Constant Rotational Acceleration

Section 9.5 Constant Rotational Acceleration

Constant rotational acceleration provides a good way to do the average case analysis for rotation motion. That makes learning to use this important type of rotation very important. To make it clear that we are using a constant angular acceleration formula, I will denote the constant angular acceleration with a line above the symbol, \(\bar \alpha\text{.}\)

🔗

If a body is rotating with a constant acceleration, its angular velocity will change at a constant rate. For instance, if the angular accleration is \(3\text{ rad/sec}^2\text{,}\) then, \(3\text{ rad/sec}\) of counterclockwise velocity will be added to the angular velocity every second. Thus, if the angular velocity was \(-5\text{ rad/sec}\) at some instant, then, one second later, angular velocity will be \(-2\text{ rad/sec}\text{,}\) and two seconds later, it will be \(1\text{ rad/sec}\text{,}\) etc.

🔗

Thus, a fundamental equation of constant angular acceleration, \(\bar\alpha \text{,}\) would be the following. The change in angular velocity from \(\omega_i \) to \(\omega_f \) during the interval from \(t_i \) to \(t_f\) will be

\begin{equation} \omega_f =\omega_i + \bar\alpha\,\Delta t,\tag{9.13} \end{equation}

where \(\Delta t = t_f - t_i \text{.}\)

🔗

When the angular acceleration is constant, the angular velocity increases or decreases at a steady pace. That means the average angular velocity is just the arithematic average of the initial and final angular velocities.

\begin{equation} \omega_{\text{av}} = \dfrac{\omega_i + \omega_f}{2}\ \ \ \ ( \text{if }\alpha \text{ is consant}.)\tag{9.14} \end{equation}

🔗

The final equation for constant angular acceleration motion is based on using the average angular velocity to get the net angle rotated. Let \(\Delta \theta\) denote the net angle rotated in duration \(\Delta t\text{.}\)

\begin{equation} \Delta \theta = \omega_{\text{av}}\, \Delta t.\tag{9.15} \end{equation}

🔗

From these three equations, we can deduce the following two equations, which are particularly useful in solving problems.

\begin{align} \amp \Delta \theta = \omega_i\, \Delta t + \dfrac{1}{2}\, \bar\alpha\, (\Delta t )^2,\tag{9.16}\\ \amp \omega_f^2 = \omega_i^2 + 2\,\bar\alpha\, \Delta \theta.\tag{9.17} \end{align}

Summary: We sumarize the equations for constant angular acceleration motion now. Recall that for a counterclockwise rotation has \(\Delta \theta\) positive, \(\omega\) positive, but \(\alpha\) positive if speeding up and negative if slowing down. For clockwise rotation, the signs change.

\begin{align*} \amp \omega_f =\omega_i + \bar\alpha\,\Delta t, \\ \amp \omega_{\text{av}} = \dfrac{\omega_i + \omega_f}{2}, \\ \amp \Delta \theta = \omega_{\text{av}}\, \Delta t, \\ \amp \Delta \theta = \omega_i\, \Delta t + \dfrac{1}{2}\, \bar\alpha\, (\Delta t )^2,\\ \amp \omega_f^2 = \omega_i^2 + 2\,\bar\alpha\, \Delta \theta. \end{align*}

🔗

Remark 9.30. Analogy with One-Dimensional Translational Motion.

Looking at the equations for constant angular acceleration you might have the sense of Déjà vu. Yes, constant angular acceleration, just as the one-dimensional constant acceleration are kinematic equations for constant acceleration case along one axis. In the case of rotation, they are components along the axis of rotation and in the case of motion on \(x\) axis, they are \(x\) components. The following list shows the corresponding quantities.

\begin{align*} \amp \Delta x \leftrightarrow \Delta \theta, \\ \amp v_{i,x} \leftrightarrow \omega_i, \\ \amp v_{f,x} \leftrightarrow \omega_f, \\ \amp a_{x} \leftrightarrow \bar{\alpha}, \\ \amp \Delta t \leftrightarrow \Delta t. \end{align*}

🔗

Example 9.31. Rotation of a Wheel with Constant Angular Acceleration.

A wheel is mounted on an axel and rotated from rest with a uniform angular acceleration of \(3\ \text{rad/s}^2\text{.}\) (a) Find the angular velocity at \(t = 10\ \text{s}\text{.}\) (b) Find the total rotation in \(10\ \text{s}\text{.}\)

🔗

Answer.

(a) \(30\,\text{rad/s}\text{,}\) (b) \(150\ \text{rad}\) or \(23\ \text{turns}\) and \(315^\circ\)

🔗

Solution. a,b

Since the angular acceleration is constant, we can use the constant angular acceleration formulas given in this section.

