Year-13 Physics — Rotational Dynamics & Banked Curves

Banked-Curve & Circular-Motion Applications

  • Design question 5

    • Required banking angle for a roadway when
    • Design speed v=13.2ms1v = 13.2\,\text{m\,s}^{-1}
    • Curve radius r=48.0mr = 48.0\,\text{m}
    • Friction-free condition implies
      tanθ=v2rg\tan\theta = \frac{v^{2}}{rg}
    • Used in road safety, racetrack design, aviation (tilted wings).
  • Design question 6 (race-track)

    • Two limiting radii on a concrete, friction-free, banked oval
    • Inner lane rmin=112mr_{\min}=112\,\text{m}
    • Outer lane rmax=165mr_{\max}=165\,\text{m}
    • Wall height h=18mh = 18\,\text{m} allows determination of banking angle by simple trigonometry (rise/run). Once θ\theta is fixed the usable speed range is
      v<em>min=rgtanθ,v</em>max=rgtanθ<em>r=r</em>maxv<em>{\min}=\sqrt{rg\tan\theta}\,,\qquad v</em>{\max}=\sqrt{rg\tan\theta}\,\Big|<em>{r=r</em>{\max}}
    • Shows how a single surface can be safe for different radii.
  • Design question 7 (Ferris wheel)

    • Child mass m=40kgm=40\,\text{kg}, radius r=10mr=10\,\text{m}, constant speed v=3ms1v=3\,\text{m\,s}^{-1}.
    • Forces on rider
    • Bottom: Nbottom=mg+mv2rN_{\text{bottom}}=mg+\frac{mv^{2}}{r} (heavier feeling)
    • Top: Ntop=mgmv2rN_{\text{top}} = mg-\frac{mv^{2}}{r} (lighter feeling).
    • Demonstrates interplay of apparent weight and centripetal requirement.

Rotational Dynamics — Motivation

  • Extended bodies (wheels, planets, motors) cannot be analysed as point masses because their particles have differing linear speeds.
  • Need rotational analogues of displacement, velocity, acceleration, force.

Angular Variables

Angular Displacement θ\theta

  • Measured in radians: θ=sr\theta = \frac{s}{r} (arc length over radius).
  • One revolution =2πrad=2\pi\,\text{rad}.

Angular Velocity ω\omega

  • Rate of change of angular displacement: ω=dθdt\omega = \frac{d\theta}{dt} (units rads1\text{rad\,s}^{-1}).
  • Linear connection: v=rωv = r\omega.
  • Frequency relation: ω=2πf\omega = 2\pi f.

Angular Acceleration α\alpha

  • Rate of change of angular velocity: α=dωdt\alpha=\frac{d\omega}{dt} (units rads2\text{rad\,s}^{-2}).
  • If ω\omega changes uniformly: α=ω<em>fω</em>iΔt\alpha = \frac{\omega<em>{f}-\omega</em>{i}}{\Delta t}.
  • Linear link: at=rαa_{t}=r\alpha (tangential acceleration).

Uniform Angular Acceleration — Kinematic Equations

  • Exact analogues of linear s,u,v,a,ts,u,v,a,t equations.
    • ω<em>f=ω</em>i+αt\omega<em>{f}=\omega</em>{i}+\alpha t
    • θ=ωit+12αt2\theta = \omega_{i}t + \tfrac12 \alpha t^{2}
    • ω<em>f2=ω</em>i2+2αθ\omega<em>{f}^{2}=\omega</em>{i}^{2}+2\alpha\theta

Linear↔Angular Variable Table

  • Speed: u,v    ω<em>i,ω</em>fu,v\;\leftrightarrow\;\omega<em>{i},\omega</em>{f}
  • Displacement: s    θs\;\leftrightarrow\;\theta
  • Acceleration: a    αa\;\leftrightarrow\;\alpha

Worked Example 1.13 — Photographic Record

  • Starts from rest, reaches 45revmin145\,\text{rev\,min}^{-1} (i.e. 4.71rads14.71\,\text{rad\,s}^{-1}) in 5s5\,\text{s}.
    • Angular acceleration α=0.94rads2\alpha=0.94\,\text{rad\,s}^{-2}.
    • Revolutions covered in this startup =11.75rev=11.75\,\text{rev}.

Practice Problems (Exercise 1.5)

  • Convert Earth’s orbital period T=3.156×107sT=3.156\times10^{7}\,\text{s} to ωearth\omega_{\text{earth}}.
  • Wheel speed on a moving bicycle, v=7.5ms1,r=0.30mv=7.5\,\text{m\,s}^{-1}, r=0.30\,\text{m}.
  • Flywheel deceleration from 500Hz500\,\text{Hz} to rest in 5s5\,\text{s}.
  • Car wheel slowdown: r=0.36m,v=18ms1,N=25turnsr=0.36\,\text{m}, v=18\,\text{m\,s}^{-1}, N=25\,\text{turns}.
  • etc.

Moment of Inertia II

  • Rotational analogue of mass: I=m<em>ir</em>i2I=\sum m<em>{i} r</em>{i}^{2}.
  • Depends on mass distribution relative to axis.
  • Units kgm2\text{kg\,m}^{2}.

