ထာꩻရုဲင်ꩻတကလွီႏနေးဖုံႏရာ

ယိုနောဝ်ꩻ ထွာဒျာႏ ၁၂ တန်ꩻ ရူပဗေဒ အခန်ꩻ ၁ ခရာႏသွူ။

CHAPTER 1

ROTATIONAL MOTION

In Grade 11, the angular displacement, angular velocity and angular acceleration of an object in circular motion have been learnt. In this chapter, the concepts and equations for rotational motion under constant angular acceleration and centripetal acceleration will be studied.

Learning Outcomes

• It is expected that students will differentiate between the circular motion and rotational motion.

• examine rotational motion under constant angular acceleration.

• deduce the relations between angular and linear quantities.

• understand the centripetal acceleration.

• distinguish between the uniform circular motion and nonuniform circular motion.

1.1 ROTATIONAL MOTION UNDER CONSTANT ANGULAR ACCELERATION

In a circular motion, the object just moves in a circle. For example, artificial satellites going around the earth at a constant height.

In rotational motion, the object rotates about an axis. Some of the examples of rotational motion are; rotation of earth about its own axis which creates the cycle of day and night, motion of the blades of a fan, and motion of a Ferris wheel in an amusement park as shown in Figure 1.1 (a), (b) and (c).

A rigid body is a body that does not deform or change shape. That is, no matter how the body moves, the distance between any two particles within the body remains constant. If a rigid body is in rotational motion, all the particles constituting it undergo circular motion about a common axis. When an object (rigid body) rotates, it can speed up (or) slow down. During these time intervals, its angular velocity is changing therefore it has an angular acceleration. If its angular velocity changes at a costant rate, then we can say that the angular acceleration is constant and the motion is called the rotational motion under constant angular acceleration. Rotational variables; such as angular displacement, angular velocity, and angular acceleration have been introduced previously. In this section, these variables are applied to analyze the rotational motion for a rigid body about a fixed axis under a constant angular acceleration. Foran object that rotates with a constant angular acceleration, the kinematic equations for rotational motion can be derived in terms of the angular displacement, angular velocity and angular acceleration of the object at any instant of time.

If the initial angular speed (magnitude of angular velocity of the object at t = 0) is ${\omega }_0$, its angular speed $\omega$ at time t can be derived from the following constant angular acceleration.

$a=\frac{\omega -{\omega }_0}{t}$

We get,

$\omega ={\omega }_0-at(1.1)$

Average angular speed is,

$\overline{\omega }=\frac{\omega +{\omega }_0}{2}(1.2)$

Angular displacement for time interval t is $\theta =\overline{\omega }t$. Using Eq. (1.1) and Eq. (1.2) angular displacement $\theta$ and angular speed $\omega$ can be expressed by the following equations.

$\theta ={\omega }_{0}t+\frac{1}{2}a{t}^2(1.3)$

${\omega }^2={\omega }_0^2+2a\theta (1.4)$

In Grade 10, we derived the kinematic equations that relate acceleration, velocity, distance, and time for linear motion with constant acceleration. We can see that the equations of motion for constant angular acceleration are the same as those for linear motion, with the angular quantities replaced with corresponding linear ones.

Therefore, the equations for constant angular acceleration in rotational motion are analogous to equations for constant acceleration in linear motion as shown in Table 1.1.

Table 1.1 Analogies between kinematic equations of linear and angular motion

Linear motion	Angular motion
$v=v_0+at$	$\omega ={\omega }_0+at$
$s=v_{0}t+\frac{1}{2}a{t}^2$	$\theta ={\omega }_{0}t+\frac{1}{2}a{t}^2$
$v^2=v_0^2+2as$	${\omega }^2={\omega }_0^2+2\alpha \theta$
$\overline{v}=\frac{v+v_0}{2}$	$\overline{\omega }=\frac{\omega +{\omega }_0}{2}$

Example 1.1 The Ferris wheel starts from rest and reaches an angular velocity of 1.5 rad $s^{-1}$ over a 10 s period under the constant angular acceleration. (i) Find the angular acceleration of Ferris wheel. (ii) How many revolutins does it make during 10 s?

Initial angular velocity ${\omega }_0=0$, angular velocity $\omega =1.5rad{s}^{-1},t=10s$

(i) The angular acceleration of Ferris wheel can be calculated by using equations of motion for constant angular acceleration.

$\omega ={\omega }_0+at$

$1.5=0+a\times 10$

$a=0.15rad{s}^{-2}$

(ii) To get number of revolutions in 10 s, first we must find angular displacement.

