Chapter 5 Homework

Homework H5.O – Sp26

9 Comments
Problem statement
Solution video

DISCUSSION THREAD

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DISCUSSION
As always, we should follow the four-step plan:

STEP 1: FBD
Draw an FBD of the bar. Since the support at B is smooth, the reaction on the bar at B will be perpendicular to the bar.

STEP 2: Kinetics
You should write down the two Newton equations, and the Euler equation. As always, take care in choosing your reference point for Euler’s equation. Since we have no fixed points for the bar, you should choose the center of mass G as your reference point.

STEP 3: Kinematics
Note that the path of the center of mass G is tangent to the surface of the bar (that is, the bar can move only along the support, not into the support). With the bar being released from rest, the acceleration of G is tangent to the path. Specifically, the acceleration of G is directed along the direction of the bar. Write down the rigid body kinematics equation that relates the accelerations of A and G. This vector equation represents two scalar equations (x- and y-components).

STEP 4: Solve
At this point, you will have three kinetics equations and two kinematics equations, for a total of five equations. You will have five unknowns. Solve!

9 thoughts on “Homework H5.O – Sp26”

  1. At the instant of release, why must aG point directly along the bar? Since point A is sliding on the floor and the bar is rotating, shouldn’t G be accelerating partially downward or outward?

    1. The tangent to the path of G is along the edge of the bar as the bar cannot penetrate the pin at B. The bar is released from rest, so there can be no normal component of acceleration. All that is left is having the acceleration tangent to the path; that is, it must be along the length of the bar.

  2. I’m confused on how to relate the acceleration of point A to the acceleration of G in the x direction in the kinematics step. I know that point A is constrained in the horizontal direction so that was helpful to simplify my kinematics step in the y direction, but I believe there is some important step I am missing to relate the two in the x direction. Could this have to do with the instant center approach?

    1. Use the standard rigid body acceleration equation (as we did back in Chapter 2):

      a_A = a_G + alpha x r_A/G – omega^2*r_A/G

      From the discussion above, the acceleration of G is along the length of the bar, and the acceleration of A is along the straight path of A.

  4. I am slightly confused. After finding the five equations, I seem to have 6 unknowns, the reaction components of the bar, the angular acceleration, the acceleration components of G, and the acceleration of point A. Am I able to cancel out a variable I am not aware of?

    1. I am having this same issue. It seems as though we are missing something to relate the acceleration of the center of mass to the acceleration of one side such as an instant center approach.

      1. Ah, making aGx and aGy into aGcos(theta) and aGsin(theta) eliminates one unknown as our professor was trying to point out above. We can do this since we know the path of the center of mass which is constrained to the joint at B.

        2

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