Standard Shapes (about stated axes)

  • Hoop / thin cylindrical shell: I=MR2I = MR^{2}.
  • Solid cylinder/disc: I=12MR2I = \tfrac12 MR^{2}.
  • Hollow cylinder: I=12M(R<em>12+R</em>22)I = \tfrac12 M\left(R<em>{1}^{2}+R</em>{2}^{2}\right).
  • Thin rod (centre): I=112ML2I = \tfrac1{12} ML^{2}.
  • Thin rod (end): I=13ML2I = \tfrac13 ML^{2}.
  • Solid sphere: I=25MR2I = \tfrac25 MR^{2}.
  • Thin spherical shell: I=23MR2I = \tfrac23 MR^{2}.
  • Rectangular plate (axis through centre, perpendicular): I=112M(a2+b2)I = \tfrac1{12}M(a^{2}+b^{2}).

Examples

  1. Point mass m=2kgm=2\,\text{kg} on string r=0.6mr=0.6\,\text{m}I=0.72kgm2I=0.72\,\text{kg\,m}^{2}.
  2. Two-mass dumb-bell: m<em>1=2kg,m</em>2=3kg,L=1mm<em>{1}=2\,\text{kg}, m</em>{2}=3\,\text{kg}, L=1\,\text{m}. About midpoint ⟂ to rod:
    I=m<em>1(12L)2+m</em>2(12L)2=1.25kgm2I = m<em>{1}\left(\tfrac12L\right)^{2}+m</em>{2}\left(\tfrac12L\right)^{2}=1.25\,\text{kg\,m}^{2}.

Torque τ\tau

  • Rotational effectiveness of a force: τ=rF=Iα\tau = rF_{\perp} = I\alpha (units Nm\text{N\,m}).
  • Produced by string tension in falling-mass apparatus, figure-of-eight belts, engine crankshafts.

Torque–Acceleration Derivation (Example 1.15)

  • Uniform disc I=12mr2I=\tfrac12 m r^{2} hanging by wrapped string.
  • Translational mgT=mamg - T = ma; rotational τ=Tr=Iα\tau = Tr = I\alpha with a=rαa=r\alpha.
  • Solve ⇒ linear acceleration a=23ga = \tfrac23 g(i.e., two-thirds of free fall).
  • Demonstrates energy sharing: some gravitational PE goes to rotational KE.

Angular Momentum LL

  • For rigid body: L=IωL = I\omega.
  • For point mass: L=rmv=mr2ωL = r m v = m r^{2} \omega (direction via right-hand rule).
  • Units kgm2s1\text{kg\,m}^{2}\,\text{s}^{-1}.

Conservation Principle

  • If external torque τ<em>ext=0\sum \tau<em>{\text{ext}} = 0, then L</em>initial=LfinalL</em>{\text{initial}} = L_{\text{final}}.
  • Enables ice-skater spin-ups, neutron-star dynamo predictions, merging-disk problems.

Worked Examples (1.16)

  1. Wheel accelerated by τ=100Nm\tau=100\,\text{N\,m} from rest to ω=20rads1\omega=20\,\text{rad\,s}^{-1} over θ=10rad\theta=10\,\text{rad}.
    • Using ω2=2αθ\omega^{2}=2\alpha\thetaα=20rads2\alpha=20\,\text{rad\,s}^{-2}.
    • Moment of inertia from τ=Iα\tau=I\alphaI=5kgm2I=5\,\text{kg\,m}^{2}.
    • Angular momentum gained ΔL=Iω=100kgm2s1\Delta L = I\omega = 100\,\text{kg\,m}^{2}\text{s}^{-1}.
  2. Two coaxial disks couple (clutch problem).
    • Disk A: I<em>1=2kgm2,ω</em>1=10rads1I<em>{1}=2\,\text{kg\,m}^{2}, \omega</em>{1}=10\,\text{rad\,s}^{-1}.
    • Disk B: I<em>2=3kgm2,ω</em>2=0I<em>{2}=3\,\text{kg\,m}^{2}, \omega</em>{2}=0.
    • After drop: ω<em>f=I</em>1ω<em>1+I</em>2ω<em>2I</em>1+I2=4rads1\omega<em>{f}=\tfrac{I</em>{1}\omega<em>{1}+I</em>{2}\omega<em>{2}}{I</em>{1}+I_{2}}=4\,\text{rad\,s}^{-1}.

Additional Exercises (1.6)

  • Turntable–record coupling, I<em>1=0.09,I</em>2=0.03kgm2I<em>{1}=0.09, I</em>{2}=0.03\,\text{kg\,m}^{2}.
  • Cylindrical spacecraft attitude control via expelled gas: each puff m=0.4kg,v=100ms1m=0.4\,\text{kg}, v=100\,\text{m\,s}^{-1} at r=2mr=2\,\text{m}.
    • Linear momentum per puff p=mvp=mv.
    • Angular momentum delivered L=rpL=rp, conserve to find craft’s Δω\Delta\omega.

Conceptual & Real-World Connections

  • Banking removes dependence on tyre friction ⇒ higher safety margins.
  • Ferris-wheel apparent-weight changes underpin design of safe lap-bars.
  • Disk-clutch coupling is mechanical analogue of perfectly inelastic collision.
  • Moment of inertia critical in flywheels (energy storage) & figure-skating (artistic spins).
  • Torque link to power P=τωP=\tau\omega governs engine ratings.
  • Conservation of angular momentum fundamental in astrophysics (planet formation, pulsars).

Ethical & Practical Notes

  • Proper banking saves lives by reducing skidding accidents.
  • Amusement-ride designers must account for human tolerance to varying normal forces.
  • Rocket attitude thrusters must conserve propellant while delivering required angular impulse.