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Rotational Dynamics and Angular Momentum

Moment of inertia, rotational dynamics, angular momentum, and conservation

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🔄 Rotational Dynamics and Angular Momentum

Moment of Inertia

Moment of inertia II is the rotational equivalent of mass. It measures resistance to rotational acceleration.

I=∑miri2I = \sum m_i r_i^2

For continuous objects: I=∫r2 dmI = \int r^2 \, dm

where:

  • II = moment of inertia (kg·m²)
  • mim_i = mass of particle ii
  • rir_i = distance of particle ii from axis of rotation

💡 Key Idea: Moment of inertia depends on both mass AND how that mass is distributed relative to the axis. Same object can have different moments of inertia for different axes!


Common Moments of Inertia

ObjectAxisMoment of Inertia
Point massDistance rr from axisI=mr2I = mr^2
Thin rodThrough center, perpendicularI=112mL2I = \frac{1}{12}mL^2
Thin rodThrough end, perpendicularI=13mL2I = \frac{1}{3}mL^2
Solid disk/cylinderThrough center, along axisI=12mr2I = \frac{1}{2}mr^2
Hollow cylinder (thin)Through center, along axisI=mr2I = mr^2
Solid sphereThrough centerI=25mr2I = \frac{2}{5}mr^2
Hollow sphere (thin)Through centerI=23mr2I = \frac{2}{3}mr^2

Note: These formulas are given on AP Physics 1 formula sheet!


Properties of Moment of Inertia

Depends on Axis

Same object, different axis → different II

Example: Rod

  • Center axis: I=112mL2I = \frac{1}{12}mL^2
  • End axis: I=13mL2I = \frac{1}{3}mL^2 (larger!)

Depends on Mass Distribution

  • Mass farther from axis → larger II
  • Mass closer to axis → smaller II

Why figure skaters spin faster with arms in: Pulling arms in decreases II, so ω\omega increases to conserve angular momentum!


Newton's Second Law for Rotation

τnet=Iα\tau_{net} = I\alpha

where:

  • τnet\tau_{net} = net torque (N·m)
  • II = moment of inertia (kg·m²)
  • α\alpha = angular acceleration (rad/s²)

Analogy: F=maF = ma for linear motion

Interpretation:

  • Larger torque → larger angular acceleration
  • Larger moment of inertia → harder to accelerate rotationally

Rotational Kinetic Energy

KErot=12Iω2KE_{rot} = \frac{1}{2}I\omega^2

Analogy: KE=12mv2KE = \frac{1}{2}mv^2 for linear motion

Total kinetic energy of rolling object: KEtotal=KEtrans+KErot=12mvcm2+12Iω2KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2

For rolling without slipping (vcm=rωv_{cm} = r\omega): KEtotal=12mvcm2+12I(vcmr)2KE_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\left(\frac{v_{cm}}{r}\right)^2


Angular Momentum

Angular momentum LL is the rotational equivalent of linear momentum:

L=IωL = I\omega

where:

  • LL = angular momentum (kg·m²/s)
  • II = moment of inertia (kg·m²)
  • ω\omega = angular velocity (rad/s)

Analogy: p=mvp = mv for linear motion

For Point Particle

L⃗=r⃗×p⃗=m(r⃗×v⃗)\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})

Magnitude: L=mvrsin⁡θL = mvr\sin\theta


Conservation of Angular Momentum

In an isolated system (no external torques):

Linitial=LfinalL_{initial} = L_{final}

Iiωi=IfωfI_i \omega_i = I_f \omega_f

💡 Fundamental Law: Like linear momentum, angular momentum is conserved when no external torques act on the system.


Applications of Conservation

Figure Skating Spins

  • Arms out: large II, small ω\omega
  • Arms in: small II, large ω\omega
  • Iiωi=IfωfI_i \omega_i = I_f \omega_f

If II decreases by factor of 3, then ω\omega increases by factor of 3!

