HINTS FOR PROBLEM SET #20

Problem Set #20 – Problem 1

Draw the FBD of the rod AB.

The forces will include the weight, the given force P acting at B, and the x and y 

components of the reaction at A. 

Set the FBD equal to a diagram showing the effective forces acting at G (= mass x 

acceleration vector) and the effective moment (= I x alpha).  

The acceleration vector at G can be shown using tangential and normal components since 

all of the points on the rod are in pure rotation about point A.  Furthermore, the normal 

component of this acceleration vector is zero since the rod starts from rest.

Notice that the angular acceleration, alpha, of the rigid body and the tangential 

component of acceleration are related by this equation:

(tangential component of acceleration) = r x alpha

Now get the x and y components of the equation: F = ma

Now get the k component of the equation: (Sum of the moments about G) = (Sum of the 

effective moments about G)  

These three equations will allow you to solve for the unknowns, alpha and the 

components of the reaction force acting at A.

Alternate Problem

Draw the FBD of the rod AB.

The forces will include the weight, the given force P and only the y reaction at.

Set the FBD equal to a diagram showing the effective force (= mass x acceleration vector) acting at G and the effective moment (= I x alpha).

Note that the acceleration vector at G only has a tangential component since it starts at rest.

Now get the tangential component of the equation: F = ma

Now get the k component of the equation: (Sum of the moments about A) = (Sum of the effective moments about A)  This will include (I x alpha) but also have another term.

Notice that the angular acceleration and the tangential component of acceleration are related: 

(tangential component of acceleration) = r x alpha

These three equations will allow you to solve for the unknowns, alpha, h and the tangential acceleration.

Problem Set #20 – Problem 2

We have two bodies (disks) in this problem that are attached by a cable.  Draw the FBD of each disk. 
The forces will include the tension in cable, the weight of the disks, and the surface forces, both frictional and normal.

Set the FBD of each disk equal to a diagram showing the effective force (= mass x acceleration vector) acting at G and the effective moment (I x alpha). 

Since there is no slip, we know that:

              (acceleration of disk B) = r x (alpha for disk B)

              (acceleration of disk A) = r x (alpha for disk A)

Call the point of contact of each disk with the surface, point P. 

For each disk, write the equation: 

(Sum of the moments about P) = (Sum of the effective moments about P)

These give 4 equations in 5 unknowns: 

1)      alpha for disk A

2)      alpha for disk B

3)      acceleration of disk A

4)      acceleration of disk B

5)      the tension in the cable

Note:  It is tempting to assume that the alpha for disk A and the alpha for disk B are the same,  but this is not necessarily true!  Nor are the acceleration of disk A and the acceleration of disk B necessarily the same.

So we need one more equation. 

We will have to use the equation that relates the accelerations of two points on a rigid body:

              aA = aB +  aA/B

(OK.  In this specific case, it DOES turn out that the alpha for disk A and the alpha for disk B are equal, and the acceleration of disk A and the acceleration of disk B are also the same.  But this is not true for the general case.  Only for the angles given!)

Problem Set #20 – Problem 3

First do a kinematic analysis of the problem:  aA = aG +  aA/G

This will result in a relationship between the alpha of the rod and aG. 

In fact you should get aG = (alpha x L) /2  

Use this in the kinetic analysis.

Draw the FBD of the rod AB.  The forces will include the reaction from the rod at G, the reaction from the rod at A and the weight of the rod. (Neither reaction force has a frictional component.)

Set the FBD equal to a diagram showing the effective force (= mass x acceleration vector) acting at G and the effective moment (I x alpha).  This acceleration vector is aG and we have it in terms of alpha.  (The first thing that we did above.)

Now get the x and y components of the equation: F = ma

Now get the k component of the equation: (Sum of the moments about Q) = (Sum of the effective moments about Q) where Q is the point of intersection for the lines of action of the reaction forces at A and G.  This will include (I x alpha) but also have another term.

These will allow you to solve for the unknowns.