Rolling - AP Physics C Mechanics

  1. What is the condition for rolling motion in terms of angular displacement?
    s=rθs = r \theta. Linear displacement equals radius times angular displacement.
  2. What is aa down an incline for a hoop rolling without slipping?
    a=12gsin⁡θa=\frac{1}{2}g\sin\theta. Substitute Icm=MR2I_{\mathrm{cm}}=MR^2 into general formula.
  3. What is the formula for the moment of inertia of a solid sphere?
    I=25mr2I = \frac{2}{5} m r^2. For a solid sphere, mass distributed throughout volume.
  4. What is the rolling-without-slipping condition relating vcmv_{\mathrm{cm}} and ω\omega for radius RR?
    vcm=ωRv_{\mathrm{cm}}=\omega R. No slipping means the contact point is instantaneously at rest.
  5. State the formula for the moment of inertia of a hoop (thin ring) about its central axis.
    Icm=MR2I_{\mathrm{cm}}=MR^2. All mass is concentrated at distance RR from the axis.
  6. State the formula for the moment of inertia of a thin spherical shell about a diameter.
    Icm=23MR2I_{\mathrm{cm}}=\frac{2}{3}MR^2. All mass is at distance RR from center, integrated over sphere surface.
  7. Calculate the mechanical energy of a rolling object.
    Emech=KEtrans+KErot+PEE_{mech} = KE_{trans} + KE_{rot} + PE. Conservation of energy includes all energy forms.
  8. What is aa down an incline for a solid disk rolling without slipping?
    a=23gsin⁡θa=\frac{2}{3}g\sin\theta. Substitute Icm=12MR2I_{\mathrm{cm}}=\frac{1}{2}MR^2 into general formula.
  9. Identify which has larger aa down the same incline: hoop (I=MR2I=MR^2) or solid disk (I=12MR2I=\frac{1}{2}MR^2).
    Solid disk (larger aa). Smaller II gives larger aa in the acceleration formula.
  10. What is the acceleration of the contact point for rolling without slipping when acm=0a_{\mathrm{cm}}=0 and ω\omega is constant?
    acontact=ω2Ra_{\mathrm{contact}}=\omega^2R (upward). Contact point has centripetal acceleration toward the center.
  11. What is the equation for angular displacement in rolling?
    θ=sr\theta = \frac{s}{r}. Arc length equals radius times angle in radians.
  12. State the formula for the moment of inertia of a solid sphere about a diameter.
    Icm=25MR2I_{\mathrm{cm}}=\frac{2}{5}MR^2. Mass distribution in 3D gives factor 25\frac{2}{5} for solid sphere.
  13. Calculate the acceleration of a disk rolling down an incline.
    a=23gsinθa = \frac{2}{3} g \text{sin}\theta. For disk, I=12mr2I = \frac{1}{2}mr^2 gives this acceleration.
  14. What is the formula for the moment of inertia of a disk?
    I=12mr2I = \frac{1}{2} m r^2. For uniform disk, mass distributed over area.
  15. For rolling without slipping, what is the static friction direction for a body rolling down an incline under gravity only?
    Up the incline. Friction provides torque to maintain rolling constraint.
  16. State the formula for the moment of inertia of a solid cylinder (disk) about its central axis.
    Icm=12MR2I_{\mathrm{cm}}=\frac{1}{2}MR^2. Mass is distributed uniformly throughout the solid cylinder.
  17. What is the speed of the top point of a wheel rolling without slipping, in the ground frame?
    vtop=2vcmv_{\mathrm{top}}=2v_{\mathrm{cm}}. Top point has both translational and rotational velocities adding up.
  18. What is the formula for the moment of inertia of a hoop?
    I=mr2I = m r^2. All mass concentrated at rim, distance rr from center.
  19. State the equation for the translational kinetic energy of a rolling object.
    KEtrans=12mv2KE_{trans} = \frac{1}{2} m v^2. Standard kinetic energy formula using linear velocity.
  20. What is the parallel-axis theorem relating IcmI_{\mathrm{cm}} and II about an axis a distance dd away?
    I=Icm+Md2I=I_{\mathrm{cm}}+Md^2. Adds Md2Md^2 when axis is displaced by distance dd.
  21. What is II for a rolling body about the instantaneous contact point (use radius RR)?
    Icontact=Icm+MR2I_{\mathrm{contact}}=I_{\mathrm{cm}}+MR^2. Apply parallel-axis theorem with d=Rd=R.
  22. Calculate the final speed of a rolling object down an incline.
    v=2gh1+Imr2v = \sqrt{\frac{2gh}{1 + \frac{I}{mr^2}}}$. Energy conservation with rotational inertia term.
  23. Determine the ratio of rotational to translational kinetic energy for a solid sphere.
    KErotKEtrans=25\frac{KE_{rot}}{KE_{trans}} = \frac{2}{5}. For solid sphere, I=25mr2I = \frac{2}{5}mr^2 gives this ratio.
  24. Identify the formula for the frictional force in rolling without slipping.
    f=Iar2f = \frac{I a}{r^2}. Friction provides torque for rolling without slipping.
  25. Find the total kinetic energy of a rolling object.
    KEtotal=KEtrans+KErotKE_{total} = KE_{trans} + KE_{rot}. Sum of translational and rotational kinetic energies.
  26. What is the center-of-mass acceleration down an incline for rolling without slipping, using IcmI_{\mathrm{cm}}?
    a=gsin⁡θ1+IcmMR2a=\frac{g\sin\theta}{1+\frac{I_{\mathrm{cm}}}{MR^2}}. Derived from F=MaF=Ma and τ=Iα\tau=I\alpha with rolling constraint.
  27. What is aa down an incline for a solid sphere rolling without slipping?
    a=57gsin⁡θa=\frac{5}{7}g\sin\theta. Substitute Icm=25MR2I_{\mathrm{cm}}=\frac{2}{5}MR^2 into general formula.
  28. Determine the linear acceleration of a rolling object down an incline.
    a=gsin⁡θ1+Imr2a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}}. Derived from applying Newton's laws to rolling motion.
  29. What is the total kinetic energy of a rigid body rolling without slipping on a surface?
    K=12Mvcm2+12Icmω2K=\frac{1}{2}Mv_{\mathrm{cm}}^2+\frac{1}{2}I_{\mathrm{cm}}\omega^2. Sum of translational and rotational kinetic energies.
  30. What is the rolling-without-slipping condition relating acma_{\mathrm{cm}} and α\alpha for radius RR?
    acm=αRa_{\mathrm{cm}}=\alpha R. Linear acceleration equals angular acceleration times radius for no slip.
  31. What is the speed of the contact point of a wheel rolling without slipping, in the ground frame?
    vcontact=0v_{\mathrm{contact}}=0. No-slip condition requires contact point to be instantaneously at rest.

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