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Chris Fox's Engineering Section


Plane Kinematics of Particles

Consider a particle moving in a straight line. At any time, it has a position, denoted by P. The vector running to P from a reference point O describes the displacement of the particle. Displacement (x) may be a function of time, i.e. x=x(t). The SI units of displacement and time are the metre (m) and second (s) respectively. Distance is the magnitude of displacement and is a scalar quantity.

Velocity (v) is defined as the rate of change of displacement. It is therefore a vector quantity and is measured in ms-1 (m/s; metres per second). Velocity will generally be a function of time. Speed is the magnitude of velocity and is a scalar quantity.

Acceleration (a) is defined as the rate of change of velocity. It is therefore a vector quantity and is measured in ms-2. Like velocity, acceleration will generally be a function of time. The rate of change of speed is also called acceleration, so the term also applies to a scalar quantity.

On a displacement-time graph, the velocity at any point is equal to the gradient. Velocity is also equal to tan ψ, where ψ is the angle between the tangent to the graph and the horizontal. Similarly, acceleration is equal to the gradient of a velocity-time graph. Conversely, velocity is equal to the integral of an acceleration-time graph, and displacement is equal to the integral of a velocity-time graph.

One may be given the displacement and asked to calculate the velocity and acceleration. Alternatively, the acceleration may be given and the velocity and displacement needed. In both cases, one needs to know the initial conditions.

Consider first a case of constant acceleration. Because v = ∫ a dt, for constant acceleration v=at+c1. When t=0, v equals the initial velocity v0. Therefore, c1=v0 and v=v0+at.

Integrating both sides of the velocity equation with respect to t gives x=v0t+0.5at²+c2. When t=0, x=0. Therefore, c2=0 and x=v0t+0.5at².

Let us now turn to one situation where acceleration changes with velocity: that of a particle falling through a resisting medium. Here, the acceleration is given by a=g-kv, where g and k are constants. Because dv/dt=g-kv, dt/dv=1/(g-kv). Multiplying both sides of the latter equation by dv then integrating both sides gives t = -(1/k) ln (g-kv) + c1. When t=0, v=0 and it can be seen that c1 = (1/k) ln g. Therefore, t = -(1/k){ln (g-kv) - ln g}. Simplifying this gives ln (1-kv/g) = -kt, and taking e to the power of both sides gives 1-kv/g=e-kt. Finally, rearranging gives v=(g/k)(1-e-kt).

Because x = ∫ v dt, x=(g/k){t+(1/k)e-kt}+c2. When t=0, x=0 and the equation becomes 0=g/k²+c2. Therefore, c2=-g/k². Substituting for c2 in the displacement equation and rearranging gives x=gt/k-(g/k²)(1-e-kt).

Now consider a case of simple harmonic motion, where a particle's acceleration is directly proportional to its displacement from a given point and is directed towards that point. The acceleration is given by a=-ω²x, where ω denotes the angular velocity, measured in rad s-1.

Because dv/dt=-ω²x, (dx/dt)(dv/dx)=-ω²x. This leads to v dv/dx = -ω²x. Multiplying both sides by dx and integrating gives 0.5v²=-0.5ω²x²+c1. When x=0, v=v0. Therefore, c1=0.5v0² and v²=v0²-ω²x².

Because dx/dt=√(v0²-ω²x²), or (v0²-ω²x²)0.5, dt/dx=(v0²-ω²x²)-0.5. Rearranging this gives ω-1(v0²/ω²-x²)-0.5 dx = dt, integrating gives ω-1 sin-1 (ωx/v0) = t + c2, and rearranging gives ωx/v0 = sin {ω(t+c2)}. When t=0, x=0 and 0 = sin ωc2. From this we get ωc2=0 (That is not the only solution, of course; but bear in mind that the particle passes through its initial position at regular intervals). It follows that x = (v0/ω) sin ωt.

Now let us move to two dimensions. The position P of a particle is given in the form (x,y). The displacement of this point from the origin is given by the position vector r. Now move the particle to P'. The vector running from P to P' is denoted by Δr.

The velocity of the particle is given by v = limΔt=0rt). Velocity has the components dx/dt and dy/dt and is directed along the tangent to the path. Speed is given by v=√{(dx/dt)²+(dy/dt)²}.

Acceleration is given by a=dv/dt, and its components are d²x/dt² and d²y/dt². Acceleration will not generally be directed along the tangent to the path; it has one component directed along the tangent and one directed along the normal.


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