🔗

(a) The angular velocity changes linearly for a constant angular acceleration (suppressing units)

\begin{equation*} \omega = \omega_0 + \alpha t = 0+3 \times 10 = 30\,\text{rad/s}. \end{equation*}

🔗

(b) The net rotation for a constant angular acceleration motion is given by

\begin{equation*} \theta - \theta_0 = \omega_0 t +\frac{1}{2} \alpha t^2 = 0+\frac{1}{2}\times3\times 10^2 = 150\ \text{rad}, \end{equation*}

which is \(150/2\pi\,\text{turns}\) or \(23\ \text{turns}\) and \(315^\circ\text{.}\)

🔗

Example 9.32. Spinning a Centrifuge to Full Speed.

An ultracentrifuge can spin at \(150,000\text{ rpm}\text{,}\) where \(\text{rpm}\) stands for revolutions per minute. Suppose the centrifuge starts from rest and takes \(5\text{ minutes}\) to reach it maximum spin rate. Assume constant rotational acceleration.

🔗

(a) What is the average angular acceleration of the centrifuge?

🔗

(b) How many revolutions did it take to get to the final angular speed?

🔗

Answer.

(a) \(30,000\text{ rev/min}^2\) or \(52.6\text{ rad/sec}^2\text{,}\) (b) \(375,000\text{ revs} \) or \(2,356,194.5\text{ rad}\text{.}\)

🔗

Solution 1. a

Let us work in units of revolutions and minutes. From the given data we get

\begin{equation*} 150,000 = 0 + \bar\alpha\times 5. \end{equation*}

Solving this we get

\begin{equation*} \bar\alpha = 30,000\text{ rev/min}^2. \end{equation*}

We can convert this to \(\text{rad/sec}^2\) by appropriate unit conversions.

\begin{equation*} 30,000\text{ rev/min}^2 \times 2\pi \text{ rad/rev} \times (1 \text{ min}/60\text{ sec})^2 = 52.6\text{ rad/sec}^2. \end{equation*}

🔗

Solution 2. b

Let us work in units of revolutions and minutes. We found \(\bar\alpha = 30,000\text{ rev/min}^2 \text{.}\) Now, we can use it to get \(\Delta \theta\text{.}\)

\begin{align*} \Delta\theta \amp = \omega_i\Delta t + \dfrac{1}{2}\,\bar\alpha\, (\Delta t)^2 \\ \amp = 0 + \dfrac{1}{2}\times 30000 \times 5^2 = 375,000\text{ revs} \end{align*}

We can also write this as radians

\begin{equation*} 375,000\text{ revs}\times 2\pi\text{ rad/rev} = 2,356,194.5\text{ rad}. \end{equation*}

🔗

Example 9.33. Using Vector Form of Angular Quantities for Constant Acceleration Problems.

Starting from rest a wheel rotates an angle of \(200\) radians in \(20\) seconds at a constant angular acceleration. The axis of rotation is pointed towards the North. (a) Find the magnitude and direction of the angular velocity at \(t=20\) sec. (b) Find the magnitude and direction of angular acceleration.

🔗

Answer.

(a) \(20\ \text{rad/s}\text{,}\) North, (b) \(1\ \texxt{rad/s}^2\text{,}\) North.

🔗

Solution. a,b

Here, I will illustrate the use of vector form of angular quantities. Let the \(z\)-axis be pointed towards the North. Note that the magnitude of the angular angular velocity is not \(200\,\text{rad}/20\,\text{s}\text{,}\) i.e, 10 rad/s. This would be the average angular velocity. The angular velocity in this question is the instantaneous angular velocity. We would need to first figure out the angular acceleration. All rotational vector quantities are along the \(z\) axis with constant angular acceleration. Therefore, we us the following equation.

\begin{equation*} \Delta \theta = \omega_{0z} t + \frac{1}{2} \alpha_z t^2. \end{equation*}

With \(\omega_{0z} = 0\text{,}\) \(\Delta\theta = 200\) rad, and \(t = 20\) s. we obtain.

\begin{equation*} \alpha_z = \frac{2\Delta \theta}{t^2} = \frac{2\times 200}{20^2} = 1\ \textrm{rad/s}^2. \end{equation*}

Now, we can obtain the \(z\)-component of the angular velocity at \(t=20\) s.

\begin{equation*} \omega_z = \omega_{0z} + \alpha_z t = 20\ \textrm{rad/s}. \end{equation*}

This result could also be obtained from the average angular velocity \(\omega_{ave} = 10\, \text{rad/s}\) towards the \(z\)-axis. Since the acceleration is constant, the average angular velocity is also equal to the average of the instantaneous velocity at the start and at the end.

\begin{equation*} \omega_z^{ave} = \frac{\omega_{0z} + \omega_{z} }{2} \ \Longrightarrow\ \omega_{z} = 2\omega_z^{ave} - \omega_{0z} = 20\ \textrm{rad/s}. \end{equation*}

We found that the angular acceleration has the magnitude \(1\, \text{rad/s}^2\) and is pointed due North, and the angular velocity at \(20\, \text{sec}\) mark has the magnitude \(20\, \text{rad/s}\) and is pointed due North.

