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5.2 - Connecting Linear and Rotational Motion

Building the Bridge

We’re now connecting the rotational quantities from 5.1 (θ\theta, ω\omega, α\alpha) to the familiar linear quantities from Unit 1 (xx, vv, aa). The key insight is that radius provides the conversion factor between angular and linear descriptions of the same motion. A point at distance rr from the rotation axis travels a longer linear distance than a point closer to the axis, even though both sweep through the same angle.

Arc Length and Angular Displacement

Arc length is the linear distance a point travels along a circular path.
Δs=rΔθor equivalentlyΔx=rΔθ\Delta s = r \Delta \theta \qquad \text{or equivalently} \qquad \Delta x = r \Delta \theta
where Δs\Delta s (or Δx\Delta x) is the linear distance along the arc in meters, rr is the radius (distance from the rotation axis) in meters, and Δθ\Delta \theta is the angular displacement, which must be in radians for this formula to work.
5.2 - Connecting Linear and Rotational Motion
The farther from the axis, the greater the linear distance traveled for the same angular displacement. For example, two points on a spinning disk both rotate through Δθ=π/2\Delta\theta = \pi/2 rad (a quarter turn): Point A at rA=0.1r_A = 0.1 m travels ΔsA=(0.1)(π/2)=0.157\Delta s_A = (0.1)(\pi/2) = 0.157 m, while Point B at rB=0.3r_B = 0.3 m travels ΔsB=(0.3)(π/2)=0.471\Delta s_B = (0.3)(\pi/2) = 0.471 m, three times farther, even though both swept the same angle.
The formula only works with radians because that’s literally how a radian is defined: one radian is the angle that produces an arc length equal to the radius, θ=1 rad  ⟹  s=r\theta = 1 \text{ rad} \implies s = r. That’s why the formula is so clean, it’s built right into the definition.

Tangential Velocity and Angular Velocity

Tangential velocity is the linear speed of a point moving along a circular path.
v=rωv = r\omega
where vv is tangential velocity in m/s, rr is radius in meters, and ω\omega is angular velocity in rad/s. This velocity always points tangent to the circle, perpendicular to the radius, at the point’s current position.
Deriving it: starting from Δs=rΔθ\Delta s = r\Delta\theta, dividing both sides by time gives ΔsΔt=rΔθΔt\frac{\Delta s}{\Delta t} = r\frac{\Delta\theta}{\Delta t}, and recognizing the definitions of linear and angular velocity gives v=rωv = r\omega.
Points farther from the axis move faster linearly, even with the same angular velocity. Take a merry-go-round: Child A at r=1r = 1 m and Child B at r=3r = 3 m both complete one rotation in 10 seconds, giving them the same angular velocity, ω=2π10≈0.628\omega = \frac{2\pi}{10} \approx 0.628 rad/s. But their linear velocities differ: Child A moves at vA=(1)(0.628)=0.628v_A = (1)(0.628) = 0.628 m/s, while Child B moves at vB=(3)(0.628)=1.88v_B = (3)(0.628) = 1.88 m/s, three times as fast. This is exactly why the outer horses on a merry-go-round move faster.
5.2 - Connecting Linear and Rotational Motion

Tangential Acceleration and Angular Acceleration

Tangential acceleration is the linear acceleration along the circular path, speeding up or slowing down the rotation.
at=rαa_t = r\alpha
where ata_t is tangential acceleration in m/s², rr is radius in meters, and α\alpha is angular acceleration in rad/s². The subscript “t” is a reminder this is tangential (along the circle), not centripetal (toward the center), which exists even at constant angular velocity and gets covered separately.
Deriving it: starting from v=rωv = r\omega, taking the rate of change with respect to time, ΔvΔt=rΔωΔt\frac{\Delta v}{\Delta t} = r\frac{\Delta\omega}{\Delta t}, and recognizing the definitions gives at=rαa_t = r\alpha.
For a bicycle wheel spinning up at α=2\alpha = 2 rad/s²: the hub (r=0.05r = 0.05 m) accelerates at at=(0.05)(2)=0.1a_t = (0.05)(2) = 0.1 m/s², while the rim (r=0.35r = 0.35 m) accelerates at at=(0.35)(2)=0.7a_t = (0.35)(2) = 0.7 m/s², seven times faster linearly than the hub.

