Unit 5 · Topic 5.2

5.2 Connecting Linear and Rotational Motion

A point on a spinning object moves along a circle, so its linear motion is tied to the object's rotation: $s = r\theta$ , $v = r\omega$ and $a_t = r\alpha$ . Every point on a rigid object shares the same ω and α, but points farther from the axis move faster.

Key terms

arc length
tangential velocity
tangential acceleration
centripetal acceleration
rigid system

What changed in the 2024 update

Before the Fall 2024 course update, this topic was part of one big Rotation unit (old Unit 5). That unit is now split in two: Unit 5 covers spinning motion, torque and rotational inertia, and Unit 6 covers rotational energy, angular momentum and rolling. Older videos and practice labeled "Unit 5: Rotation" still fit, but they may mix in Unit 6 ideas.

From angles to distances

Picture a point stuck to a wheel at distance r from the axle. When the wheel turns through an angle θ (in radians), the point travels an arc of length $s = r\theta$ . This only works with θ in radians, because that's how the radian is defined.

Take the time derivative of both sides. Since r is constant, the point's speed is $v = r\omega$ . Differentiate again and you get the tangential acceleration $a_t = r\alpha$ . The tangential acceleration points along the circle and changes the point's speed.

Same ω, different speeds

Every point on a rigid object turns through the same angle in the same time. So every point has the same ω and the same α. But the linear quantities grow with r. A point on the rim of a merry-go-round moves faster than a point halfway out, even though both make one turn in the same time.

The axis itself doesn't move at all: r = 0 there, so v = 0.

Quantity	Same for every point?	Depends on r?
θ, ω, α	yes	no
s, v, a_t	no	yes, proportional to r
centripetal acceleration	no	yes, equal to ω²r

Two parts of the acceleration

A point on a spinning object always has a centripetal acceleration toward the axis, because it moves in a circle (topic 2.10). Its size is $a_c = \dfrac{v^2}{r} = \omega^2 r$ .

If the object is also speeding up or slowing down, the point has a tangential acceleration $a_t = r\alpha$ as well. The two parts are perpendicular, so the total acceleration has magnitude $\sqrt{a_t^2 + a_c^2}$ . For a wheel turning at a steady rate, α = 0 and only the centripetal part is left.

Strings, pulleys and gears

When a string wraps around a pulley or spool without slipping, the string moves as fast as the rim. So a hanging block's speed and acceleration match the rim: $v = R\omega$ and $a = R\alpha$ , where R is the radius the string wraps around. This link is how you connect the linear and rotational equations in 5.6.

Two gears or wheels touching at their rims (or joined by a belt) share the same rim speed. So the smaller one spins faster: r₁ω₁ = r₂ω₂.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Speeds and accelerations on a disk
A disk of radius 0.40 m spins at 30 rad/s and is speeding up with α = 5.0 rad/s². For a point on the rim, find the speed, the tangential acceleration, the centripetal acceleration and the total acceleration. What's the speed of a point 0.20 m from the axis?
Show the solutionHide the solution
1. Step 1: Speed: $v = r\omega = (0.40)(30) = 12$ m/s.
2. Step 2: Tangential acceleration: $a_t = r\alpha = (0.40)(5.0) = 2.0$ m/s².
3. Step 3: Centripetal acceleration: $a_c = \omega^2 r = (30)^2(0.40) = 360$ m/s².
4. Step 4: Total: $\sqrt{2.0^2 + 360^2} \approx 360$ m/s², almost all centripetal.
5. Step 5: The point at 0.20 m has the same ω, so its speed is (0.20)(30) = 6.0 m/s, half the rim speed.
Answer: Rim: 12 m/s, a_t = 2.0 m/s², a_c = 360 m/s², total ≈ 360 m/s²; 6.0 m/s at 0.20 m
Example 2Calculator allowed
Bucket on a well
A bucket hangs from a rope wrapped around a drum of radius 0.25 m. The bucket accelerates downward at 2.0 m/s² and the rope doesn't slip. Find the drum's angular acceleration. How much rope unwinds while the drum turns 15 revolutions?
Show the solutionHide the solution
1. Step 1: The rope moves with the drum's rim, so $a = R\alpha$ and $\alpha = \dfrac{a}{R} = \dfrac{2.0}{0.25} = 8.0$ rad/s².
2. Step 2: 15 revolutions is 15 × 2π = 30π rad. The rope length is the arc length: $s = R\theta = (0.25)(30\pi) \approx 23.6$ m.
Answer: α = 8.0 rad/s²; about 23.6 m of rope
Example 3Calculator allowed
Forgetting to convert rpm (classic trap)
A vinyl record turns at 33⅓ rpm. How fast is a point 0.15 m from the center moving?
Show the solutionHide the solution
1. Step 1: The trap is writing v = (0.15)(33.3) = 5.0 m/s. That treats rpm as rad/s.
2. Step 2: Convert: $\omega = 33\tfrac{1}{3}\,\dfrac{\text{rev}}{\text{min}} \times \dfrac{2\pi\text{ rad}}{1\text{ rev}} \times \dfrac{1\text{ min}}{60\text{ s}} \approx 3.49$ rad/s.
3. Step 3: Then $v = r\omega = (0.15)(3.49) \approx 0.52$ m/s.
Answer: About 0.52 m/s

Common mistakes

Thinking points farther from the axis have a larger ω. All points share ω and α; only v, a_t and a_c depend on r.
Calling rα the whole acceleration. A point on a spinning object also has centripetal acceleration ω²r, even when α = 0.
Using s = rθ with θ in degrees or revolutions. It only works in radians.
Using the pulley's outer radius when the string wraps around a smaller inner hub. Use the radius where the string actually leaves.

On the exam

Ratio questions are common: if one point is twice as far from the axis, it has twice the speed and twice the centripetal acceleration, with the same ω.
In pulley problems, write the constraint a = Rα explicitly. It's often the step that links your two equations and earns a point.

Connected topics

Videos

Topic 5.2 - Connecting Linear and Rotational Motion
Lessons With LondotWatch on YouTube (opens in a new tab)
Rotational Motion Made Easy | AP Physics 1 - Unit 5 Lesson 1
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Differences between Angular, Tangential, & Centripetal Acceleration / A Tale of 3 Accelerations
Flipping PhysicsWatch on YouTube (opens in a new tab)
Relating angular and regular motion variables | Physics | Khan Academy
Khan Academy PhysicsWatch on YouTube (opens in a new tab)
Rotational Motion Physics, Basic Introduction, Angular Velocity & Tangential Acceleration
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)
8.2 Circular Motion: Position and Velocity Vectors
MIT OpenCourseWareWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 5.2 Connecting Linear and Rotational Motion. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A disk spins at a constant 15 rad/s about its center. How fast is a point 0.20 m from the center moving?

Question 2 of 4Calculator allowed

A wheel of radius 0.50 m is speeding up. At one instant its angular velocity is 4.0 rad/s and its angular acceleration is 6.0 rad/s². What is the magnitude of the acceleration of a point on the rim at that instant?

Question 3 of 4Calculator allowed

Point P is 0.10 m from the axis of a spinning rigid disk, and point Q is 0.20 m from the axis. The disk turns at constant angular velocity. What is the ratio of Q's centripetal acceleration to P's?

Question 4 of 4Calculator allowed

A light string is wrapped around a pulley of radius 0.10 m. A block tied to the string falls 1.5 m without the string slipping. Through what angle does the pulley turn?

0 of 4 answered