Machine Mechanics: Linear and Angular Motion Kinetics

Linear Displacement, Velocity, and Acceleration

Linear displacement is defined as the distance moved by a body with respect to a certain fixed point. This displacement may occur along a straight or a curved path. In the context of a reciprocating steam engine, particles on the piston, piston rod, and cross-head trace a straight path. Conversely, particles on the crank and crank pin trace circular paths centered on the axis of the crankshaft. It is specifically noted that particles on the connecting rod trace neither a straight nor circular path, but an oval path where the radius of curvature changes over time. Displacement is a vector quantity, possessing both magnitude and direction, and can be represented graphically by a straight line.

Linear velocity is defined as the rate of change of linear displacement of a body with respect to time. As velocity is expressed in a specific direction, it is a vector quantity. Mathematically, linear velocity is expressed as $v = \frac{ds}{dt}$. If the displacement occurs along a circular path, the direction of the linear velocity at any given instant is along the tangent at that specific point. In contrast, speed is defined as the rate of change of linear displacement with respect to time without consideration of direction, making it a scalar quantity.

Linear acceleration is defined as the rate of change of linear velocity of a body with respect to time and is also a vector quantity. It is expressed mathematically as $\alpha = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{ds}{dt}\right) = \frac{d^2s}{dt^2}$. Additionally, acceleration can be expressed as $a = \frac{dv}{dt} = \frac{dv}{ds} \times \frac{ds}{dt} = v \times \frac{dv}{ds}$. Negative acceleration is referred to as deceleration or retardation.

Equations of Linear Motion

There are four critical equations for bodies moving with uniform linear acceleration, where $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration, $s$ is displacement, and $t$ is time in seconds. These are:

1. $v = u + a \times t$
2. $s = u \times t + \frac{1}{2} \times a \times t^2$
3. $v^2 = u^2 + 2 \times a \times s$
4. $s = \frac{(u + v)}{2} \times t = v_{av} \times t$, where $v_{av}$ represents the average velocity.

In Example 2.1, a car starts from rest ($u = 0$) and accelerates uniformly to a speed of $72\,km/h$ ($20\,m/s$) over a distance of $500\,m$. Using $v^2 = u^2 + 2 \times a \times s$, calculating $(20)^2 = 0 + 2 \times a \times 500$ yields an acceleration of $a = 0.4\,m/s^2$. The time taken is found via $v = u + a \times t$, resulting in $t = \frac{20}{0.4} = 50\,s$. If speed is then raised to $90\,km/h$ ($25\,m/s$) in $10\,s$, the new acceleration is $a = \frac{(25 - 20)}{10} = 0.5\,m/s^2$ and the distance moved is $s = 20 \times 10 + \frac{1}{2} \times 0.5 \times (10)^2 = 225\,m$. If brakes bring the car to rest in $5\,s$, the braking distance is $s = \frac{(25 + 0)}{2} \times 5 = 62.5\,m$.

In Example 2.5, a planing machine has a cutting stroke of $500\,mm$ completed in $1\,s$. The cycle consists of uniform acceleration for $125\,mm$, constant speed for $250\,mm$, and uniform retardation for $125\,mm$. Let $v$ be the maximum cutting speed. The average velocity during acceleration/retardation is $\frac{v}{2}$. Thus, time for acceleration is $\frac{125}{v/2} = \frac{250}{v}$, time for constant speed is $\frac{250}{v}$, and time for retardation is $\frac{125}{v/2} = \frac{250}{v}$. Since the total time is $1\,s$, $\frac{250}{v} + \frac{250}{v} + \frac{250}{v} = 1$, solving to $v = 750\,mm/s$.

Scalars and Vectors

Scalar quantities are those that have magnitude only, such as mass, time, volume, and density. Vector quantities are those that possess both magnitude and direction, such as velocity, acceleration, and force. When adding or subtracting vector quantities, their directions must be taken into account. Vectors can be added by drawing them tip-to-tail. For two vectors $P$ and $Q$, one draws a line segment $AB$ representing $P$, then a segment $BC$ from point $B$ representing $Q$. The resultant vector is the line segment $AC$.

Angular Displacement, Velocity, and Acceleration

Angular displacement is represented by a vector that must satisfy three conditions. First, the direction of the axis of rotation is fixed by drawing a line perpendicular to the plane of rotation. Second, the magnitude is represented by the length of the vector along the axis of rotation to a suitable scale. Third, the sense of the angular displacement is determined by the right-hand screw rule: if a screw rotates clockwise in a fixed nut, the vector points away from the observer; if anti-clockwise, it points toward the observer.