Sign Convention

The standard convention: counterclockwise is positive (positive angular displacement, angular velocity, and angular acceleration), while clockwise is negative. Think of a clock’s hands moving clockwise (the negative direction) and the opposite being counterclockwise (positive), matching the unit circle convention from pre-calculus, where positive angles are measured counterclockwise.
5.2 - Connecting Linear and Rotational Motion

Same Angular Quantities at Every Radius

Every point on a rigid rotating object shares the same angular displacement, angular velocity, and angular acceleration, no matter how far it is from the axis. But linear quantities, distance, velocity, acceleration, all depend on the radius.
Quantity
Same for all radii?
Formula
Angular displacement (Δθ\Delta\theta)
Yes
doesn’t depend on rr
Angular velocity (ω\omega)
Yes
doesn’t depend on rr
Angular acceleration (α\alpha)
Yes
doesn’t depend on rr
Linear distance (Δs\Delta s)
No
Δs=rΔθ\Delta s = r\Delta\theta
Linear velocity (vv)
No
v=rωv = r\omega
Tangential acceleration (ata_t)
No
at=rαa_t = r\alpha
So when analyzing a rotating system, use angular quantities to describe the rotation as a whole, and linear quantities to describe the motion of specific points.
Example, a spinning disk with two clay lumps: Lump A near the center (r=0.1r = 0.1 m) and Lump B near the edge (r=0.5r = 0.5 m) both share the same angular quantities, rotating through Δθ=2π\Delta\theta = 2\pi rad (one full turn) with ω=10\omega = 10 rad/s and α=3\alpha = 3 rad/s². But their linear quantities differ: Lump A travels Δs=(0.1)(2π)=0.628\Delta s = (0.1)(2\pi) = 0.628 m per rotation while Lump B travels Δs=(0.5)(2π)=3.14\Delta s = (0.5)(2\pi) = 3.14 m per rotation, five times farther; and Lump A moves at v=(0.1)(10)=1v = (0.1)(10) = 1 m/s while Lump B moves at v=(0.5)(10)=5v = (0.5)(10) = 5 m/s, five times as fast.

Worked Example: Fishing Reel

A fishing reel with radius r=4.5r = 4.5 cm =0.045= 0.045 m accelerates at α=110\alpha = 110 rad/s² for t=2.0t = 2.0 s, starting from rest, previously found to have final angular velocity ω=220\omega = 220 rad/s and angular displacement Δθ=220\Delta\theta = 220 rad. Find the linear speed of the fishing line after 2 seconds, and how much line came out.
Linear speed of the line: the line unwraps from the edge of the reel, so use the reel’s radius with v=rωv = r\omega:
v=(0.045)(220)=9.9 m/sv = (0.045)(220) = 9.9 \text{ m/s}
After 2 seconds of acceleration, the line pulls out at 9.9 meters per second.
Length of line released: using arc length, Δs=rΔθ\Delta s = r\Delta\theta:
Δs=(0.045)(220)=9.9 m\Delta s = (0.045)(220) = 9.9 \text{ m}
During those 2 seconds, 9.9 meters of line unwraps from the reel.
Interestingly, both values come out to 9.9 here, purely coincidental for these particular numbers. In general, v=rωv = r\omega (a speed) and Δs=rΔθ\Delta s = r\Delta\theta (a distance) are different quantities that will usually have different numerical values, don’t assume they’ll always match.
As for context, is 9.9 meters a lot of fishing line? Not really, recreational spools typically carry 100 to 300 meters, and deep-sea spools carry 500 to 1500 meters, so 9.9 meters is a small fraction of a typical spool, leaving the fish plenty of line to swim with.