Diving

  • Tucked position: small II → fast rotation
  • Layout position: large II → slow rotation
  • Controls rotation rate mid-air

Planetary Orbits

  • Closer to sun: smaller rr, larger vv
  • Farther from sun: larger rr, smaller vv
  • Kepler's 2nd Law (equal areas in equal times)

Gyroscopes

  • Large angular momentum resists changes in orientation
  • Used in navigation, stabilization

Relationship: Torque and Angular Momentum

τnet=dLdt\tau_{net} = \frac{dL}{dt}

Analogy: F=dpdtF = \frac{dp}{dt} (Newton's 2nd Law)

If no external torque: dLdt=0\frac{dL}{dt} = 0 → LL is constant ✓


Problem-Solving Strategy

For Rotational Dynamics:

  1. Identify moment of inertia II (use formula or calculate)
  2. Find net torque: τnet=∑τi\tau_{net} = \sum \tau_i
  3. Apply: τnet=Iα\tau_{net} = I\alpha
  4. Solve for unknown (τ\tau, II, or α\alpha)
  5. Use kinematics if needed (to find ω\omega, θ\theta, etc.)

For Angular Momentum Conservation:

  1. Check: Is system isolated? (No external torques?)
  2. Write initial state: Li=IiωiL_i = I_i \omega_i
  3. Write final state: Lf=IfωfL_f = I_f \omega_f
  4. Set equal: Iiωi=IfωfI_i \omega_i = I_f \omega_f
  5. Solve for unknown

⚠️ Common Mistakes

Mistake 1: Forgetting II Depends on Axis

Same object, different axis → different II!

Mistake 2: Using Wrong II Formula

Check axis location and object shape carefully!

Mistake 3: Confusing L=IωL = I\omega with p=mvp = mv

These are analogous but not the same!

  • Angular momentum has different units (kg·m²/s)
  • Depends on axis choice

Mistake 4: Assuming II is Constant

In conservation problems, II often changes (figure skater pulling arms in)!

Mistake 5: Forgetting Rolling Kinetic Energy

Rolling objects have BOTH translational and rotational KE!


Energy in Rotational Motion

Work Done by Torque

W=τθW = \tau \theta

(when τ\tau is constant)

Analogy: W=FdW = Fd for linear motion

Work-Energy Theorem (Rotational)

Wnet=ΔKErot=12Iωf2−12Iωi2W_{net} = \Delta KE_{rot} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2


Rolling Motion

For object rolling without slipping down incline:

Energy conservation: mgh=12mvcm2+12Iω2mgh = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2

Using vcm=rωv_{cm} = r\omega and I=βmr2I = \beta mr^2 (where β\beta is shape factor):

gh=12vcm2(1+β)gh = \frac{1}{2}v_{cm}^2\left(1 + \beta\right)

vcm=2gh1+βv_{cm} = \sqrt{\frac{2gh}{1 + \beta}}

Different objects, different speeds:

  • Hollow cylinder (β=1\beta = 1): slowest
  • Solid cylinder (β=1/2\beta = 1/2): medium
  • Solid sphere (β=2/5\beta = 2/5): fastest

All faster than sliding (no friction) box!


Comparing Linear and Rotational

LinearRotationalRelationship
Mass mmMoment of inertia III=∑mr2I = \sum mr^2
Velocity vvAngular velocity ω\omegav=rωv = r\omega
Acceleration aaAngular acceleration α\alphaat=rαa_t = r\alpha
Force FFTorque τ\tauτ=rFsin⁡θ\tau = rF\sin\theta
F=maF = maτ=Iα\tau = I\alphaNewton's 2nd Law
Momentum p=mvp = mvAngular momentum L=IωL = I\omegaL=r×pL = r \times p
KE=12mv2KE = \frac{1}{2}mv^2KErot=12Iω2KE_{rot} = \frac{1}{2}I\omega^2Kinetic energy
pp conservedLL conservedNo external force/torque

