm2​+21​Icm​ω2

Conservation Of Energy (Ktrans+Krot+UK_{trans} + K_{rot} + UKtrans​+Krot​+U)

Conservation Of Linear And Angular Momentum

$L = \mathbf{r} \times \mathbf{p}$ (Particle)

$L = I\omega$ (Rigid Body About Fixed Axis)

Total Angular Momentum: $L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)$ Or $L_{total} = I_{total} \omega$ (For Rigid Bodies)

Net External Torque: $\Sigma\tau_{ext} = 0$ (Or Negligible).

$L_{initial} = L_{final}$ (If $\Sigma\tau_{ext} = 0$ )

Integral Form: $W = \int \tau d\theta$

Constant Torque: $W = \tau \Delta\theta$ (If Torque Is Constant)

Work-energy Theorem For Rotation: $W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2$

Instantaneous Power: $P = \frac{dW}{dt} = \tau \omega$

Relationship To Linear Power: Analogous To $P = \mathbf{F} \cdot \mathbf{v}$ In Linear Motion.

Rolling Without Slipping Condition: $v_{cm} = R\omega$ (Where $R$ Is Radius)

Total Kinetic Energy: $\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^2$

Conservation Of Energy ( $K_{trans} + K_{rot} + U$ )

$L = \mathbf{r} \times \mathbf{p}$ (Particle)

$L = I\omega$ (Rigid Body About Fixed Axis)

Total Angular Momentum: $L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)$ Or $L_{total} = I_{total} \omega$ (For Rigid Bodies)

Net External Torque: $\Sigma\tau_{ext} = 0$ (Or Negligible).

$L_{initial} = L_{final}$ (If $\Sigma\tau_{ext} = 0$ )

Integral Form: $W = \int \tau d\theta$

Constant Torque: $W = \tau \Delta\theta$ (If Torque Is Constant)

Work-energy Theorem For Rotation: $W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2$

Instantaneous Power: $P = \frac{dW}{dt} = \tau \omega$

Relationship To Linear Power: Analogous To $P = \mathbf{F} \cdot \mathbf{v}$ In Linear Motion.

Rolling Without Slipping Condition: $v_{cm} = R\omega$ (Where $R$ Is Radius)

Total Kinetic Energy: $\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^2$

Conservation Of Energy ( $K_{trans} + K_{rot} + U$ )

$L = \mathbf{r} \times \mathbf{p}$ (Particle)

$L = I\omega$ (Rigid Body About Fixed Axis)

Total Angular Momentum: $L_{total} = \Sigma (\mathbf{r}_i \times \mathbf{p}_i)$ Or $L_{total} = I_{total} \omega$ (For Rigid Bodies)

Net External Torque: $\Sigma\tau_{ext} = 0$ (Or Negligible).

$L_{initial} = L_{final}$ (If $\Sigma\tau_{ext} = 0$ )

Integral Form: $W = \int \tau d\theta$

Constant Torque: $W = \tau \Delta\theta$ (If Torque Is Constant)

Work-energy Theorem For Rotation: $W_{net} = \Delta K_{rot} = \frac{1}{2}I \omega_f^2 - \frac{1}{2}I \omega_i^2$

Instantaneous Power: $P = \frac{dW}{dt} = \tau \omega$

Relationship To Linear Power: Analogous To $P = \mathbf{F} \cdot \mathbf{v}$ In Linear Motion.

Rolling Without Slipping Condition: $v_{cm} = R\omega$ (Where $R$ Is Radius)

Total Kinetic Energy: $\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm} \omega^2$

Conservation Of Energy ( $K_{trans} + K_{rot} + U$ )