5.2 Connecting Linear and Rotational Motion

Connecting linear and rotational motion means translating angular quantities into the linear motion of a point on a rotating rigid body. For a point a distance $r$ from a fixed axis, the core relationships are $s = r\theta$, $v = r\omega$, and $a_T = r\alpha$.

When an object rotates, different points on that object can move with different linear speeds, but all points in the same rigid body share the same angular velocity and angular acceleration. That difference between shared angular motion and radius-dependent linear motion is the main idea of AP Physics C: Mechanics Topic 5.2.

Linear Motion of Rotating Points

more resources to help you study

practice multiple choiceFRQ practice & scoringcheatsheetsscore calculator

Distance and Angle Relationship

When a point rotates around a fixed axis, its linear motion along the circular path directly relates to the angle it sweeps through. This fundamental relationship connects the linear world to the rotational world.

• The linear distance (arc length) $s$ traveled by a point is calculated using $\Delta s = r \Delta \theta$
• This means the distance traveled equals the radius multiplied by the angle (in radians)
• For a complete circle (2π radians), the distance traveled equals the circumference ($2\pi r$)

Velocity and Acceleration Relationships

Points at different distances from the rotation axis move at different linear speeds, even though they share the same angular velocity. This creates important relationships between linear and angular quantities.

Linear velocity $v$ relates to angular velocity $\omega$ through:

• $v = r\omega$ (linear velocity equals radius times angular velocity)
• Points farther from the axis move faster linearly but rotate at the same angular rate
• Direction of linear velocity is always tangent to the circular path

Linear acceleration has two components in rotational motion:

• Tangential acceleration: $a_T = r\alpha$ (relates to changes in speed)
• Centripetal acceleration: $a_C = r\omega^2 = \frac{v^2}{r}$ (relates to changes in direction)

For example:

• A point on a CD at radius 6 cm rotating at 33.3 rpm (≈ 3.5 rad/s) has a linear velocity of: $v = (0.06 \text{ m})(3.5 \text{ rad/s}) = 0.21 \text{ m/s}$
• If the CD speeds up with angular acceleration 2 rad/s², this point experiences a tangential acceleration of: $a_T = (0.06 \text{ m})(2 \text{ rad/s}^2) = 0.12 \text{ m/s}^2$

Angular Motion in Rigid Bodies

In a rigid body, all points move together in a coordinated way, creating consistent angular motion throughout the object.

• Every point in a rigid body has the same angular velocity $\omega$ and angular acceleration $\alpha$
• Points at different distances from the axis have different linear velocities and accelerations
• The rigid structure ensures all points rotate through the same angle in the same time interval
• A wheel rolling without slipping is a perfect example: all points rotate with the same angular velocity, but their linear motions vary dramatically

This relationship allows us to analyze complex rotational systems by focusing on the angular quantities that remain constant throughout the rigid body.

Practice Problem 1: Linear and Angular Velocity

Solution

For the linear speed of a point on the rim, we use the relationship $v = r\omega$: $v = (0.25 \text{ m})(12 \text{ rad/s}) = 3 \text{ m/s}$

For the tangential acceleration, we use $a_T = r\alpha$: $a_T = (0.25 \text{ m})(-3 \text{ rad/s}^2) = -0.75 \text{ m/s}^2$

The negative sign indicates the acceleration is in the opposite direction of the velocity, causing the wheel to slow down.

Practice Problem 2: Distance Traveled During Rotation

Solution

To find the linear distance (arc length) traveled by the point, we use the relationship $s = r\theta$: $s = (0.10 \text{ m})(\pi/2 \text{ rad})$ $s = (0.10 \text{ m})(1.57 \text{ rad}) = 0.157 \text{ m}$

The point travels approximately 15.7 cm along its circular path when the rigid body rotates through a quarter turn (π/2 radians).