10.3 Relating Angular and Translational Quantities

Angular vs. Linear Variables

In Rotational Variables, we introduced angular variables. If we compare the rotational definitions with the definitions of linear kinematic variables from Motion Along a Straight Line and Motion in Two and Three Dimensions, we find that there is a mapping of the linear variables to the rotational ones. Linear position, velocity, and acceleration have their rotational counterparts, as we can see when we write them side by side:

Let’s compare the linear and rotational variables individually. The linear variable of position has physical units of meters, whereas the angular position variable has dimensionless units of radians, as can be seen from the definition of $\theta =\frac{s}{r}$, which is the ratio of two lengths. The linear velocity has units of m/s, and its counterpart, the angular velocity, has units of rad/s. In Rotational Variables, we saw in the case of circular motion that the linear tangential speed of a particle at a radius r from the axis of rotation is related to the angular velocity by the relation $v_{\text{t}}=r\omega$. This could also apply to points on a rigid body rotating about a fixed axis. Here, we consider only circular motion. In circular motion, both uniform and nonuniform, there exists a centripetal acceleration (Motion in Two and Three Dimensions). The centripetal acceleration vector points inward from the particle executing circular motion toward the axis of rotation. The derivation of the magnitude of the centripetal acceleration is given in Motion in Two and Three Dimensions. From that derivation, the magnitude of the centripetal acceleration was found to be

where r is the radius of the circle.

Thus, in uniform circular motion when the angular velocity is constant and the angular acceleration is zero, we have a linear acceleration—that is, centripetal acceleration—since the tangential speed in Figure is a constant. If nonuniform circular motion is present, the rotating system has an angular acceleration, and we have both a linear centripetal acceleration that is changing (because $v_{\text{t}}$ is changing) as well as a linear tangential acceleration. These relationships are shown in Figure, where we show the centripetal and tangential accelerations for uniform and nonuniform circular motion.

The centripetal acceleration is due to the change in the direction of tangential velocity, whereas the tangential acceleration is due to any change in the magnitude of the tangential velocity. The tangential and centripetal acceleration vectors ${\overrightarrow{\mathbf{a}}}_{\text{t}}$ and ${\overrightarrow{\mathbf{a}}}_{\text{c}}$ are always perpendicular to each other, as seen in Figure. To complete this description, we can assign a total linear acceleration vector to a point on a rotating rigid body or a particle executing circular motion at a radius r from a fixed axis. The total linear acceleration vector $\overrightarrow{\mathbf{a}}$ is the vector sum of the centripetal and tangential accelerations,

The total linear acceleration vector in the case of nonuniform circular motion points at an angle between the centripetal and tangential acceleration vectors, as shown in Figure. Since ${\overrightarrow{\mathbf{a}}}_{\text{c}}\perp {\overrightarrow{\mathbf{a}}}_{\text{t}}$, the magnitude of the total linear acceleration is

Note that if the angular acceleration is zero, the total linear acceleration is equal to the centripetal acceleration.

Relationships between Rotational and Translational Motion

We can look at two relationships between rotational and translational motion.

1. Generally speaking, the linear kinematic equations have their rotational counterparts. Figure lists the four linear kinematic equations and the corresponding rotational counterpart. The two sets of equations look similar to each other, but describe two different physical situations, that is, rotation and translation.
   

   Rotational and Translational Kinematic Equations

   Rotational	Translational
   ${\theta }_{\text{f}}={\theta }_0+\bar{\omega }t$	$x=x_0+\bar{v}t$
   ${\omega }_{\text{f}}={\omega }_0+\alpha t$	$v_{\text{f}}=v_0+at$
   ${\theta }_{\text{f}}={\theta }_0+{\omega }_{0}t+\frac{1}{2}\alpha t^2$	$x_{\text{f}}=x_0+v_{0}t+\frac{1}{2}a{t}^2$
   ${\omega }_{\text{f}}^2={\omega }_0^2+2\alpha (\mathrm{\Delta }\theta )$	$v_{\text{f}}^2=v_0^2+2a(\mathrm{\Delta }x)$

2. The second correspondence has to do with relating linear and rotational variables in the special case of circular motion. This is shown in Figure, where in the third column, we have listed the connecting equation that relates the linear variable to the rotational variable. The rotational variables of angular velocity and acceleration have subscripts that indicate their definition in circular motion.
   

