7.11 Kinematics of rotational motion about a fixed axis

We know the angular velocity \(\boldsymbol{\omega}\) plays the same role in rotation as the linear velocity \(\mathbf{v}\) in translation. We wish to take this analogy further. In doing so we shall restrict the discussion only to rotation about a fixed axis.

For specifying the angular displacement of the rotating body we take any particle 11ke P (FIg.7.33) of the body. Its angular displacement \(\theta\) in the plane it moves is the angular displacement of the whole body; \(\theta\) is measured from a fixed direction in the plane of motion of P, which we take to be the \(x^{\prime}\)-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. F1g. 7.33 also shows \(\theta_0\), the angular displacement at \(t=0\).

We also recall that the angular velocity is the time rate of change of angular displacement, \(\omega=\mathrm{d} \theta / \mathrm{d} t\). Note since the axis of rotation is fixed, there is no need to treat angular velocity as a vector. Further, the angular acceleration, \(\alpha=\) \(\mathrm{d} \omega / \mathrm{d} t\).

The kinematical quantities in rotational motion, angular displacement \((\theta)\), angular velocity \((\omega)\) and angular acceleration \((\alpha)\) 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:
\(
\begin{aligned}
& v=v_0+a t \dots(a)\\
& x=x_0+v_0 t+\frac{1}{2} a t^2 \dots(b)\\
& v^2=v_0^2+2 a x \dots(c)
\end{aligned}
\)

where \(x_0=\) inttal 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:
\(
\begin{aligned}
& \omega=\omega_0+\alpha t \dots(7.38) \\
& \theta=\theta_0+\omega_0 t+\frac{1}{2} \alpha t^2 \dots(7.39) \\
& \text { and } \omega^2=\omega_0^2+2 \alpha\left(\theta-\theta_0\right) \dots(7.40)
\end{aligned}
\)
where \(\theta_0=\) initial angular displacement of the rotating body, and \(\omega_0=\) initial angular velocity of the body.

Example 7.25: Obtain the following equation \(\omega=\omega_0+\alpha t\) from first principles.

Solution:

The angular acceleration is uniform, hence
\(
\frac{\mathrm{d} \omega}{\mathrm{d} t}=\alpha=\text { constant } \dots(i)
\)
Integrating this equation,
\(
\begin{aligned}
\omega & =\int \alpha \mathrm{d} t+c \\
& =\alpha t+c \quad \text { (as } \alpha \text { is constant) }
\end{aligned}
\)
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.

Example 7.26: The angular speed of a motor wheel is increased from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) 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?

Solution:

(i) We shall use \(\omega=\omega_0+\alpha t\)
\(
\begin{aligned}
\omega_0= & \text { initial angular speed in rad } / \mathrm{s} \\
& =2 \pi \times \text { angular speed in rev } / \mathrm{s} \\
& =\frac{2 \pi \times \text { angular speed in rev } / \mathrm{mm}}{60 \mathrm{~s} / \mathrm{min}} \\
& =\frac{2 \pi \times 1200}{60} \mathrm{rad} / \mathrm{s} \\
& =40 \pi \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
Similarly \(\omega=\) final angular speed in \(\mathrm{rad} / \mathrm{s}\)
\(
\begin{aligned}
& =\frac{2 \pi \times 3120}{60} \mathrm{rad} / \mathrm{s} \\
& =2 \pi \times 52 \mathrm{rad} / \mathrm{s} \\
& =104 \pi \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
\(\therefore \quad\) Angular acceleration
\(
\alpha=\frac{\omega-\omega_0}{t} \quad=4 \pi \mathrm{rad} / \mathrm{s}^2
\)
The angular acceleration of the engine \(=4 \pi \mathrm{rad} / \mathrm{s}^2\)

(ii) The angular displacement in time \(t\) is given by
\(
\begin{aligned}
& \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\
= & \left(40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2\right) \mathrm{rad} \\
= & (640 \pi+512 \pi) \mathrm{rad} \\
= & 1152 \pi \mathrm{rad}
\end{aligned}
\)
Number of revolutions \(=\frac{1152 \pi}{2 \pi}=576\)

Example 7.27: The motor of an engine is rotating about its axis with an angular velocity of \(100 \mathrm{~rev} /\) minute. It comes to rest in \(15 \mathrm{~s}\), after being switched off. Assuming constant angular deceleration, calculate the number of revolutions made by it before coming to rest.

