HINTS FOR PROBLEM SET #16

 

Problem Set #16 – Problem 1

Part a)
To find the instant center of the pulley, draw (to scale) the velocity vectors of points B and D. 

Now draw the lines perpendicular to these vectors. 

The instant center of the pulley must be on both of these lines and so you just need to get their intersection. 

However, in this case the lines are the same line.  So there is no intersection!

So you need to draw a line passing through the tips of the two velocity vectors.  The intersection of this line with the one perpendicular to the velocity vectors is the instant center.

(Do you know why?) 

 

Part b)
Now that you have the instant center, C, of the pulley, you can easily find the angular velocity of the pulley. 

                   vB/C =  vC +  w x rB/C 

                   vB/C =   0  +  w x rB/C

                   vB/C =  w rB/C

                   w = vB / rB/C

 

To get the velocity of the slider block you just need to get the velocity of point A on the pulley.  Easy! 

                   vA =  w x rC/A

                    vA =  w rC/A

 

Part c)
To get the millimeters of cord wrapped or unwrapped on each pulley per second, you need the relative velocity of B with respect to A and the relative velocity of D with respect to A. 

(This is not obvious. Think about it and hopefully, you will see it.) 

                   vB/A = w x rB/A

                   vD/A = w x rD/A

 

 
 

Problem Set #16 – Problem 2

Part a)

First consider the link AB. 

Calculate the velocity of point B. 

(You are given the angular velocity of AB and since AB is just rotating about the fixed point A, this should be very easy.  If it isn’t, talk to me!)

 

Now find the instant center, C of link DB.

Draw a line perpendicular to the velocity of B and another line perpendicular to the velocity of D.  (You don’t know the magnitude of the velocity of D but you know the direction!) 

 The intersection of these two lines is the instant center, C.

 

Now that you know the instant center of DB and the velocity of point B on it, you can easily find the angular velocity of DB. 

                   vB = vC + vB/C  

                   vB = vC + wDB x rB/C

                   vB =  0  + wDB x rB/C

                   vB =  wDB rB/C

                   wDB = vB / rB/C 

 

 

Part b)

You are asked for the velocity of the midpoint of rod BD.

Call this point F.

You know the angular velocity of BD and the velocity of a point on it.

So you can find the velocity of point F using:

                  vF = vB + wDB x rF/B

Alternatively you can use the instant center:

                  vF = vC + wDB x rF/C

Note:

If you use B, you will need to do a vector addition. 

If you use the instant center, C, the geometry gets a bit tricky. 

Six of one, half dozen of the other. 

Either way it is work!
 

Problem Set #16 – Problem 3

Part a)

First consider rod AB. 

Find its instant center, C. 

To do this, draw a line perpendicular to the velocity of A and another line perpendicular to the velocity of B. 

The intersection of these two lines is the instant center, C. 

 

Since you are given the velocity of A, you can now get the angular velocity of AB. 
                   wAB = vA / rA/C 

Note:  To get the distance from C to A is a bit tricky but you can do it!

 

Now you need to get the angular velocity of BD. 

First find its instant center. 

Call it K.  (You already have a point C in this problem.)

 

Now find the velocity of B as a point on AB. 

                   vB =  wAB rB/C

Now find the angular velocity of BD:

                   vB = vK + vB/K  

                   vB = vK + wDB x rB/K

                   vB =  0  + wDB x rB/K

                   vB =  wDB rB/K

                   wDB = vB / rB/K 

 

Part b)

To get the velocity of the collar D:

                   vD = vK + vD/K  

                   vD = vK + wDB x rD/K

                   vD =  0  + wDB x rD/K

                   vD=  wDB rD/K

 

Problem Set #16 – Problem 4

Part a)

First find the instant center, C, of link ADB. 

(You should know how by now! Just draw two lines that are perpendicular to two velocities on the body.  The intersection of these two lines is the instant center of the body.)

 

Next find the angular velocity of ADB.

                   wADB = vB / rB/C 

 

Now find the velocity of point D as a point on ADB. 

                   vD =  wADB x rD/C

 

Now find the instant center of link DE, call it K. 

 

Next find the angular velocity of DE.

                   wDE = vD / rD/K 

 

Part b)

Finally, the velocity of point E:

                   vE=  wDE rE/K