Real-World Applications

Spinning Top/Gyroscope

  • Large LL → resists tipping
  • Precession when external torque applied
  • Used in navigation systems

Bicycle Wheels

  • Spinning wheels have angular momentum
  • Harder to tip over when moving
  • Gyroscopic stability

Earth's Rotation

  • Huge moment of inertia
  • Angular momentum conserved
  • Day length nearly constant

Tornadoes and Hurricanes

  • Air spiraling inward
  • rr decreases → ω\omega increases
  • Conservation of angular momentum

Key Formulas Summary

ConceptFormulaUnits
Moment of inertiaI=∑mr2I = \sum mr^2kg·m²
Rotational 2nd Lawτ=Iα\tau = I\alphaN·m
Rotational KEKErot=12Iω2KE_{rot} = \frac{1}{2}I\omega^2J
Angular momentumL=IωL = I\omegakg·m²/s
ConservationIiωi=IfωfI_i\omega_i = I_f\omega_f(isolated)
Torque-momentumτ=dLdt\tau = \frac{dL}{dt}N·m
Work by torqueW=τθW = \tau\thetaJ

Common II values:

  • Point mass: mr2mr^2
  • Disk: 12mr2\frac{1}{2}mr^2
  • Sphere: 25mr2\frac{2}{5}mr^2
  • Rod (center): 112mL2\frac{1}{12}mL^2
  • Rod (end): 13mL2\frac{1}{3}mL^2

📚 Practice Problems

1Problem 1easy

❓ Question:

A solid disk with mass 2 kg and radius 0.4 m rotates about its center. A constant torque of 3 N·m is applied. Find: (a) the moment of inertia, (b) the angular acceleration, and (c) the angular velocity after 5 seconds (starting from rest).

💡 Show Solution

Given Information:

  • Mass: m=2m = 2 kg
  • Radius: r=0.4r = 0.4 m
  • Torque: τ=3\tau = 3 N·m
  • Initial angular velocity: ωi=0\omega_i = 0 rad/s (starts from rest)
  • Time: t=5t = 5 s

(a) Find moment of inertia


Step 1: Use formula for solid disk

For a solid disk rotating about its center:

I=12mr2I = \frac{1}{2}mr^2

I=12(2)(0.4)2I = \frac{1}{2}(2)(0.4)^2

I=12(2)(0.16)I = \frac{1}{2}(2)(0.16)

I=0.16 kg⋅m2I = 0.16 \text{ kg·m}^2


Answer (a): Moment of inertia = 0.16 kg·m²


(b) Find angular acceleration


Step 2: Apply Newton's 2nd Law for rotation

τ=Iα\tau = I\alpha

α=τI=30.16\alpha = \frac{\tau}{I} = \frac{3}{0.16}

α=18.75 rad/s2\alpha = 18.75 \text{ rad/s}^2


Answer (b): Angular acceleration = 18.75 rad/s²


(c) Find angular velocity after 5 seconds


Step 3: Use rotational kinematics

ωf=ωi+αt\omega_f = \omega_i + \alpha t

ωf=0+(18.75)(5)\omega_f = 0 + (18.75)(5)

ωf=93.75 rad/s\omega_f = 93.75 \text{ rad/s}


Answer (c): Angular velocity after 5 s = 93.75 rad/s (about 94 rad/s)

Check:

  • In revolutions: 93.752π≈14.9\frac{93.75}{2\pi} \approx 14.9 rev/s
  • Tangential speed at rim: v=rω=0.4(93.75)=37.5v = r\omega = 0.4(93.75) = 37.5 m/s ✓

2Problem 2medium

❓ Question:

A figure skater is spinning at 2 rev/s with arms extended (moment of inertia = 3 kg·m²). She pulls her arms in, reducing her moment of inertia to 1.5 kg·m². Find: (a) her new angular velocity, and (b) the ratio of her final kinetic energy to initial kinetic energy.

