Elasticity of Collision Coefficient of Restitution

The examples of collision discussed in the previous sections of this chapter had two common features: (a) the impulse of collision occurred in the direction of common normal to the contact area, and (b) after the collision the velocity at the point of contact was the same in both solids. There are many cases in real life where these assumptions do not hold. For example, in running, the impulse of collision has both a tangential and normal component at the surface of contact. Condition b is not satisfied when a ball dropped onto a hard surface rebounds from the surface. In this section we relax these two assumptions and consider impulse in more general terms.

Let A and B denote two rigid bodies colliding in the time interval tf — ti. The times ti and tf represent the initial and final instants of a time interval during which the spatial positions of the two bodies remain essentially the same. Let P and P' designate points that come into contact with each other during the collision of two bodies A and B. Let n be the unit vector in the direction of the common normal to the contact surface. How do we determine the unit normal n? Let T denote the common tangent plane to the surfaces of A and B at the point of contact. The vector n is a unit vector that is normal to T. When a runner hits the asphalt with one foot, the unit vector n could be chosen as the unit normal vector to the asphalt pointing outward from the ground. This is because both the foot and the running shoe are much more deformable than the asphalt, and therefore, at the point of contact, the shoe would assume the curvature of the asphalt ground. Note that although the direction of n is uniquely determined by the tangent plane T, the sense of direction of n is arbitrary.

Let us next resolve the velocities at the point of contact into two components, one in the direction of the normal n and the other in the tangent plane T:

where uP and uP' denote the projections of the velocities of the contact points P and P' along the n direction. The velocities wP and wP' are in the tangent plane T.

A parameter e called coefficient of restitution is introduced as a measure of the capacity of colliding solids to rebound from each other:

where the subscripts i and f refer to the times ti and tf. Thus, for example, uPf denotes the value of uP at time tf. The parameter e takes values from 0 to 1. When two solids stick at the point of impact, the collision is said to be plastic and e = 0. The coefficient of restitution is very nearly zero if one of the colliding solids is made of soft clay. In the opposite end, when e = 1, the velocity of approach has the same magnitude of velocity of separation in the n direction. The collision is said to be elastic when e = 1. In this case the impact of collision does not lead to dissipation of mechanical energy. A ball dropped from a height h onto a hard planar surface will rebound to the same height after the collision if e = 1. It must be emphasized that although parameter e is relatively easy to measure, its interpretation as a measure of deformability of the colliding bodies is difficult because it depends upon the materials, geometry, and initial velocities.

What happens during collision to the velocities of contact points in the tangential plane of contact? To address this question, let us resolve the impulse £ into two components:

where v is parallel to the common normal n and t is the projection of the impulse in the tangent plane at the point of contact. We assume that there is no slipping at time tf if and only if

where f, the coefficient of friction, is assumed constant for the colliding solids. Equation 7.14e is satisfied during running where the impacting foot does not slip on the ground. Most accidental falls occur, on the other hand, when this inequality cannot be satisfied. In such cases there is slip at time tf and the impulse of collision in the tangential plane is related to the impulse in the normal direction by the following equation:

Note that the sense of direction of the tangential impulse t must be opposite to the tangential component of the velocity of separation.

Example 7.7. Kicking a Rock. An angry young man kicks a spherical rock (Fig. 7.12). The mass of his leg is M and the mass of the rock is m. At the instant the man hits the rock, his leg makes an angle of 6 with the vertical axis. The angular velocity wo of his leg just before he hits the rock is in the counterclockwise direction. The length of his leg is L and the radius of the rock is R. Assuming e = 0.5, determine the subsequent motion and the impulse.

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Figure 7.12a-b. A young man kicks a rock in anger (a). The kicking leg is represented by a rod of mass M and length L and the rock by a sphere of mass m and radius R (b).

Figure 7.12a-b. A young man kicks a rock in anger (a). The kicking leg is represented by a rod of mass M and length L and the rock by a sphere of mass m and radius R (b).

Solution: Let us represent the lower limb involved in the kicking with a uniform rod of mass M and length L. We will assume that the rod is fixed at point O representing the hip joint. We will determine the location of the impact as well as its magnitude and direction. Let B and B' denote the points of contact on the foot and on the rock during the course of impact. The collision occurs at B, a distance R (1 — sin 6) from the ground. The geometry shown in the figure dictates that the point B is at a distance L — R (1 — sin 6) from point O.

The direction of collision n is given by the equation n = cos 6 e1 + sin 6 e2 The velocity of point B before the collision is vo n = [L — R (1 — sin 6)] mo n

The velocity of B' before the impact is zero. Then, according to the equation of restitution (Eqn 7.14c), we have e {[L — R (1 — sin 6)] Mo — 0} = (u — v) (7.15a)

in which u and v denote, respectively, the velocity of the rock and the rod at points B' and B immediately after the collision. Because the rod is fixed at point O even after the impact, the velocity v is related to the angular velocity of the rod after the collision by the following relation:

Because the impulsive force between the leg and the rock is an internal force, the moment of momentum for this system (rod plus sphere) should not be affected by the collision, and thus

(M L2/3) Mo = (M L2/3) m + m [L — R (1 — sin 6)] u (7.15c)

Solving Eqns. 7.15a, 7.15b, and 7.15c for m, v, and u, we find:

m = Mo{(M L2/3) — e m [L — R (1 — sin 6)]2}/{(ML2/3) + m [L — R (1 — sin 6)]2}

v = [L — R (1 — sin 6)] m u = [L — R (1 — sin 6)][e Mo + m]

Assuming e = 0.5, 6 = k/6, mo = 15 rad/s, M = 12 kg, m = 7 kg, L = 1.08 m, and R = 0.16 m, we find m = —1.5 rad/s v = —1.5 m/s u = 9 m/s

As the man hits the rock in the direction of common normal at the point of contact, neither the impulse nor the velocity of approach had a com ponent in the tangential plane. The impulse £ acting on the leg (rod) is given by the equation:

[L - R (1 - sin d)] £ = (M L2/3) K - o>) £ = 4.67 (15 - 1.5) = 63 N-s

Getting Started With Dumbbells

Getting Started With Dumbbells

The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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