Connecting Linear and Rotational Motion: Study Guide (Physics 1 5.2)

This guide is for the algebra-based Physics 1 course. Topic 5.1 described a rotating body as a whole, using θ, ω and α. This topic zooms in on one point of the body and asks: how far does it travel, how fast does it move and how quickly does its speed change? No calculus is needed.

One body, many paths

Watch a spinning wheel. Every point goes round in a circle centred on the axis. In one full turn:

• every point turns through the same angle, 2π rad;
• a point near the axle travels round a small circle;
• a point on the rim travels round a much larger circle.

So the angular quantities are shared, but the linear quantities (distance travelled, speed, tangential acceleration) depend on how far the point is from the axis. The link between them is the radius r.

Arc length and the radian

The radian is defined so that the link is as simple as possible. For a point at distance r from a fixed axis, turning through an angle Δθ (in radians) carries it along an arc of length

Δs = rΔθ

This is the definition of the radian rearranged: an angle in radians is arc length divided by radius. One full turn is 2π rad, and the arc is then 2πr, the circumference, as it should be.

Example. A wheel of radius 0.30 m turns through 2.0 rad. A point on its rim travels 0.30 m × 2.0 = 0.60 m along its circular path.

Because radians are a ratio of two lengths, they have no physical dimension. That is why “m × rad” simply becomes “m”. It also means Δs = rΔθ is false if Δθ is in degrees or revolutions. Convert first.

Linear speed and tangential acceleration

Divide both sides of Δs = rΔθ by the time interval Δt. Since r is fixed for a point on a rigid body:

Δs / Δt = r(Δθ / Δt), so v = rω

Here v is the point’s linear speed along its path. Its velocity is always tangent to the circle, perpendicular to the radius.

Now look at how v changes. If ω changes by Δω, then v changes by rΔω. Divide by Δt:

Δv / Δt = r(Δω / Δt), so a_T = rα

a_T is the tangential acceleration: the part of the acceleration along the path that changes the point’s speed. (Topic 2.9 wrote it as a_t; it is the same quantity.)

Rotational quantity (shared by every point)	Linear quantity for a point at radius r
angular displacement Δθ (rad)	distance along the arc, Δs = rΔθ (m)
angular velocity ω (rad/s)	linear speed, v = rω (m/s)
angular acceleration α (rad/s²)	tangential acceleration, a_T = rα (m/s²)

The relationships work in both directions. If you know how fast a string comes off a spool, you can find the spool’s ω from ω = v / r.

Directions in this course are described only as clockwise or counterclockwise. If a disc turns counterclockwise, each point’s velocity is tangent to its circle in the counterclockwise sense.

Same ω, different v: comparing points

For a rigid body, all points have the same ω and the same α at any instant. Then:

• v = rω: linear speed is proportional to r. A point twice as far from the axis moves twice as fast.
• a_T = rα: tangential acceleration is also proportional to r.
• a_c = v²/r = (rω)²/r = ω²r: the centripetal acceleration from Topic 2.9 is also proportional to r when ω is shared.

That last result surprises many students, who remember a_c = v²/r and expect a smaller r to give a bigger a_c. That is true only when v is held fixed. On one rigid body, ω is the same for every point, so a_c = ω²r grows with r.

Things that share a linear speed

When two rotating parts are linked so that nothing slips, the linear speed at the contact is shared:

• Belt or chain drives. The belt moves at one speed, so the rims of both pulleys move at that speed: r₁ω₁ = r₂ω₂. The smaller pulley turns faster.
• String on a spool or winch. A string unwinding without slipping moves at v = rω of the drum, and its acceleration equals a_T = rα. The length unwound equals rΔθ.
• Gears in mesh. The teeth at the contact point move together, so r₁ω₁ = r₂ω₂ again (meshing gears turn in opposite senses; a belt keeps the same sense).

So when you see a belt, chain or string, think “same v”. When you see one rigid body, think “same ω”.

Worked example 1: a winch lifting a load

Question. A builder’s winch has a drum of radius 0.15 m. A cable wrapped round it lifts a bucket. The drum starts from rest and has a constant angular acceleration of 2.0 rad/s² for 3.0 s. The cable does not slip or stretch. Find (a) the drum’s angular velocity and angle turned after 3.0 s, and (b) the bucket’s speed, acceleration and the height it has risen.

