14.2: Rigid Body Surface Collisions

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In instances where a moving rigid body impacts a solid and immovable surface, the resulting impact can seem chaotic. However, we can still use the ideas of impulse and momentum to predict this impact behavior.

Just as with particles, an important first step in solving problems where a rigid body impacts a surface is to identify the normal and tangential directions. The normal direction will be perpendicular to the surface being impacted, while the tangential direction will be parallel to the surface being impacted. Also important for rigid body impact is identifying both the center of mass for the body and the point of impact between the body and the surface.

A dropped wrench at the instant of its hitting the floor. It is oriented so that its lowest point is A, the edge of one of the jaws, and this is the only part of the wrench currently in contact with the floor. Point C on the wrench handle, upwards and to the right of A, is the object's center of mass. The t-direction points from point A to the right along the floor, and the n-direction points upwards from point A. — Figure \(\PageIndex{1}\): Dropping a wrench on a concrete floor is a good example of rigid body impact with a solid surface. The first step in solving these problems is to identify the normal and tangential directions, which will be perpendicular and parallel to the surface, respectively. Also important is identifying the center of mass (C) and point of impact (A).

To predict the linear velocities after impact in the \(n\) and \(t\) directions, as well as the angular velocity after the impact (a total of three unknowns), we will need three equations. These equations will come in the form of the conservation of momentum in the \(t\) direction, as well as two equations based on the coefficient of restitution for the velocity in the \(n\) direction and the angular velocity after the impact.

For momentum, we will notice that the normal forces during impact will always be in the normal direction. Assuming negligible friction forces (which would be in the tangential direction), we will have no change in momentum in the tangential direction and therefore no change in velocity in the tangential direction for the center of mass of the body. This gives us the first equation we can use.

\[ v_{t,C,f} = v_{t,C,i} \]

Next, examining the coefficient of restitution will give us another equation. Specifically, the coefficient of restitution relates the velocities before an after the collision in the normal direction at the point of impact.

\[ \epsilon = - \frac{v_{n,A,f}}{v_{n,A,f}} \]

If the collision is elastic we would also conserve the kinetic energy of the body, giving us our third equation. More generally, however, the coefficient of restitution can be used to quantify the amount of energy lost on the collision, with more elastic collisions conserving a higher percentage of kinetic energy and more inelastic equations conserving less kinetic energy. The equation below can be used for any elastic or semi elastic collision with a surface. Using the third equation below, we simply set \(\epsilon\) equal to one for elastic collisions or equal to the coefficient of restitution for semi-elastic collisions.

\[ \epsilon^2 = \frac{KE_f}{KE_i} \]

Between the three equations above and whatever relevant kinematics relationships are necessary, we should be able to solve for up to three unknowns. This allows us to completely predict the velocities after an impact, assuming we know the pre-impact velocities and the coefficient of restitution for the impact itself.

Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/LQKk2WLMe4A.

Practice \(\PageIndex{1}\)

An 80-centimeter-long metal bar with a mass of 1 kilogram, falling at 2 meters per second, strikes the edge of a table as shown below. Assuming a coefficient of restitution of 0.9, what is the expected velocity and angular velocity of the bar after impact?

A horizontal bar is falling straight downwards at an initial velocity of 2 m/s, when its very leftmost end strikes the flat top of a table. — Figure \(\PageIndex{2}\): problem diagram for Example Problem \(\PageIndex{2}\). A horizontal bar falls straight down, until its leftmost end strikes the top of a flat table.

Solution (not yet available): Not yet available.