Worked Example: Linear to Rotational (Crane Drum)

Going the other way works the same. A crane lifts a crate by winding its cable onto a drum of radius 0.25 m. The crate rises at a steady 0.80 m/s. Find the drum’s angular velocity and the number of revolutions it makes while the crate rises 6.0 m.
Angular velocity: the cable comes off the edge of the drum, so the cable’s speed equals the tangential speed of the drum’s rim. Solving v=rωv = r\omega for ω\omega:
ω=vr=0.80 m/s0.25 m=3.2 rad/s\omega = \frac{v}{r} = \frac{0.80\text{ m/s}}{0.25\text{ m}} = 3.2\text{ rad/s}
Revolutions: 6.0 m of cable winds onto the drum, which is the arc length traveled by a point on the rim. Solving Δs=rΔθ\Delta s = r\Delta\theta for Δθ\Delta\theta:
Δθ=Δsr=6.0 m0.25 m=24 rad=242π≈3.8 revolutions\Delta\theta = \frac{\Delta s}{r} = \frac{6.0\text{ m}}{0.25\text{ m}} = 24\text{ rad} = \frac{24}{2\pi} \approx 3.8\text{ revolutions}
Since the crate rises at a steady speed, ω\omega is constant and α=0\alpha = 0. If the crate were speeding up at 0.40 m/s² instead, the drum’s angular acceleration would be α=atr=0.400.25=1.6\alpha = \frac{a_t}{r} = \frac{0.40}{0.25} = 1.6 rad/s².

Summary of Conversion Formulas

Linear Quantity
Formula
Rotational Quantity
Distance
Δs=rΔθ\Delta s = r\Delta\theta
Angular displacement
Velocity
v=rωv = r\omega
Angular velocity
Tangential acceleration
at=rαa_t = r\alpha
Angular acceleration
The radius rr appears in all three, acting as the conversion factor between the two descriptions. Checking units: meters equals meters times radians, m/s equals meters times rad/s, and m/s² equals meters times rad/s², which all work out correctly since radians are dimensionless (a ratio of two lengths).

Real-World Applications

Car speedometers: a speedometer actually measures wheel rotation (ω\omega) but displays linear speed (vv), calculated using v=rωv = r\omega with the wheel’s radius. This is exactly why changing your tire size can throw off speedometer accuracy.
Drill bits: high-speed drills are rated in RPM (angular velocity), but the actual cutting speed at the bit’s edge (linear velocity) is what determines cutting effectiveness. For a 1 cm radius bit at 3000 RPM: ω=3000×2π60=314\omega = 3000 \times \frac{2\pi}{60} = 314 rad/s, giving v=(0.01)(314)=3.14v = (0.01)(314) = 3.14 m/s.
Record players: vinyl records spin at a constant 33⅓ RPM (constant ω\omega), but the needle moves slower near the center and faster near the edge, which is why sound quality can vary slightly across a record, the needle encounters different linear speeds depending on where it is.
Wind turbines: blade tips move much faster than the roots. For a 50 m blade rotating at ω=2\omega = 2 rad/s, the tip speed is v=(50)(2)=100v = (50)(2) = 100 m/s while the root speed is nearly zero, meaning blade tips can reach dangerous speeds even when the rotation looks slow.

Common Mistakes

Forgetting to convert radius to meters: don’t use r=4.5r = 4.5 cm directly in the formulas, convert to meters first, r=0.045r = 0.045 m.
Using degrees instead of radians: don’t plug Δθ=90°\Delta\theta = 90° into Δs=rΔθ\Delta s = r\Delta\theta, convert first, Δθ=π/2\Delta\theta = \pi/2 rad.
Assuming the same linear velocity at every radius: don’t assume two points spinning at the same ω\omega move at the same speed, same ω\omega still gives different vv depending on rr.
Confusing tangential and centripetal: at=rαa_t = r\alpha is tangential (along the circle), not centripetal (toward the center), those are two separate quantities.

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