Angular velocity, usually expressed by the Greek letter $\omega$ (omega), is the rate of change of angular displacement with respect to time: $\omega = \frac{d\theta}{dt}$. Since it involves magnitude and direction, it is a vector quantity. If direction is constant, the rate of change of magnitude is termed angular speed.

Angular acceleration, expressed by $\alpha$ (alpha), is the rate of change of angular velocity with respect to time: $\alpha = \frac{d\omega}{dt} = \frac{d}{dt}\left(\frac{d\theta}{dt}\right) = \frac{d^2\theta}{dt^2}$. It is a vector quantity, though its direction may differ from that of angular displacement and velocity.

Equations of Angular Motion and Unit Relations

The equations for angular motion mirror those of linear motion:

1. $\omega = \omega_0 + \alpha \times t$
2. $\theta = \omega_0 \times t + \frac{1}{2} \times \alpha \times t^2$
3. $\omega^2 = (\omega_0)^2 + 2 \times \alpha \times \theta$
4. $\theta = \frac{(\omega_0 + \omega)}{2} \times t$ In these equations, $\omega_0$ is initial angular velocity ($rad/s$), $\omega$ is final angular velocity ($rad/s$), $t$ is time ($s$), $\theta$ is angular displacement ($rad$), and $\alpha$ is angular acceleration ($rad/s^2$). For a body rotating at $N\,r.p.m.$ (revolutions per minute), the angular velocity is $\omega = \frac{2\pi N}{60}\,rad/s$.

Example 2.6 concerns a wheel accelerating from rest ($\omega_0 = 0$) to $2000\,r.p.m.$ in $20\,s$. The final angular velocity is $\omega = \frac{2\pi \times 2000}{60} = 209.5\,rad/s$. Using $\omega = \omega_0 + \alpha \times t$, we find $209.5 = 0 + \alpha \times 20$, resulting in $\alpha = 10.475\,rad/s^2$. The displacement is $\theta = \frac{(0 + 209.5)}{2} \times 20 = 2095\,rad$. The number of revolutions is $n = \frac{\theta}{2\pi} = \frac{2095}{2\pi} = 333.4\,revolutions$.

Relation Between Linear and Angular Motion

If $r$ is the radius of the circular path, the relations are as follows:

• Linear displacement: $s = r \times \theta$
• Linear velocity: $v = r \times \omega$
• Linear acceleration: $a = r \times \alpha$

Acceleration of a Particle along a Circular Path

Consider a particle moving from position $A$ to $B$ across an angle $\delta \theta$ in time $\delta t$. The velocity at $A$ is $v$ and at $B$ is $v + dv$. By drawing a vector triangle, the change in velocity $ab$ can be resolved into two mutually perpendicular components: tangential and normal components.

The tangential component of acceleration ($a_t$) represents the rate of change of the magnitude of velocity. In the limit as $\delta t$ approaches zero, $a_t = \frac{dv}{dt} = \alpha \times r$.

The normal, radial, or centripetal component of acceleration ($a_n$) is directed toward the center of the path. It is defined by $a_n = v \times \frac{d\theta}{dt} = v \times \omega = \frac{v^2}{r} = \omega^2 \times r$.

The total acceleration ($a$) is the resultant of these two perpendicular vectors: $a = \sqrt{(a_t)^2 + (a_n)^2}$. The angle of inclination $\theta$ with the tangential acceleration is given by $\tan(\theta) = \frac{a_n}{a_t}$.

Specifically, if a particle moves along a straight path, the radius of curvature is infinite, meaning $a_n = \frac{v^2}{r} = 0$; thus only tangential acceleration exists. If a particle moves with uniform velocity (constant speed), $a_t = \frac{dv}{dt} = 0$, meaning only normal (centripetal) acceleration exists.

In Example 2.7, a $1.5\,m$ bar rotates about a vertical axis at one end, accelerating from $1200\,r.p.m.$ ($\omega_0 = 125.7\,rad/s$) to $1500\,r.p.m.$ ($\omega = 157\,rad/s$) in $5\,s$. Initial linear velocity at the end is $v_0 = 1.5 \times 125.7 = 188.6\,m/s$. Final linear velocity is $v_5 = 1.5 \times 157 = 235.5\,m/s$. For the midpoint of the bar ($r = 0.75\,m$), the angular acceleration is $\alpha = \frac{(157 - 125.7)}{5} = 6.26\,rad/s^2$. The tangential acceleration is $a_t = 6.26 \times 0.75 = 4.7\,m/s^2$. The radial acceleration is $a_n = (157)^2 \times 0.75 = 18487\,m/s^2$.