Solution:

The initial angular velocity \(=100 \mathrm{rev} / \mathrm{minute}\) \(=(10 \pi / 3) \mathrm{rad} / \mathrm{s}\).
Final angular velocity \(=0\).
Time interval \(=15 \mathrm{~s}\).
Let the angular acceleration be \(\alpha\). Using the equation \(\omega=\omega_0+\alpha t\), we obtain \(\alpha=(-2 \pi / 9) \mathrm{rad} / \mathrm{s}^2\).

The angle rotated by the motor during this motion is
\(
\begin{aligned}
\theta & =\omega_0 t+\frac{1}{2} \alpha t^2 \\
& =\left(\frac{10 \pi}{3} \frac{\mathrm{rad}}{\mathrm{s}}\right)(15 \mathrm{~s})-\frac{1}{2}\left(\frac{2 \pi}{9} \frac{\mathrm{rad}}{\mathrm{s}^2}\right)(15 \mathrm{~s})^2 \\
& =25 \pi \mathrm{rad}=12.5 \text { revolutions. }
\end{aligned}
\)
Hence the motor rotates through 12.5 revolutions before coming to rest.

Example 7.28: Starting from rest, a fan takes five seconds to attain the maximum speed of \(400 \mathrm{rpm}\) (revolutions per minute). Assuming constant acceleration, find the time taken by the fan in attaining half the maximum speed.

Solution:

Let the angular acceleration be \(\alpha\). According to the question,
\(
400 \mathrm{rev} / \mathrm{min}=0+\alpha  5 \mathrm{~s}
\)
Let \(t\) be the time taken in attaining the speed of 200 \(\mathrm{rev} / \mathrm{min}\) which is half the maximum.
Then,
\(
200 \mathrm{rev} / \mathrm{min}=0+\alpha t
\)
Dividing (i) by (ii), we get,
\(
2=5 \mathrm{~s} / t \text { or, } \quad t=2.5 \mathrm{~s} \text {. }
\)

Relation between the Linear Motion of a Particle of a Rigid Body and its Rotation

Consider a point \(P\) of the rigid body rotating about a fixed axis as shown in the figure above. As the body rotates, this point moves on a circle. The radius of this circle is the perpendicular distance of the particle from the axis of rotation. Let it be \(r\). If the body rotates through an angle \(\theta\), so does the radius joining the particle with the centre of its circle. The linear distance moved by the particle is \(s=r \theta\) along the circle.
The linear speed along the tangent is
\(
v=\frac{d s}{d t}=r \cdot \frac{d \theta}{d t}=r \omega
\)
and the linear acceleration along the tangent, i.e., the tangential acceleration, is
\(
a=\frac{d v}{d t}=r \cdot \frac{d \omega}{d t}=r \alpha
\)
The relations \(v=r \omega\) and \(a=r \alpha\) are very useful and their meanings should be clearly understood. For different particles of the rigid body, the radius \(r\) of their circles has different values, but \(\omega\) and \(\alpha\) are same for all the particles. Thus, the linear speed and the tangential acceleration of different particles are different. For \(r=0\), i.e., for the particles on the axis, \(v=r \omega=0\) and \(a=r \alpha=0\), consistent with the fact that the particles on the axis do not move at all.

Example 7.29: A bucket is being lowered down into a well through a rope passing over a fixed pulley of radius \(10 \mathrm{~cm}\). Assume that the rope does not slip on the pulley. Find the angular velocity and angular acceleration of the pulley at an instant when the bucket is going down at a speed of \(20 \mathrm{~cm} / \mathrm{s}\) and has an acceleration of \(4.0 \mathrm{~m} / \mathrm{s}^2\).

Solution:

Since the rope does not slip on the pulley, the linear speed \(v\) of the rim of the pulley is same as the speed of the bucket. The angular velocity of the pulley is then
\(
\omega=v / r=\frac{20 \mathrm{~cm} / \mathrm{s}}{10 \mathrm{~cm}}=2 \mathrm{rad} / \mathrm{s}
\)
and the angular acceleration of the pulley is
\(
\alpha=a / r=\frac{4 \cdot 0 \mathrm{~m} / \mathrm{s}^2}{10 \mathrm{~cm}}=40 \mathrm{rad} / \mathrm{s}^2 .
\)