MAE2323 Dynamics
(275pts)
Homework #9
Due November 7, 2024
MAE 2323 DYNAMICS Homework #9 with complete solution (275pts) University of Texas, Arlington Grade problems P5.29 and P5.32 below. S5.25 (5pts) What is the system level sum of forces? The summation of all forces acting on a system of rigid bodies. S5.26 (5pts) Under what conditions is linear momentum conserved? When the sum of forces on a system is equal to zero. S5.27 (5pts) When is it useful to sum moments about a point other than the mass center? If the arbitrary point has zero velocity. If this is true, we may also be able to eliminate some moments from the equations of motion. S5.28 (5pts) What is a key advantage of summing moments about the mass center? Only rotational quantities appear in Euler’s Second Law, which does not occur when moments are summed about an arbitrary point. S5.30 (5pts) Under what circumstances is angular momentum conserved? When the sum of moments about some point is equal to zero. P5.18 (93pts) Figure 5.65 shows a sphere that impacts a triangular block. The block can slide on the ground, but it does not rotate. All surfaces are frictionless. The sphere travels in the horizontal direction at a speed of v1 = 10ft/s when it strikes the block, which is at rest. Model the sphere as a particle. The mass of the sphere is m and the mass of the block is 1.5m. Find the horizontal speed of the block just after the collision where the sphere is moving at a speed of v2 = 1ft/s toward the right. Define any quantities you need to solve the problem. (answer: 6ft/s.) Position Orientation: (43pts) N
YB
R
A
gravity
B
B m
XN
n
Figure 5.65: Frames (6pts), points (3pts), coordinates (3pts), parameters (1pts), forces (12pts).
Simplifications: All surfaces are frictionless. We will use conservation of linear momentum to solve this problem. Rigid Bodies: This problem involves on rigid body, the wedge, and one particle, the sphere. (1pts) Inertial Reference Frame and Point: The inertial reference point is labeled N and the inertial reference frame is N = (XN, YN). (1pts) Other Points and Frames: The body-attached points are A and B, which represent the mass centers of each body. The body-attached frame is B = (XB, YB). (1pts) Location Descriptions: (12pts) LA = (2.
(1)
32)
{PNA}
PNA = (2
.9)
x XN + y YN
LB
=
(2.
26)
(PNB, N BR} + geom.
PNB = (2
.9)
x XN + h YN
(3pts) (4pts)