💡 Show Solution

Given Information:

  • Initial: Ii=3I_i = 3 kg·m², ωi=2\omega_i = 2 rev/s
  • Final: If=1.5I_f = 1.5 kg·m², ωf=?\omega_f = ?

Step 0: Convert units

ωi=2 rev/s×2π rad1 rev=4π rad/s≈12.57 rad/s\omega_i = 2 \text{ rev/s} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 4\pi \text{ rad/s} \approx 12.57 \text{ rad/s}


(a) Find new angular velocity


Step 1: Apply conservation of angular momentum

No external torques, so angular momentum is conserved:

Li=LfL_i = L_f

Iiωi=IfωfI_i \omega_i = I_f \omega_f

(3)(12.57)=(1.5)ωf(3)(12.57) = (1.5)\omega_f

37.7=1.5ωf37.7 = 1.5\omega_f

ωf=37.71.5=25.1 rad/s\omega_f = \frac{37.7}{1.5} = 25.1 \text{ rad/s}


Convert to rev/s:

ωf=25.12π≈4 rev/s\omega_f = \frac{25.1}{2\pi} \approx 4 \text{ rev/s}


Answer (a): New angular velocity = 4 rev/s (or 25.1 rad/s)

Note: She spins twice as fast when she halves her moment of inertia!


(b) Find ratio of kinetic energies


Step 2: Calculate initial kinetic energy

KEi=12Iiωi2KE_i = \frac{1}{2}I_i\omega_i^2

KEi=12(3)(12.57)2KE_i = \frac{1}{2}(3)(12.57)^2

KEi=12(3)(158.0)KE_i = \frac{1}{2}(3)(158.0)

KEi=237 JKE_i = 237 \text{ J}


Step 3: Calculate final kinetic energy

KEf=12Ifωf2KE_f = \frac{1}{2}I_f\omega_f^2

KEf=12(1.5)(25.1)2KE_f = \frac{1}{2}(1.5)(25.1)^2

KEf=12(1.5)(630.0)KE_f = \frac{1}{2}(1.5)(630.0)

KEf=472.5 JKE_f = 472.5 \text{ J}


Step 4: Calculate ratio

KEfKEi=472.5237≈2\frac{KE_f}{KE_i} = \frac{472.5}{237} \approx 2


Answer (b): The ratio is 2:1 (kinetic energy doubles!)

Explanation: Where does the extra energy come from? The skater does work pulling her arms in against centrifugal effects. This work becomes rotational kinetic energy!

General formula: If If=IinI_f = \frac{I_i}{n}, then:

  • ωf=nωi\omega_f = n\omega_i (angular velocity multiplies by nn)
  • KEf=nKEiKE_f = nKE_i (kinetic energy multiplies by nn)

In this case, n=2n = 2 ✓

3Problem 3hard

❓ Question:

A solid sphere (mass 5 kg, radius 0.2 m) and a hollow cylinder (mass 5 kg, radius 0.2 m) both start from rest at the top of a 3 m high incline. They roll without slipping. Find: (a) the speed of each at the bottom, and (b) which one reaches the bottom first.

💡 Show Solution

Given Information:

  • Both: mass m=5m = 5 kg, radius r=0.2r = 0.2 m
  • Height: h=3h = 3 m
  • Start from rest: vi=0v_i = 0, ωi=0\omega_i = 0
  • Roll without slipping: v=rωv = r\omega

Moments of inertia:

  • Solid sphere: Is=25mr2I_s = \frac{2}{5}mr^2
  • Hollow cylinder: Ic=mr2I_c = mr^2

