6.6 Non-uniform circular motion  (Page 3/4)

Angular acceleration

The magnitude of angular acceleration is the ratio of angular speed and time interval.

If the ratio is evaluated for finite time interval, then the ratio is called average angular acceleration and If the ratio is is evaluated for infinitesimally small period (Δ t →0), then the ratio is called instantaneous angular acceleration. Mathematically, the instantaneous angular acceleration is :

The angular acceleration is measured in “ $\mathrm{rad}/s^2$ ”. It is important to emphasize here that this angular acceleration is associated with the change in angular speed (ω) i.e. change in the linear speed of the particle (v = ωr) - not associated with the change in the direction of the linear velocity ( v ). In the case of uniform circular motion, ω = constant, hence angular acceleration is zero.

Relationship between linear and angular acceleration

We can relate angular acceleration (α) with tangential acceleration ( $a_T$ ) in non – uniform circular motion as :

$$\begin{array}{l}{a}_T=\frac{đv}{đt}=\frac{{đ}^{2}s}{đt^2}=\frac{{đ}^2}{đt^2}(r\theta )=r\frac{{đ}^2\theta }{đt^2}\\ \Rightarrow a_T=\alpha r\end{array}$$

We see here that angular acceleration and tangential acceleration are representation of the same aspect of motion, which is related to the change in angular speed or the equivalent linear speed. It is only the difference in the manner in which change of the magnitude of motion is described.

The existence of angular or tangential acceleration indicates the presence of a tangential force on the particle.

Note : All relations between angular quantities and their linear counterparts involve multiplication of angular quantity by the radius of circular path “r” to yield to corresponding linear equivalents. Let us revisit the relations so far arrived to appreciate this aspect of relationship :

Linear and angular acceleration relation in vector form

We can represent the relation between angular acceleration and tangential acceleration in terms of vector cross product :

$$\begin{array}{l}{\mathbf{a}}_{\mathbf{T}}=\mathbf{\alpha }\mathbf{X}\mathbf{r}\end{array}$$

The order of quantities in vector product is important. A change in the order of cross product like ( $\mathbf{r}\mathbf{X}\mathbf{\alpha }$ ) represents the product vector in opposite direction. The directional relationship between thee vector quantities are shown in the figure. The vectors “ ${\mathbf{a}}_{\mathbf{T}}$ ” and “ r ” are in “xz” plane i.e. in the plane of motion, whereas angular acceleration ( α ) is in y-direction i.e. perpendicular to the plane of motion. We can know about tangential acceleration completely by analyzing the right hand side of vector equation. The spatial relationship among the vectors is automatically conveyed by the vector relation.

We can evaluate magnitude of tangential acceleration as :

$$\begin{array}{l}{\mathbf{a}}_{\mathbf{T}}=\mathbf{\alpha }\mathbf{X}\mathbf{r}\\ \Rightarrow a_T=\left|{\mathbf{a}}_{\mathbf{T}}\right|=\alpha r\sin \theta \end{array}$$

where θ is the angle between two vectors α and r . In the case of circular motion, θ = 90°, Hence,

$$\begin{array}{l}{a}_T=\left|\mathbf{\alpha }\mathbf{X}\mathbf{r}\right|=\alpha r\end{array}$$

Uniform circular motion

In the case of the uniform circular motion, the speed (v) of the particle in uniform circular motion is constant (by definition). This implies that tangential acceleration, $a_T$ , is zero. Consequently, angular acceleration ( $\frac{a_T}{r}$ ) is also zero.