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5.2 Connecting Linear and Rotational Motion

Syllabus
2024
Topic
5.2
Level
—

Learning objectives

5.2A—Describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of…Describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa.• For a point at a distance r from a fixed axis of rotation, the linear distance s traveled by the point as the system rotates through an angle is given by the equation• Derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are given by the following equations:• For a rigid system, all points within that system have the same angular velocity and angular acceleration. BOUNDARY STATEMENT Descriptions of the directions of rotation for a point or object are limited to clockwise and counterclockwise with respect to a given axis of rotation. | AP Physics 1: Algebra-Based Course and Exam Description TOPIC 5.3 Torque | AP Physics 1: Algebra-Based Course and Exam Description

Radius converts angular motion into linear motion

Connect radius to the path

For a point a perpendicular distance rr from a fixed rotation axis, radius converts angular motion into motion along the circular path. Angular displacement must be in radians.

Δs=rΔθvt=rωat=rα\begin{aligned}\Delta s&=r\Delta\theta\\[3pt]v_t&=r\omega\\[3pt]a_t&=r\alpha\end{aligned}

Separate shared and radius-dependent quantities

For points on one rigid system Same for every point Increases with radius rr
Position change Δθ\Delta\theta Arc length Δs\Delta s
Rate of motion ω\omega Tangential speed vtv_t
Change of rate α\alpha Tangential acceleration ata_t

Compare two radii

Two points on one wheel: at one instant, ω=3.0 rad/s\omega=3.0\,\text{rad/s} and α=2.0 rad/s2\alpha=2.0\,\text{rad/s}^2. Point A is at rA=0.20 mr_A=0.20\,\text{m}; point B is at rB=0.50 mr_B=0.50\,\text{m}.

vA=rAω=(0.20)(3.0)=0.60 m/s,vB=(0.50)(3.0)=1.5 m/sv_A=r_A\omega=(0.20)(3.0)=0.60\,\text{m/s},\quad v_B=(0.50)(3.0)=1.5\,\text{m/s},

at,A=rAα=(0.20)(2.0)=0.40 m/s2,at,B=(0.50)(2.0)=1.0 m/s2a_{t,A}=r_A\alpha=(0.20)(2.0)=0.40\,\text{m/s}^2,\quad a_{t,B}=(0.50)(2.0)=1.0\,\text{m/s}^2.

Both points share ω\omega and α\alpha, but the farther point has larger tangential speed and acceleration.

Keep radians and acceleration components clear

Use radians in Δs=rΔθ\Delta s=r\Delta\theta. The relation at=rαa_t=r\alpha gives only the tangential component of acceleration; it does not by itself describe every acceleration component of a point moving in a circle. Rotation direction here is described only as clockwise or counterclockwise about the stated axis.