1. Drum: ω = ω₀ + αt = 0 + 2.0 × 3.0 = 6.0 rad/s. Δθ = ½αt² = ½ × 2.0 × 3.0² = 9.0 rad (about 1.4 turns).
2. The cable moves with the rim, and the bucket moves with the cable.
3. Speed: v = rω = 0.15 × 6.0 = 0.90 m/s (upward).
4. Acceleration: a = rα = 0.15 × 2.0 = 0.30 m/s² (upward).
5. Height: Δs = rΔθ = 0.15 × 9.0 = 1.35 m (about 1.4 m).

Check. Treat the bucket with Unit 1 kinematics: from rest with a = 0.30 m/s² for 3.0 s, v = 0.30 × 3.0 = 0.90 m/s and Δy = ½ × 0.30 × 3.0² = 1.35 m. ✓ The rotational and linear pictures give the same answer, as they must.

Worked example 2: a belt drive

Question. A motor pulley of radius 4.0 cm turns at 1500 rpm. A belt connects it to a second pulley of radius 12 cm. The belt does not slip. Find (a) the belt’s speed and (b) the angular velocity of the large pulley, in rad/s and rpm.

1. Convert: ω₁ = 1500 × 2π ÷ 60 = 50π ≈ 157 rad/s.
2. Belt speed = rim speed of the motor pulley: v = r₁ω₁ = 0.040 m × 157.1 rad/s ≈ 6.3 m/s.
3. The large pulley’s rim moves at the same v: ω₂ = v / r₂ = 6.283 ÷ 0.12 ≈ 52 rad/s.
4. In rpm: 52.36 × 60 ÷ 2π = 500 rpm.

Check. r₁ω₁ = r₂ω₂ gives ω₂ = ω₁ × (r₁ / r₂) = 1500 rpm × (4.0 ÷ 12) = 500 rpm. ✓ The pulley three times larger turns three times more slowly. Notice that the ratio method works in rpm directly, because the conversion factor cancels; but the belt speed itself needs rad/s.

Worked example 3: two points on a fan blade

Question. A ceiling fan is speeding up. At one instant its angular velocity is 8.0 rad/s and its angular acceleration is 2.5 rad/s². Point P is 0.20 m from the axis; point Q, at the blade tip, is 0.60 m from the axis. Compare their linear speeds, tangential accelerations and centripetal accelerations.

Quantity	P (r = 0.20 m)	Q (r = 0.60 m)
ω	8.0 rad/s	8.0 rad/s
α	2.5 rad/s²	2.5 rad/s²
v = rω	1.6 m/s	4.8 m/s
a_T = rα	0.50 m/s²	1.5 m/s²
a_c = ω²r	12.8 m/s²	38.4 m/s²

Q is three times as far from the axis, so v, a_T and a_c are all three times larger at Q. The angular quantities are equal.

Interpretation. At this instant a_c is much bigger than a_T at both points. The total acceleration of the tip has size √(38.4² + 1.5²) ≈ 38 m/s², pointing almost straight at the axis (only about 2° from it). As ω keeps growing, a_c grows as ω², while a_T stays fixed as long as α does.

Check. a_c at Q using v²/r: 4.8² ÷ 0.60 = 38.4 m/s². ✓

Common misconceptions

• “Points farther out have a bigger angular velocity.” They have a bigger linear speed. ω is the same for the whole rigid body.
• “Pulleys joined by a belt turn at the same rate.” They share the belt’s linear speed. The smaller pulley has the larger ω.
• Using degrees, revolutions or rpm in v = rω. Use rad and rad/s. Otherwise ω is off by a factor of about 57.3 (degrees), 2π (revolutions) or 9.5 (rpm), and so is v.
• “a_T = rα is the acceleration of the point.” It is only the tangential part. There is also a centripetal part ω²r, even when α = 0.
• “Smaller radius means bigger centripetal acceleration.” Only when v is fixed. For points on one rigid body ω is fixed, so a_c = ω²r is larger farther out.
• Mixing up r and the length of string. In Δs = rΔθ, r is the radius of the drum, not the length of cable wound on it.

Where this leads

You now have the full kinematic picture of rotation. Topic 5.3, Torque, asks what makes a body’s angular velocity change. Later, Topic 6.5 uses v = rω for bodies that roll without slipping. Try the practice questions now, then use the revision notes and the checklist. You can also go back to Topic 5.1, Rotational Kinematics, or return to the course roadmap.