Final answer:
To show that the system of moment a for a rigid slab in plane motion reduces to a single vector, the angular momentum for each mass segment is calculated and shown to align along the axis of rotation. The distance from mass center G to the line of action of this vector is equal to the centroidal radius of gyration k, which is related to the velocity v of G and the angular velocity ω.
Step-by-step explanation:
The question pertains to the system of moment a for a rigid slab in plane motion and how it reduces to a single vector. When a rigid body is constrained to rotate about an axis (z-axis in this case), all its mass segments—represented as Δmi—undergo circular motion with the same angular velocity ω.
The angular momentum (ℑi) of a mass segment with position vector αi from the origin and radius Ri to the z-axis is given by the product of the position vector and linear momentum, where the linear momentum is the product of the mass segment and its tangential velocity (vi = Riω).
The angular momentum is expressed as ℑi = ri (Δmi vi)sin 90°, where the sine of 90 degrees is 1 since the position vector and velocity vector are perpendicular. This brings us to the moment of inertia I, which for a point particle is given by mr².
To determine the distance from the mass center G to the line of action of the resultant angular momentum vector, we consider the relationship between the velocity v of G (the velocity of the slab's center of mass), and the angular velocity ω.
The corresponding equation is UCM = Rω, where UCM is the velocity of the center of mass and R is the radius of gyration k, related to I by I = Mk². Therefore, the distance from G to the line of action of the angular momentum vector is the radius of gyration k.