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Connecting Linear and
Rotational Motion

Discover how translational quantities like velocity and acceleration map onto their rotational counterparts through the radius of circular motion.

AP Physics 1

What Is Rotational Motion?

When an object moves along a straight line, we describe it with displacement, velocity, and acceleration. But when an object spins or moves in a circle, we need a parallel set of rotational (angular) quantities to describe the same physics from a new perspective.

The Bridge Variable: Radius \(r\)

\(r\) \(v\) (tangential) \(\omega\) center

The radius \(r\) is the key that converts between linear and angular worlds. A point on a spinning wheel has a tangential speed that depends on how far it sits from the center.

Key Angular Quantities

\(\theta\) : Angular Displacement
The angle (in radians) through which an object rotates. Analogous to displacement \(x\).
\(\omega\) : Angular Velocity
Rate of change of angle: \(\omega = \frac{d\theta}{dt}\). Analogous to velocity \(v\).
\(\alpha\) : Angular Acceleration
Rate of change of angular velocity: \(\alpha = \frac{d\omega}{dt}\). Analogous to acceleration \(a\).

The Core Conversion Equations

Arc Length
$$ s = r\theta $$
Tangential Velocity
$$ v = r\omega $$
Tangential Acceleration
$$ a_t = r\alpha $$

Visualizing the Connection

\(\theta\) (rad) \(\omega\) (rad/s) \(\alpha\) (rad/s\(^2\)) \(s\) (m) \(v\) (m/s) \(a\) (m/s\(^2\)) \(\times r\) \(\times r\) \(\times r\) ANGULAR LINEAR \(a_c = \omega^2 r = \frac{v^2}{r}\)

Multiply by \(r\) to go from angular to linear. Divide by \(r\) to go back. The centripetal acceleration \(a_c = \omega^2 r\) is the extra piece that points toward the center.

Don't Forget Centripetal Acceleration

An object in circular motion has two components of acceleration:

Tangential \(a_t\)
Changes the speed of the object along the circle: \(a_t = r\alpha\). Points along the direction of motion.
Centripetal \(a_c\)
Changes the direction of the velocity: \(a_c = \frac{v^2}{r} = \omega^2 r\). Points toward the center.

Linear vs. Rotational Kinematics

Linear Rotational
Displacement \(x\) \(\theta\)
Velocity \(v = v_0 + at\) \(\omega = \omega_0 + \alpha t\)
Displacement Eq. \(x = v_0 t + \tfrac{1}{2}at^2\) \(\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2\)
Velocity-Displacement \(v^2 = v_0^2 + 2ax\) \(\omega^2 = \omega_0^2 + 2\alpha\theta\)
Inertia \(m\) \(I\) (moment of inertia)
Newton's 2nd Law \(F = ma\) \(\tau = I\alpha\)

Rolling Without Slipping

\(R\) contact point \(v_{\text{cm}}\) \(\omega\) \(v_{\text{cm}} = R\omega\) no-slip condition

A tactile rolling race with visible mass distribution

A tabletop ramp with a hoop, disk and sphere, the latter two showing conceptual exposed interiors.For bodies rolling without slipping from the same initial center-of-mass height, mass distribution determines acceleration and the partition of kinetic energy. Equal elapsed time and equal height loss are separate comparisons.

Velocities around a rolling wheel

Velocities around a rolling wheel. In the stationary ground frame, translation and rotation combine to give zero velocity at the contact point, v at the center and 2v at the top. Zero instantaneous velocity does not imply zero acceleration.In the stationary ground frame, translation and rotation combine to give zero velocity at the contact point, v at the center and 2v at the top. Zero instantaneous velocity does not imply zero acceleration.

How to Apply the Conversions

1
Identify the axis of rotation and determine the radius \(r\) from the axis to the point of interest.
2
Choose your domain: if given angular quantities (\(\omega, \alpha, \theta\)), use rotational kinematics first. If given linear quantities (\(v, a, s\)), use linear kinematics first.
3
Convert between domains using \(s = r\theta\), \(v = r\omega\), or \(a_t = r\alpha\).
4
Check for centripetal acceleration: if the object is in circular motion, include \(a_c = v^2/r\) when computing total acceleration.
5
Verify units: angles must be in radians for the conversion formulas to work. Convert revolutions or degrees first.

Example 1: Spinning Wheel

A bicycle wheel of radius \(r = 0.35\) m rotates at \(\omega = 12\) rad/s. Find the tangential speed and centripetal acceleration of a point on the rim.