$\theta ={\omega }_{0}t+\frac{1}{2}a{t}^2$

$=0+\frac{1}{2}\times 0.15\times 100=7.5rad$

Since 1 rev = $2\pi rad$,

$\theta =\frac{7.5}{2\pi }=1.19rev$

Reviewed Exercise

1. Under what condition the angular acceleration of a rotating object is equal to zero?

2. A ball is whirled with constant angular acceleration. From rest, it attains an angular velocity of 25 $rad{s}^{-1}$ after traversing an angular displacement of 41 rad. What is angular acceleration of the ball?

Key Words: rotational motion, angular displacement, angular velocity, angular acceleration

1.2 RELATIONS BETWEEN LINEAR AND ANGULAR QUANTITIES

The relations between linear quantities (s, v, a) and angular quantities ($\theta ,\omega ,a$) for circular motion have been studied in Grade 11. In this section, the relation between these quantities for rotational motion will be studied. All particles of an object rotating around the axis of rotation move in a circular path.

Let us consider a randomly shaped rigid body undergoing a rotational motion as shown in Figure 1.2 (a). All particles of the body move in a circle lying on a plane that is perpendicular to the axis, such that the centre of rotation lies on the axis as shown in Figure 1.2 (b).

If the linear distance of a particle of rotating object measured along the arc is s and the arc subtends angle $\theta$ at the origin, the relation between arc length s and angular displacement $\theta$ , measured in radian, is $s=r\theta$ where r is the radius of the circle.

The instantaneous angular velocity $omega$ of rotational object is the same for all the particles comprising the object. The relation between the magnitude of the linear velocity v and the angular velocity $\omega$ of a particle is $v=r\omega$, where $\omega$ is measured in radian per second. This relation applies to every particle of the rotational object. Since the direction of linear velocity is tangent to the path, the linear velocity is also called tangential velocity.

If the magnitude of the angular velocity of a rotating object changes, the magnitude of linear velocity will also change and the object will have a tangential acceleration. The direction of tangential acceleration is always tangent to the circular path. The tangential acceleration $a_1$, relates to angular acceleration a (measured in radian per second squared) by the equation $a_T=r\alpha$.

The tangential acceleration of a rotating object is a measure of how fast the tangential velocity changes.

For the rotational motion, Table 1.2 expresses the relation between the linear and rotational quantities.

Table 1.2 The relation between the linear and rotational quantities

Quantity	Linear quantities	Rotational quantities	Relation
displacement	$s$	$\theta$	$s=r\theta$
tangential velocity	$v$	$\omega$	$v=r\omega$
tangential acceleration	$a_T$	$\alpha$	$a_T=r\alpha$

The direction of the angular velocity vector is perpendicular to the plane of rotation, and along the axis of rotation. The direction of the angular velocity is susally specified by the right-hand rule. Wrap the right hand around the axis of rotation so that the fingers are pointing in the direction of rotation. The thumb points in the direction of the angular velocity vector.

When the angular velocity is increasing, the angular acceleration vector points in the same direction as the angular velocity. When the angular velocity is decreasing, the angular acceleration vector points in the opposite to the angular velocity.

Example 1.2 A boy steps on a merry-go-round which has a radius of 5 m and is at rest. It starts accelerating at a constant rate up to an angular velocity of 5 rad s¹ in 20 s. What is the distance traversed by the body?

Radius of merry-go-round r = 5 m, initial angular velocity ${\omega }_0=0$, angular velocity in 20 s, $\omega =5rad{s}^{-1},t=20s$ the angular displacement $\theta ;$ that the boy passes through is , $\theta =\overline{\omega }t=(\frac{\omega +{\omega }_0}{2})t$ $=(\frac{5+0}{2})\times 20=50rad$

The distance traversed by the boy is, $s=r\theta$ $5\times 50=250m$

Example 1.3 The wheel of a car with radius 20 cm starts moving. The angular acceleration provided by the engine is 12 rad s². What is the tangential acceleration of rim of the wheel?

Radius of wheel r = 20 cm = 0.2 m,
angular acceleration α = 12 rad s^-2
The tangential acceleration of rim of the wheel is,
${\alpha }_r=r\alpha$ = 0.2 × 12 = 2.4 m s^-2
The direction of tangential acceleration is the tangent to rim of the wheel.

Reviewed Exercise
1. In circular motion, how is a tangential acceleration produced?
2. On a rotating carousel, a child sits on a horse near the outer edge and another child sits on a lion halfway out from the center. (i) Which child has the greater linear velocity? (ii) Which child has the greater angular velocity?

