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J. Peraire Dynamics 16. Fall 2004 Version 1.
REVIEW - Rectilinear Motion
We start by considering the simple motion of a particle along a straight line. The position of particle A at any instant can be specified by the coordinate s with origin at some fixed point O.
The instantaneous velocity is v = ds dt =^ v .˙^ (1) We will be using the “dot” notation, to indicate time derivative, e.g. (˙) ≡ d/dt. Here, a positive v means that the particle is moving in the direction of increasing s, whereas a negative v, indicates that the particle is moving in the opposite direction. The acceleration is
a = dv^ = v˙ = d^2 s^ = s .¨ (2) dt dt^2
The above expression allows us to calculate the speed and the acceleration if s and/or v are given as a function of t, i.e. s(t) and v(t). In most cases however, we will know the acceleration and then, the velocity and the position will have to be determined from the above expressions by integration.
Determining the velocity from the acceleration
From a(t)
If the acceleration is given as a function of t, a(t), then the velocity can be determined by simple integration of equation (2), (^) � (^) t v(t) = v 0 + a(t) dt. (3) t 0 Here, v 0 is the velocity at time t 0 , which is determined by the initial conditions.
From a(v)
If the acceleration is given as a function of velocity a(v), then, we can still use equation (2), but in this case we will solve for the time as a function of velocity, � (^) v (^) dv t(v) = t 0 + (^) a(v). (4) v 0
Once the relationship t(v) has been obtained, we can, in principle, solve for the velocity to obtain v(t). A typical example in which the acceleration is known as a function of velocity is when aerodynamic drag forces are present. Drag forces cause an acceleration which opposes the motion and is typically of the form a(v) ∝ v^2 (the sign “∝” means proportional to, that is, a(v) = κv^2 for some κ, which is not a function of velocity).
From a(s)
When the acceleration is given as a function of s then, we need to use a combination of equations (1) and (2), to solve the problem. From a dt = dv and v dt = ds we eliminate dt and write
a ds = v dv. (5)
This equation can now be used to determine v as a function of s, � (^) s v 2 (s) = v^20 + 2 a(s) ds. (6) s 0 where, v 0 , is the velocity of the particle at point s 0. Here, we have used the fact that, v v (^) v 2 v 2 v 2 v dv = d( 2 ) = 2 − 20. v 0 v 0 A classical example of an acceleration dependent on the spatial coordinate s, is that induced by a deformed linear spring. In this case, the acceleration is of the form a(s) ∝ s.
Of course, when the acceleration is constant, any of the above expressions (3, 4, 6), can be employed. In this case we obtain, v = v 0 + a(t − t 0 ), or v 2 = v 02 + 2a(s − s 0 ).
Determining the position from the velocity
Once we know the velocity, the position can be found by integrating equation (1). Thus, when the velocity is known as a function of time we have, � (^) t s = s 0 + v(t) dt. (7) t 0 If the velocity is known as a function of position, then � (^) s (^) ds t = t 0 + (^) v(s). (8) s 0 Here, s 0 is the position at time t 0.
It is worth pointing out that equation (5), can also be used to derive an expression for v(s), given a(v), � (^) s � (^) v (^) v s − s 0 = ds = (^) a(v)dv. (9) s 0 v 0
We see that for, say, v = 0. 95 vf , s = 427.57m. This is the distance travelled by the payload in 11.21s, which can be compared with the distance that would be travelled in the same time if we were to neglect air resistance, sno drag = gt^2 /2 = 615.75m.
Example Spring-mass system
Here, we consider a mass allowed to move without friction on a horizontal slider and subject to the force exerted by a linear spring. Initially the system is in equilibrium (no force on the spring) at s = 0. Suddenly, the mass is given a velocity v 0 and then the system is left free to oscillate. We know that the effect of the spring is to cause an acceleration to the body, opposing the motion, of the form a = −κs, where κ > 0 is a constant.
Using equation (6), we have v^2 = v^20 − κs^2.
The displacement can now be obtained using expression (8), s (^) ds 1 √κs t = �v 0 −^ κs^2 2 =^ √κ arcsin^ v 0 , 0
which gives, s = √^ v^0 κ sin^
√κt.
Finally, the velocity as a function of time is simply, v = v 0 cos √κt. We recognize this motion as that of an undamped harmonic oscillator.
ADDITIONAL READING
J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 2/