Example 2: Accelerating Disk

A disk of radius \(r = 0.20\) m starts from rest and reaches \(\omega = 30\) rad/s in \(t = 5\) s. Find the angular acceleration, the tangential acceleration of a rim point, and the total distance traveled by a point on the rim.

1
Angular acceleration: \(\alpha = \frac{\omega - \omega_0}{t} = \frac{30 - 0}{5} =\) \(6\) rad/s\(^2\)
2
Tangential acceleration: \(a_t = r\alpha = (0.20)(6) =\) \(1.2\) m/s\(^2\)
3
Angular displacement: \(\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = 0 + \tfrac{1}{2}(6)(25) = 75\) rad
4
Arc length: \(s = r\theta = (0.20)(75) =\) \(15\) m

Example 3: Rolling Ball

A bowling ball of radius \(R = 0.11\) m rolls without slipping down a lane at \(v_{\text{cm}} = 6.0\) m/s. What is its angular velocity?

Points at Different Radii

\(v_1 = r_1\omega\) \(v_2 = r_2\omega\) \(r_1\) \(r_2\) \(\omega\) Same \(\omega\) Smaller \(r \rightarrow\) slower \(v\) Larger \(r \rightarrow\) faster \(v\)

Comparing two radii

In this separate illustration, both points share angular velocity and angular acceleration. Their tangential speeds and accelerations depend on radius; the companion diagram shows both acceleration values.

At the illustrated instant, ω = 4 rad/s. Points at radii 0.05 m and 0.10 m move at 0.20 m/s and 0.40 m/s. A shared 90° rotation traces arcs of 0.0785 m and 0.1571 m. The required companion adds the instantaneous tangential accelerations.
At the illustrated instant, ω = 4 rad/s. Points at radii 0.05 m and 0.10 m move at 0.20 m/s and 0.40 m/s. A shared 90° rotation traces arcs of 0.0785 m and 0.1571 m. The required companion adds the instantaneous tangential accelerations.
At the instant shown, ω = 4 rad/s and α = 2 rad/s². Both points share angular motion; v = rω and a_t = rα, so the outer point has twice the speed and twice the tangential acceleration. The 90° arc lengths are 0.0785 m and 0.1571 m.
At the instant shown, ω = 4 rad/s and α = 2 rad/s². Both points share angular motion; v = rω and a_t = rα, so the outer point has twice the speed and twice the tangential acceleration. The 90° arc lengths are 0.0785 m and 0.1571 m.

Common Mistakes

✕ Using Degrees
The formulas \(s = r\theta\) and \(v = r\omega\) only work in radians. Convert degrees to radians first: multiply by \(\frac{\pi}{180}\).
✕ Confusing \(a_t\) and \(a_c\)
Tangential acceleration changes speed; centripetal acceleration changes direction. They are perpendicular, not the same.
✕ Same \(v\) for All Points
All points on a rigid body share \(\omega\), but not \(v\). A point at twice the radius has twice the tangential speed.

Summary of Connections

Quantity Angular Relation Linear
Displacement \(\theta\) \(s = r\theta\) \(s\)
Velocity \(\omega\) \(v = r\omega\) \(v\)
Tangential Accel. \(\alpha\) \(a_t = r\alpha\) \(a_t\)
Centripetal Accel. \(\omega^2\) \(a_c = \omega^2 r\) \(v^2/r\)
Rolling Constraint \(\omega\) \(v_{\text{cm}} = R\omega\) \(v_{\text{cm}}\)

Question 1 of 5

A fan blade of radius 0.50 m spins at \(\omega = 20\) rad/s. What is the tangential speed of the tip?

Question 2 of 5

A wheel accelerates uniformly from rest to \(\omega = 40\) rad/s in 8 seconds. If the wheel has radius 0.25 m, what is the tangential acceleration of a point on the rim?

Question 3 of 5

Two points, A and B, are on the same rotating disk. Point A is at radius \(r\) and Point B is at radius \(2r\). How do their tangential speeds compare?

Question 4 of 5

A car tire of radius 0.30 m rolls without slipping. If the car moves at 15 m/s, what is the angular velocity of the tire?

Question 5 of 5

A point on the edge of a disk moves with tangential speed \(v = 3\) m/s at radius \(r = 0.5\) m. What is the centripetal acceleration?