Key Words: linear velocity, tangential acceleration

1.3 CENTRIPETAL ACCELERATION

An object that move in a circle at constant speed v is said to experience uniform circular motion. The magnitude of the velocity remains constant in this case, but the direction of the velocity continuously changes as the object moves around the circle as shown in figure 1.3 (a). The change in direction of velocity constitute an acceleration. Thus an object revolving in a circle is continuously accelerating, even when the speed remain constant (v₁ = v₂ = v). We now investigate this acceleration quantitatively.
Average acceleration has been defined as
$\overline{\overrightarrow{a}}=\frac{{\overrightarrow{v}}_2-{\overrightarrow{v}}_1}{\mathrm{\Delta }t}=\frac{\mathrm{\Delta }\overrightarrow{v}}{\mathrm{\Delta }t}(1.5)$
where $\mathrm{\Delta }\overrightarrow{v}$ is the change in velocity during the time interval $\mathrm{\Delta }t$. Consider the situation in which $\mathrm{\Delta }t$ approaches zero and thus the instantaneous acceleration can be obtained. During the time interval $\mathrm{\Delta }t$, the object moves from point A to point B, covering a distance $\mathrm{\Delta }x$ along the arc which subtands an angle $\mathrm{\Delta }\theta$. The change in the velocity vector is ${\overrightarrow{v}}_2-{\overrightarrow{v}}_1=\mathrm{\Delta }\overrightarrow{v}$, and is shown in Figure 1.3 (b).

If $\mathrm{\Delta }t$ is very small (approaching zero), then $\mathrm{\Delta }x$ and $\mathrm{\Delta }\theta$ are also very small, and , ${\overrightarrow{v}}_2$ will be almost parallel to ${\overrightarrow{v}}_1$. $\mathrm{\Delta }\overrightarrow{v}$ will be essentially perpendicular to both of them as shown in Figure 1.3 (c) and points toward the center of the circle. Acceleration $\overrightarrow{a}$ is in the same direction as $\mathrm{\Delta }\overrightarrow{v}$, it also points toward the center of the circle. Therefore, this acceleration is centripetal acceleration or radial acceleration (since it is directed along the radius towards the center of the circle), and denoted by ${\overrightarrow{a}}_c$.

The magnitude of the centripetal (radial) acceleration can be determined as follow. The triangle CAB of figure 1.3(a) is geometrically similar to vector triangle of Figure 1.3(b).
Since $v_1=v_2=v$
$\frac{\mathrm{\Delta }v}{v}=\frac{\mathrm{\Delta }s}{r}$
When $\mathrm{\Delta }t$ approaches zero, the above equation can be expressed as an equality, so that
$\mathrm{\Delta }v=\frac{v}{r}\mathrm{\Delta }s$ (1.6)
The centripetal accelertion $a_c$ can be obtined as
$a_c=\underset{\mathrm{\Delta }t\to 0}{\lim }\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\underset{\mathrm{\Delta }t\to 0}{\lim }\frac{v}{r}\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}=\frac{v}{r}\underset{\mathrm{\Delta }t\to 0}{\lim }\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}$
But $\underset{\mathrm{\Delta }t\to 0}{\lim }\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}$ is the linear speed $v$ of the object moving in circular motion.
Therefore,
$a_c=\frac{v^2}{r}(1.7)$
Substituting $v=r\omega$ in Eq.(1.7), we get
$a_c=r{\omega }^2(1.8)$

For uniform circular motion, the centripetal acceleration vector points towards the center of the circular path while the linear velocity vector is tangential to the path.

Hence, the centripetal acceleration and linear velocity are perpendicular to each other at every point in the path as shown if Figure 1.4.

Nonuniform Circular Motion

In nonuniform circular motion, an object is moving in a circular path with a varying speed (magnitude of velocity). Therefore, the angular speed of the object is changing and the object experiences the angular acceleration. This angular acceleration gives rise to the tangential acceleration which is tangential to the circle as described previously. In this case, there is tangential acceleration in addition to centripetal acceleration.
Thus, an object in nonuniform circular motion has a resultant acceleration that is the vector sum of the centripetal and tangential accelerations as shown in Figure 1.5.

Unlike tangential acceleration, centripetal acceleration is present in both uniform and nonuniform circular motion.

Note that the two acceleration vectors ${\overrightarrow{a}}_c$ and ${\overrightarrow{a}}_r$ are perpendicular to each other, with ${\overrightarrow{a}}_c$ in the radial direction and ${\overrightarrow{a}}_r$ in the tangential direction. The resultant acceleration points at an angle between ${\overrightarrow{a}}_c$ and ${\overrightarrow{a}}_r$.

The magnitude of resultant acceleration $\overrightarrow{a}$ is $a=\sqrt{a_c^2+a_r^2}.(1.10)$

The direction of resultant acceleration $\overrightarrow{a}$ is, tan $\varphi =\frac{a_r}{a_c}(1.11)$