# Relationship 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 (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: $$ \vec{a_t} = \vec{\alpha} \times \vec{r} $$ 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: | Physical Quantity | Relationship | Notes | |---|---|---| | Displacement | $S = R\theta$ | $\theta$ must be in radians | | Velocity | $V = R\omega$ | $V$ is tangential velocity; $\omega$ must be in rad/s | | Acceleration | $a_t = R\alpha$ | $a_t$ is tangential acceleration; $\alpha$ must be in rad/s² | 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 $2\ \mathrm{rad/s^2}$. Find the tangential acceleration of a point on the rim. **Solution:** Using $a_t = R\alpha$: $$ a_t = (0.5\ \mathrm{m})(2\ \mathrm{rad/s^2}) = 1\ \mathrm{m/s^2} $$ **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 $5\ \mathrm{m/s^2}$ and is located at a distance of $0.2\ \mathrm{m}$ from the axis. Find the angular acceleration. **Solution:** Using $a_t = R\alpha$: $$ \alpha = \frac{a_t}{R} = \frac{5\ \mathrm{m/s^2}}{0.2\ \mathrm{m}} = 25\ \mathrm{rad/s^2} $$

Relation Between Linear And Angular Quantities | Rotational And Circular Motion | Physics 11