We have already indicated the analogy between rotational motion and translational motion. For example, the angular velocity ω plays the same role in rotation as the linear velocity v in translation. We wish to take this analogy further. In doing so we shall restrict the discussion only to rotation about fixed axis. This case of motion involves only one degree of freedom, i.e., needs only one independent variable to describe the motion. This in translation corresponds to linear motion. This section is limited only to kinematics. We shall turn to dynamics in later sections.
We recall that for specifying the angular displacement of the rotating body we take any particle like P (Fig.7.33) of the body. Its angular displacement θ in the plane it moves is the angular displacement of the whole body; θ is measured from a fixed direction in the plane of motion of P, which we take to be the x′-axis, chosen parallel to the x-axis. Note, as shown, the axis of rotation is the z – axis and the plane of the motion of the particle is the x - y plane. Fig. 7.33 also shows θ 0, the angular displacement at t = 0.
We also recall that the angular velocity is the time rate of change of angular displacement, ω = dθ/dt. Note since the axis of rotation is fixed, there is no need to treat angular velocity as a vector. Further, the angular acceleration, α = dω/dt.
The kinematical quantities in rotational motion, angular displacement (θ), angular velocity (ω) and angular acceleration (α) respectively are analogous to kinematic quantities in linear motion, displacement (x), velocity (v) and acceleration (a). We know the kinematical equations of linear motion with uniform (i.e. constant) acceleration:
$v=v_0+at$
$x=x_0+v_0t+\frac{1}{2}at^2$
$v^2=v_0^2+2ax$
where $x_0$ = initial displacement and $v_0$= initial velocity. The word ‘initial’ refers to values of the quantities at t = 0.
The corresponding kinematic equations for rotational motion with uniform angular acceleration are:
$\omega=\omega_0+\alpha t$
$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$
and $\omega^2=\omega_0^2+2\alpha (\theta-\theta_0)$
where $\theta_0$ = initial angular displacement of the rotating body, and $\omega_0$ = initial angular velocity of the body.
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| Fig.1 |
Fig.1: Specifying the angular position of a rigid body.
Example 1
Obtain Eq. (7.38) from first principles.
Answer
The angular acceleration is uniform, hence
$\frac{d \omega}{dt}=\alpha$ = constant
Integrating this equation,
$\omega=\int{\alpha}dt+C$
$\omega=\alpha t+C$ (as $\alpha$ is constant)
At t = 0, $\omega=\omega_0$ (given)
From (i) we get at t = 0, $\omega=c=\omega_0$
Thus, $\omega=\alpha t+\omega_0$ as required.
With the definition of ω = dθ/dt we may integrate Eq. (7.38) to get Eq. (7.39). This derivation and the derivation of Eq. (7.40) is left as an exercise.
Example 1
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the acceleration to be uniform? (ii) How many revolutions does the engine make during this time?
Answer
(i) We shall use $\omega = \omega_0 + \alpha t
$\omega_0$ = initial angular speed in rad/s
= 2π × angular speed in rev/s
= (2π × angular speed in rev/min)/(60 s/min)
= (2π × 1200)/60 rad/s
= 40π rad/s
Similarly ω = final angular speed in rad/s
= (2π × 3120)/60 rad/s
= 2π × 52 rad/s
= 104π rad/s
∴ Angular acceleration
$\alpha = \frac{\omega-\omega_0}{t}=4 \pi \ rad/s^2$
The angular acceleration of the engine = 4π $rad/s^2$
(ii) The angular displacement in time t is given by
$\theta=\omega_0t+\frac{1}{2}\alpha t$
$=(40\pi \times 16 +\frac{1}{2}\times 4\pi \times 16^2)$ rad
= 1152π rad
Number of revolutions $=\frac{1152 \pi}{2\pi}$ = 576
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