   Rotational and Translational Quantities: Circular Motion

   Rotational	Translational	Relationship ($r=\text{radius}$)
   $\theta$	s	$\theta =\frac{s}{r}$
   $\omega$	$v_{\text{t}}$	$\omega =\frac{v_{\text{t}}}{r}$
   $\alpha$	$a_{\text{t}}$	$\alpha =\frac{a_{\text{t}}}{r}$
   	$a_{\text{c}}$	$a_{\text{c}}=\frac{v_{\text{t}}^2}{r}$

Example

Linear Acceleration of a Centrifuge

A centrifuge has a radius of 20 cm and accelerates from a maximum rotation rate of 10,000 rpm to rest in 30 seconds under a constant angular acceleration. It is rotating counterclockwise. What is the magnitude of the total acceleration of a point at the tip of the centrifuge at $t=29.0\text{s?}$ What is the direction of the total acceleration vector?

Strategy

With the information given, we can calculate the angular acceleration, which then will allow us to find the tangential acceleration. We can find the centripetal acceleration at $t=0$ by calculating the tangential speed at this time. With the magnitudes of the accelerations, we can calculate the total linear acceleration. From the description of the rotation in the problem, we can sketch the direction of the total acceleration vector.

Solution

The angular acceleration is

$\alpha =\frac{\omega -{\omega }_0}{t}=\frac{0-(1.0\times {10}^4)2\pi \text{/}60.0\text{s}(\text{rad}\text{/}\text{s})}{30.0\text{s}}=-34.9\text{rad}\text{/}{\text{s}}^2.$

Therefore, the tangential acceleration is

$a_{\text{t}}=r\alpha =0.2\text{m}(-34.9\text{rad}\text{/}{\text{s}}^2)=-7.0{\text{m}\text{/}\text{s}}^2.$

The angular velocity at $t=29.0\text{s}$ is

$\begin{array}{cc}\hfill \omega & ={\omega }_0+\alpha t=1.0\times {10}^4(\frac{2\pi }{60.0\text{s}})+(-34.9\text{rad}\text{/}{\text{s}}^2)(29.0\text{s})\hfill \\ & =1047.2\text{rad}\text{/}\text{s}-1012.71=35.1\text{rad}\text{/}\text{s}.\hfill \end{array}$

Thus, the tangential speed at $t=29.0\text{s}$ is

$v_{\text{t}}=r\omega =0.2\text{m}(35.1\text{rad}\text{/}\text{s})=7.0\text{m}\text{/}\text{s}.$

We can now calculate the centripetal acceleration at $t=29.0\text{s}$:

$a_{\text{c}}=\frac{v^2}{r}=\frac{{(7.0\text{m}\text{/}\text{s})}^2}{0.2\text{m}}=245.0\text{m}\text{/}{\text{s}}^2.$

Since the two acceleration vectors are perpendicular to each other, the magnitude of the total linear acceleration is

$|\overrightarrow{\mathbf{a}}|=\sqrt{a_{\text{c}}^2+a_{\text{t}}^2}=\sqrt{{(245.0)}^2+{(-7.0)}^2}=245.1\text{m}\text{/}{\text{s}}^2.$

Since the centrifuge has a negative angular acceleration, it is slowing down. The total acceleration vector is as shown in Figure. The angle with respect to the centripetal acceleration vector is

$\theta ={\text{tan}}^{-1}\frac{-7.0}{245.0}=-{1.6}^{\circ }.$

The negative sign means that the total acceleration vector is angled toward the clockwise direction.

Significance

From Figure, we see that the tangential acceleration vector is opposite the direction of rotation. The magnitude of the tangential acceleration is much smaller than the centripetal acceleration, so the total linear acceleration vector will make a very small angle with respect to the centripetal acceleration vector.