4.5 Relation Between Linear and Angular Quantities

Introduction

When a rigid body rotates about a fixed axis, the body as a whole undergoes angular motion, described by variables such as angular displacement, angular velocity, and angular acceleration. However, each individual point on the body moves along a circular path and possesses its own linear motion. There exists a direct mathematical relationship between the angular variables of the rotating body and the linear variables of a point at a specific distance from the axis of rotation.

Relationship Between Linear and Angular Displacement

Linear Displacement ($S$): The distance a point travels along its circular path, also known as the arc length.

Angular Displacement ($\theta$): The angle through which the rotating object has turned. For these relationships to hold, $\theta$ must be measured in Supplementary Units→ (radians).

Derivation and Formula

By the definition of an angle in radians, angular displacement equals the ratio of arc length to the radius of the circular path:

$\theta =\frac{S}{R}$

Rearranging this equation gives:

$S=R\theta$

This equation states that the linear distance traveled by a point on a rotating body equals the radius multiplied by the angular displacement (in radians).

Relationship Between Linear and Angular Velocity

Linear Velocity ($V$): The instantaneous speed of a point along its circular path. Since the path is circular, this velocity is always tangent to the circle and is termed tangential velocity.

Angular Velocity ($\omega$): The rate at which the angular displacement of the object changes.

Derivation and Formula

Consider the change in displacement over a small time interval $\Delta t$:

$\Delta S=R\Delta \theta$

Dividing both sides by $\Delta t$:

$\frac{\Delta S}{\Delta t}=R\frac{\Delta \theta }{\Delta t}$

As $\Delta t$ approaches zero, these ratios become the instantaneous linear and angular velocities:

$V=R\omega$

In vector form, the relationship is expressed as: $\vec{v}=\vec{\omega }\times \vec{r}$

This result indicates that the tangential speed of a point on a rotating object is directly proportional to its distance ($R$) from the axis of rotation.

Relationship Between Linear and Angular Acceleration

Linear Acceleration ($a$): The rate at which the linear velocity of a point changes. In circular motion, we specifically consider tangential acceleration ($a_t$), which is responsible for changes in the speed of the point.

Angular Acceleration ($\alpha$): The rate at which the angular velocity of the object changes.

Derivation and Formula

Differentiating the velocity equation with respect to time:

$\Delta V=R\Delta \omega$

Dividing both sides by $\Delta t$:

$\frac{\Delta V}{\Delta t}=R\frac{\Delta \omega }{\Delta t}$

As $\Delta t$ approaches zero, these ratios become the instantaneous tangential and angular accelerations:

$a_t=R\alpha$

In vector form: $\overrightarrow{a_t}=\vec{\alpha }\times \vec{r}$

This relationship shows that the tangential acceleration of a point equals its distance from the axis multiplied by the angular acceleration of the rotating body.

Summary

The linear and angular quantities describing rotational motion are fundamentally linked through the radius of the circular path. The three essential relationships are:

These equations enable the conversion between linear and rotational frames of reference.

Frequently Asked Questions

Q: Why must angular quantities be in radians for these formulas to work?

A: The fundamental definition linking arc length and angle, $S=R\theta$, is based on the definition of a radian. If degrees were used, a conversion factor of ($\pi /180$) would be required in all three equations.

Q: Do all points on a spinning wheel have the same angular velocity?

A: Yes. For a rigid body, every point completes a full circle in the same time and rotates through the same angle in any given interval. Therefore, their angular velocity ($\omega$) is identical. However, their linear velocities ($V=R\omega$) differ because their distances from the axis ($R$) are different.

Worked Examples

Example 1: A wheel of radius $0.5\mathrm{m}$ rotates with an angular acceleration of $2rad/{\mathrm{s}}^2$. Find the tangential acceleration of a point on the rim.

Solution: Using $a_t=R\alpha$: $a_t=(0.5\mathrm{m})(2rad/{\mathrm{s}}^2)=1m/{\mathrm{s}}^2$

Example 2: A point on a rotating turntable travels an arc length of $2\mathrm{m}$ during an angular displacement of $4\mathrm{rad}$. Find the radius of the circular path.

Solution: Using $S=R\theta$: $R=\frac{S}{\theta }=\frac{2\mathrm{m}}{4\mathrm{rad}}=0.5\mathrm{m}$

Example 3: A point on a rotating object has a tangential acceleration of $5m/{\mathrm{s}}^2$ and is located at a distance of $0.2\mathrm{m}$ from the axis. Find the angular acceleration.

Solution: Using $a_t=R\alpha$: $\alpha =\frac{a_t}{R}=\frac{5m/{\mathrm{s}}^2}{0.2\mathrm{m}}=25rad/{\mathrm{s}}^2$