Timedora

5 Rotational Motion

Builds from angular position and motion to torque, rotational inertia, rotational dynamics, and the linked motion of rolling objects.

Describing rotational motion

Angular position θ\theta describes an object's orientation relative to a reference direction. Angular displacement is the change in position, Δθ=θ2−θ1\Delta\theta=\theta_2-\theta_1. Angles are commonly measured in radians; one full revolution is 2π2\pi radians.

describes how quickly angular position changes, while angular acceleration α\alpha describes how quickly changes:

ωavg=ΔθΔt,ω=dθdt,α=dωdt.\omega_{\mathrm{avg}}=\frac{\Delta\theta}{\Delta t},\qquad \omega=\frac{d\theta}{dt},\qquad \alpha=\frac{d\omega}{dt}.

For rotation about a fixed axis with constant angular acceleration, use the familiar constant-acceleration equations with angular quantities:

ω=ω0+αt\omega=\omega_0+\alpha t
θ=θ0+ω0t+12αt2\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2
ω2=ω02+2α(θ−θ0).\omega^2=\omega_0^2+2\alpha(\theta-\theta_0).

For a point a distance rr from the axis, tangential speed and acceleration are v=rωv=r\omega and at=rαa_t=r\alpha. Its inward radial acceleration is ar=rω2a_r=r\omega^2.

Takeaway: Angular variables describe rotation in much the same way that position, velocity, and acceleration describe linear motion.

and turning effect

measures how effectively a force turns an object about a chosen axis. Its magnitude depends on the force, its distance from the axis, and the angle between the position and force vectors:

τ=rFsin⁡ϕ=r⊥F.\tau=rF\sin\phi=r_{\perp}F.

Here, rr is the distance from the axis to the force's point of application, ϕ\phi is the angle between the position and force vectors, and r⊥r_{\perp} is the perpendicular lever arm. A force directed through the axis produces no . For a given force and distance, a perpendicular force produces the greatest . is measured in newton-metres, N m\mathrm{N\,m}.

Choose a sign convention, such as counterclockwise positive, and apply it consistently when adding torques.

Example: A perpendicular force of 20 N20\,\mathrm{N} applied 0.80 m0.80\,\mathrm{m} from a door's hinges produces

τ=(0.80 m)(20 N)=16 N m.\tau=(0.80\,\mathrm{m})(20\,\mathrm{N})=16\,\mathrm{N\,m}.

Applying the same force farther from the hinges increases the .

Takeaway: To increase a force's turning effect, increase the force or its perpendicular lever arm.

Rotational inertia and mass distribution

, denoted II, measures an object's resistance to angular acceleration about a particular axis. For discrete particles and a continuous body, respectively,

I=∑imiri2,I=∫r2 dm.I=\sum_i m_i r_i^2,\qquad I=\int r^2\,dm.

In these expressions, rr is each particle's or mass element's perpendicular distance from the axis. Because distance is squared, mass located farther from the axis contributes more strongly. therefore depends both on total mass and on how that mass is distributed relative to the chosen axis.

For a thin hoop of mass mm and radius RR, the about its central axis is I=mR2I=mR^2. For a uniform solid disk with the same mass and radius, it is I=12mR2I=\frac{1}{2}mR^2. The hoop has the greater because more of its mass lies farther from the axis. is measured in kg m2\mathrm{kg\,m^2}.

Takeaway: The is the rotational counterpart of mass, but it also depends on the axis and mass distribution.

For a rigid body rotating about a fixed axis, is described by the rotational form of Newton's second law:

∑τ=Iα.\sum\tau=I\alpha.

The net about the axis determines the angular acceleration. With the same net , an object with a larger has a smaller angular acceleration.

A useful problem-solving sequence is:

  1. Select the axis of rotation.

  2. Identify the forces and their lever arms.

  3. Calculate each signed and find the net .

  4. Use the about that same axis in ∑τ=Iα\sum\tau=I\alpha.

Takeaway: causes angular acceleration, while determines how much angular acceleration a given net produces.

Rolling motion

Rolling combines translation of an object's center of mass with rotation about that center. In , the contact point is instantaneously at rest relative to the surface. For rolling radius RR, the translational and angular motion are linked by

vCM=Rω,aCM=Rα.v_{\mathrm{CM}}=R\omega,\qquad a_{\mathrm{CM}}=R\alpha.

The kinetic energy of a rolling object includes both translational and rotational contributions:

K=12mvCM2+12ICMω2.K=\frac{1}{2}mv_{\mathrm{CM}}^2+\frac{1}{2}I_{\mathrm{CM}}\omega^2.

For an object down an incline at angle θ\theta, its center-of-mass acceleration is

aCM=gsin⁡θ1+ICMmR2.a_{\mathrm{CM}}=\frac{g\sin\theta}{1+\frac{I_{\mathrm{CM}}}{mR^2}}.

A solid cylinder has ICM=12mR2I_{\mathrm{CM}}=\frac{1}{2}mR^2, so its acceleration down the incline is aCM=23gsin⁡θa_{\mathrm{CM}}=\frac{2}{3}g\sin\theta. The rolling constraints apply only while the object rolls without slipping. If it slides, the contact point moves relative to the surface, so these constraints do not apply.

Takeaway: For , translation and rotation are linked, and both contribute to kinetic energy.