🔗

Exercises Exercises

1. CD-ROM Rotating at Constant Angular Acceleration.

A CD-ROM disk starts to rotate about its axis from rest at a constant angular acceleration of magnitude \(4\, \text{rad/sec}^2\text{.}\) (a) Find the angle in radians the disk rotates in \(15\, \text{sec}\text{.}\) (b) Find the angular velocity vector at the \(15\text{-sec}\) mark

🔗

Answer.

(a) \(450\ \text{rad}\text{,}\) (b) \(60\,\text{rad/s}\text{.}\)

🔗

Solution. a,b

Let \(z\) be the axis of rotation. We point the positive \(z\) axis to be the direction that corresponda to positive angular acceleration given in the problem. With this choice, we have

\begin{equation*} \alpha = 4\, \text{rad/sec}^2,\quad \omega_0 = 0,\quad \theta_0 = 0,\quad T = t - t_0 = 15\,\text{s}. \end{equation*}

Therefore,

\begin{equation*} \theta = \theta_0 + \omega_0 T + \frac{1}{2}\alpha T^2 = 450\ \text{rad}, \end{equation*}

and

\begin{equation*} \omega = \omega_0 + \alpha T = 60\,\text{rad/s}. \end{equation*}

🔗

2. A Wheel Rotating Horizzontally with Constant Angular Acceleration.

At \(t=0\) a wheel is rotating at \(3\, \text{rad/s}\) with the axis pointed up. You give it a steady torque so that the rotation of the wheel picks up angular speed at a steady rate. After some time of constant angular acceleration, you find that the rotational speed of the wheel is \(30\, \text{rad/s}\text{.}\) During this time the wheel had rotated by a total angle of \(20\,\text{rad}\text{.}\) Find the magnitude and direction of the angular acceleration.

🔗

Answer.

Magnitude: \(7.1\, \text{rad/s}^2\text{.}\)

🔗

Solution.

Let the initial angular velocity be towards the positive \(z\)-axis. That is, \(\omega_{0z} = + 3\, \text{rad/s}\text{.}\) After accelerating the \(\omega_z = 30\, \text{rad/s}\text{.}\) In this time the wheel rotates by 20 rad. Since the motion happens with a constant angular acceleration, we will use the following equation to solve for \(\alpha_z\text{.}\)

\begin{equation*} \omega_z^2 = \omega_{0z} + 2 \alpha_z\Delta \theta. \end{equation*}

This gives,

\begin{equation*} \alpha_z = \frac{30^2-3^2}{20} = \frac{894}{20} = 44.7\ \textrm{rad/s}. \end{equation*}

Therefore, the angular acceleration vector has the magnitude \(44.7\, \text{rad/s}\) and direction towards the positive \(z\)-axis.

🔗

3.

A tire of radius 25 cm rolls on a flat road without slipping such that the axis of rotation is always pointed towards the East and the tire is rolling towards the North. (a) Find the angle rotated when the tire moves a distance of \(200\) cm. (b) Set up a Cartesian coordinate system so that one axis is towards the direction of motion of the center of mass and another axis is pointed towards the axis of rotation. (c) Find the relation between the components of the velocity of the center of the wheel and the angular velocity of the wheel.

🔗

Answer.

(a) 8 radians.

🔗

Solution. a,b,c

(a) When the tire rolls, the distance covered by the tire equals the distance covered by a point on the edge of the tire, which would be moving in an the arc. when the tire has rotated one full turn of \(2\pi\) rad, the tire would have gone \(2\pi R\text{.}\)

\begin{equation*} s = R\Delta \theta \ \Longrightarrow\ \Delta \theta = \frac{s}{R} = \frac{200\ \textrm{cm}}{25\ \textrm{cm}} = 8\ \textrm{rad}. \end{equation*}

🔗

(b) See Figure 9.34.

🔗

(c) The non-zero component of the velocity of the center of the wheel is \(v_x\) and the non-zero component of the angular velocity is \(\omega_z\text{.}\) Since the relation between that of the arc length and the angle subtended by the arc at the center we have the following relation between the components here.

\begin{equation*} v_x = R \omega_z. \end{equation*}

🔗

🔗

Prev Top Next