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Rotational Kinematics and Dynamics

Angular velocity, acceleration, and torque with calculus

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Rotational Kinematics and Dynamics

Angular Quantities

Angular position: θ\theta (radians)

Angular velocity: ω=dθdt\omega = \frac{d\theta}{dt}

Angular acceleration: α=dωdt=d2θdt2\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}

Relationship to Linear Quantities

For point at distance rr from axis:

Arc length: s=rθs = r\theta

Linear velocity: v=rωv = r\omega

Tangential acceleration: at=rαa_t = r\alpha

Centripetal acceleration: ac=v2r=rω2a_c = \frac{v^2}{r} = r\omega^2

Rotational Kinematics

For constant angular acceleration α\alpha:

ω=ω0+αt\omega = \omega_0 + \alpha t

θ=θ0+ω0t+12αt2\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2

ω2=ω02+2α(θ−θ0)\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)

(Analogous to linear kinematics)

Variable Angular Acceleration

When α(t)\alpha(t) is given:

ω(t)=ω0+∫0tα(t′) dt′\omega(t) = \omega_0 + \int_0^t \alpha(t') \, dt'

θ(t)=θ0+∫0tω(t′) dt′\theta(t) = \theta_0 + \int_0^t \omega(t') \, dt'

When α(θ)\alpha(\theta) is given, use: α=dωdt=dωdθdθdt=ωdωdθ\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta}\frac{d\theta}{dt} = \omega\frac{d\omega}{d\theta}

Torque

Definition: τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

Magnitude: τ=rFsin⁡ϕ\tau = rF\sin\phi

where ϕ\phi is angle between r⃗\vec{r} and F⃗\vec{F}.

Perpendicular distance form: τ=r⊥F=rF⊥\tau = r_{\perp}F = rF_{\perp}

Rotational Dynamics

Newton's second law for rotation: τnet=Iα\tau_{net} = I\alpha

where II is moment of inertia.

Differential form: τ=Idωdt\tau = I\frac{d\omega}{dt}

Work and Power in Rotation

Work by torque: W=∫θ1θ2τ dθW = \int_{\theta_1}^{\theta_2} \tau \, d\theta

For constant torque: W=τΔθW = \tau\Delta\theta

Rotational power: P=τωP = \tau\omega

(Analogous to P=FvP = Fv for linear motion)

Rotational Kinetic Energy

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

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

Combined Translation and Rotation

For rolling object:

Total kinetic energy: KE=KEtrans+KErot=12Mvcm2+12Icmω2KE = KE_{trans} + KE_{rot} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2

Rolling without slipping: vcm=Rωv_{cm} = R\omega

where RR is radius.

Example: Rolling Down Incline

Energy conservation: Mgh=12Mv2+12Iω2Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2

Using v=Rωv = R\omega and I=βMR2I = \beta MR^2:

Mgh=12Mv2+12βMR2v2R2Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}\beta MR^2\frac{v^2}{R^2}

gh=12v2(1+β)gh = \frac{1}{2}v^2(1 + \beta)

v=2gh1+βv = \sqrt{\frac{2gh}{1 + \beta}}

where β\beta depends on shape:

  • Solid cylinder: β=1/2\beta = 1/2
  • Solid sphere: β=2/5\beta = 2/5
  • Hoop: β=1\beta = 1

Angular Impulse

∫t1t2τ dt=ΔL=IΔω\int_{t_1}^{t_2} \tau \, dt = \Delta L = I\Delta\omega

where L=IωL = I\omega is angular momentum.

📚 Practice Problems

1Problem 1easy

❓ Question:

A wheel starts from rest and accelerates uniformly at α = 2.0 rad/s² for 5.0 seconds. Find: (a) the final angular velocity, (b) the total angle rotated, and (c) the number of complete revolutions.

💡 Show Solution

Given:

  • ω₀ = 0
  • α = 2.0 rad/s²
  • t = 5.0 s

(a) Final angular velocity:

ω=ω0+αt=0+(2.0)(5.0)\omega = ω_0 + αt = 0 + (2.0)(5.0)

ω=10 rad/s\boxed{\omega = 10 \text{ rad/s}}

(b) Total angle rotated:

θ=ω0t+12αt2=0+12(2.0)(5.0)2\theta = ω_0 t + \frac{1}{2}αt^2 = 0 + \frac{1}{2}(2.0)(5.0)^2

θ=(1.0)(25)\theta = (1.0)(25)

θ=25 rad\boxed{\theta = 25 \text{ rad}}

(c) Number of revolutions:

N=θ2π=252πN = \frac{\theta}{2\pi} = \frac{25}{2\pi}

N=3.98 rev≈4 revolutions\boxed{N = 3.98 \text{ rev} \approx 4 \text{ revolutions}}

2Problem 2medium

❓ Question:

A solid disk (mass M = 5.0 kg, radius R = 0.3 m) rotates about its central axis. A tangential force F = 15 N is applied at the rim. Find: (a) the torque, (b) the angular acceleration, and (c) the angular velocity after 3.0 seconds (starting from rest).