(a) Find speeds at bottom


Step 1: Use energy conservation for solid sphere

mgh=12mvs2+12Isωs2mgh = \frac{1}{2}mv_s^2 + \frac{1}{2}I_s\omega_s^2

Using v=rωv = r\omega so ω=vr\omega = \frac{v}{r}:

mgh=12mvs2+12Is(vsr)2mgh = \frac{1}{2}mv_s^2 + \frac{1}{2}I_s\left(\frac{v_s}{r}\right)^2

mgh=12mvs2+12(25mr2)vs2r2mgh = \frac{1}{2}mv_s^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\frac{v_s^2}{r^2}

mgh=12mvs2+15mvs2mgh = \frac{1}{2}mv_s^2 + \frac{1}{5}mv_s^2

mgh=710mvs2mgh = \frac{7}{10}mv_s^2

gh=710vs2gh = \frac{7}{10}v_s^2

vs=10gh7v_s = \sqrt{\frac{10gh}{7}}

vs=10(9.8)(3)7v_s = \sqrt{\frac{10(9.8)(3)}{7}}

vs=2947v_s = \sqrt{\frac{294}{7}}

vs=42v_s = \sqrt{42}

vs=6.48 m/sv_s = 6.48 \text{ m/s}


Step 2: Use energy conservation for hollow cylinder

mgh=12mvc2+12Icωc2mgh = \frac{1}{2}mv_c^2 + \frac{1}{2}I_c\omega_c^2

mgh=12mvc2+12(mr2)vc2r2mgh = \frac{1}{2}mv_c^2 + \frac{1}{2}(mr^2)\frac{v_c^2}{r^2}

mgh=12mvc2+12mvc2mgh = \frac{1}{2}mv_c^2 + \frac{1}{2}mv_c^2

mgh=mvc2mgh = mv_c^2

vc=ghv_c = \sqrt{gh}

vc=(9.8)(3)v_c = \sqrt{(9.8)(3)}

vc=29.4v_c = \sqrt{29.4}

vc=5.42 m/sv_c = 5.42 \text{ m/s}


Answer (a):

  • Solid sphere: 6.48 m/s
  • Hollow cylinder: 5.42 m/s

(b) Which reaches bottom first?


Step 3: Compare speeds

The solid sphere has higher speed (6.48 m/s vs. 5.42 m/s).

Since both start from rest and travel the same distance, the one with higher final speed must have had higher average speed.

Higher average speed → less time → reaches bottom first


Answer (b): The solid sphere reaches the bottom first.


Explanation:

The solid sphere has a smaller moment of inertia relative to its mass (I=25mr2I = \frac{2}{5}mr^2 vs. I=mr2I = mr^2).

This means:

  • Less energy goes into rotation
  • More energy goes into translation
  • Higher linear speed
  • Shorter time to bottom

General rule: Objects with mass concentrated near center (smaller I/mr2I/mr^2) roll faster down inclines!

Ranking (fastest to slowest):

  1. Solid sphere: I=25mr2I = \frac{2}{5}mr^2 → v=10gh7v = \sqrt{\frac{10gh}{7}}
  2. Solid disk/cylinder: I=12mr2I = \frac{1}{2}mr^2 → v=4gh3v = \sqrt{\frac{4gh}{3}}
  3. Hollow cylinder: I=mr2I = mr^2 → v=ghv = \sqrt{gh}
  4. Hollow sphere: I=23mr2I = \frac{2}{3}mr^2 → v=6gh5v = \sqrt{\frac{6gh}{5}}

All are slower than a frictionless sliding block: v=2ghv = \sqrt{2gh} ✓

Explain using:

📌 Related Topics in Torque & Rotational Motion

❓ Frequently Asked Questions

What is Rotational Dynamics and Angular Momentum?▾
Moment of inertia, rotational dynamics, angular momentum, and conservation
How can I study Rotational Dynamics and Angular Momentum effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Rotational Dynamics and Angular Momentum?▾
Rotational Dynamics and Angular Momentum is part of the AP Physics 1 course on Study Mondo, specifically in the Torque & Rotational Motion section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Rotational Dynamics and Angular Momentum?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.