💡 Show Solution

Given:

  • M = 5.0 kg
  • R = 0.3 m
  • F = 15 N (tangential at rim)
  • ω₀ = 0

(a) Torque:

τ=rF=RF=(0.3)(15)\tau = rF = RF = (0.3)(15)

τ=4.5 N⋅m\boxed{\tau = 4.5 \text{ N·m}}

(b) Angular acceleration:

Moment of inertia for solid disk: I=12MR2=12(5.0)(0.3)2=0.225 kg⋅m2I = \frac{1}{2}MR^2 = \frac{1}{2}(5.0)(0.3)^2 = 0.225 \text{ kg·m}^2

α=τI=4.50.225\alpha = \frac{\tau}{I} = \frac{4.5}{0.225}

α=20 rad/s2\boxed{\alpha = 20 \text{ rad/s}^2}

(c) Angular velocity after 3.0 s:

ω=ω0+αt=0+(20)(3.0)\omega = ω_0 + αt = 0 + (20)(3.0)

ω=60 rad/s\boxed{\omega = 60 \text{ rad/s}}

Linear speed at rim: v=Rω=(0.3)(60)=18 m/sv = Rω = (0.3)(60) = 18 \text{ m/s}

3Problem 3hard

❓ Question:

A uniform rod (length L = 1.0 m, mass M = 2.0 kg) is hinged at one end and held horizontal. It is released from rest. Find: (a) the initial angular acceleration, (b) the angular velocity when the rod is vertical, and (c) the linear speed of the free end when vertical.

💡 Show Solution

Given:

  • L = 1.0 m
  • M = 2.0 kg
  • Rod hinged at one end, released from horizontal

(a) Initial angular acceleration:

Moment of inertia about end: I=13ML2=13(2.0)(1.0)2=0.667 kg⋅m2I = \frac{1}{3}ML^2 = \frac{1}{3}(2.0)(1.0)^2 = 0.667 \text{ kg·m}^2

Torque due to gravity (acts at CM, distance L/2 from hinge): τ=MgL2=(2.0)(9.8)1.02=9.8 N⋅m\tau = Mg\frac{L}{2} = (2.0)(9.8)\frac{1.0}{2} = 9.8 \text{ N·m}

α0=τI=9.80.667\alpha_0 = \frac{\tau}{I} = \frac{9.8}{0.667}

α0=14.7 rad/s2\boxed{\alpha_0 = 14.7 \text{ rad/s}^2}

(b) Angular velocity when vertical:

Using energy conservation:

Initial PE (relative to final): Ui=MgL2U_i = Mg\frac{L}{2} (CM drops by L/2)

Final KE: Kf=12Iω2K_f = \frac{1}{2}Iω^2

MgL2=12Iω2Mg\frac{L}{2} = \frac{1}{2}Iω^2

ω2=MgLI=(2.0)(9.8)(1.0)0.667ω^2 = \frac{MgL}{I} = \frac{(2.0)(9.8)(1.0)}{0.667}

ω2=29.4ω^2 = 29.4

ω=5.42 rad/s\boxed{\omega = 5.42 \text{ rad/s}}

(c) Linear speed of free end:

v=Lω=(1.0)(5.42)v = Lω = (1.0)(5.42)

v=5.42 m/s\boxed{v = 5.42 \text{ m/s}}

Check with energy: 12Mvcm2=MgL2\frac{1}{2}Mv_{cm}^2 = Mg\frac{L}{2}

vcm=gL=9.8=3.13 m/sv_{cm} = \sqrt{gL} = \sqrt{9.8} = 3.13 \text{ m/s}

Since vend=2vcmv_{end} = 2v_{cm} for rotation about end: vend≈5.4v_{end} \approx 5.4 m/s ✓

Explain using:

📌 Related Topics in Rotational Motion

❓ Frequently Asked Questions

What is Rotational Kinematics and Dynamics?▾
Angular velocity, acceleration, and torque with calculus
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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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Rotational Kinematics and Dynamics is part of the AP Physics C: Mechanics course on Study Mondo, specifically in the Rotational Motion section. You can explore the full course for more related topics and practice resources.
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.