HINTS FOR PROBLEM SET #15

Problem Set #15 – Problem 1

The velocity of any two points on a rigid body are related by this equation:

              v_B = v_A + v_B/A

In this problem we know:

• the magnitude and direction of v_A

• the direction of v_B
  It must be along the 60^o incline.
  We do not know the magnitude.

• the direction of v_B/A
  It must be perpendicular to the line AB.
  We do not know the magnitude.

Construct a force triangle of the equation above and

solve for the unknown magnitudes of v_A and v_B/A.

Once you know the magnitude of v_B/A,

you can get the angular velocity of the rod from:

              v_B/A = wr

Problem Set #15 – Problem 2

This is called a planetary gear system. 

• The sun gear is the center gear.

• The planet gears are the three gears that revolve around the sun gear.

• The ring gear is the outside gear with internal teeth.

• The spider gear is the linkage that joins the planets.

The system has four motions. (The motions of the sun, the planets, the ring and the spider)

The sun, ring and spider gears are all rotating about the fixed axis.

The planets are rotating about axes that are not fixed.

The system has two degrees of freedom.

So if two of the motions (i.e. the angular velocities of two of the gears) are given,

then the angular velocities of the other two gears are determined.

Now for the problem at hand. 

The angular velocity of the ring gear is given.

The angular velocity of the sun gear is given.

You can find the angular velocities of the ring and the spider.

The velocities of the points on the rim of the ring and sun

can be calculated from knowing their angular velocities

since they are rotating about a fixed axis:

              v_ring  =  w_ring  x  r_ring  

              v_sun  =  w_sun  x  r_sun

Referring to Problem 3 in Set #14,

if two gears are meshed the velocities of the points of contact are the same.

Hence, the point on a planet gear that is in contact with the ring gear has velocity, v_ring.

Call this point E and so its velocity is v_E.

Similarly the velocity of the point (on the same planet gear) that is in contact with the sun gear is v_sun.

Call this point A and so its velocity is v_A.

Now you have the velocity of two points on a rigid body! 

Use the equation relating the velocities of two points on a rigid body in plane motion and you can solve for the angular velocity of the body (the planet gear in this case.) 

             v_E =  v_A  +  v_E/A   =  v_A  + (w  x  r_E/A)

where

             v_E     is the velocity of the point of the planet that is in contact with the sun

             v_A     is  the velocity of the point of the planet that is in contact with the ring

             r_E/A  is the diameter of the planet gear

             w     is the angular velocity of the planet gear

To get the angular velocity of the spider, use the same equation,

i.e. the equation relating the velocities of two points on a rigid body in plane motion. 

What two points do you know the velocity of? 

The first is the velocity on the tip of the spider legs because it has the same velocity as the center of the planet gears.

(The reason: pin connected parts have the same velocity at the pin.)

The second point you know the velocity for is the center of the spider gear.

(It is zero.)

Problem Set #15 – Problem 3

In this problem you should start with linkage AB.

You are given that it is rotating with constant angular velocity of 20 rad/s counterclockwise.

              w_AB = 20 k

Calculate the velocity of point B. 

B is rotating about a fixed axis through point A.

Hence its velocity can be calculated by this equation:

              v_B  =  w_AB  x  r_B/A

Since the linkage AB is pin connected to linkage BDH,

the point B on linkage BDH has the same velocity as the point B on linkage AB.

Now consider the linkage BDH (a rigid body).

Use the equation relating the velocities of two points on a rigid body in plane motion. 

              v_D =  v_B  +  v_D/B

You just calculated the velocity of point B so that is a known. 

You don’t know the velocity of point D but you do its direction.

(It must be perpendicular to linkage DE – look at the picture!) 

Furthermore, you know the direction of the velocity of D relative to B, v_D/B . 

(It must be perpendicular to the line from B to D.)

So you can find the magnitudes of  v_B  and  v_D/B from this equation.

Now you can get the angular velocity of BDH, w_BDH, using:

              v_D/B  = w_BDH  x  r_D/B

Now you need to get the velocity of point G on BDH. 

Use the same equation again! 

             v_G =  v_B  +  v_G/B   =  v_B  + (w_BDH  x  r_B/G)

You know the velocity of point B, and you know the angular velocity of BDH. 

Problem Set #15 – Problem 4

It may not be obvious, but the velocity of point C must be zero.

Why? 

Because if there is no slip between the tire and the road,

you can think of the two as meshed gears.  (Like a rack and pinion.) 

You know that the velocities of points in contact of two meshed gears are the same.

So if the ground is not moving, neither is point C on the tire.

             v_C =   0

Now use this fact with the equation relating the velocities of two points on a rigid body in plane motion to get the angular velocity of the tire:

             v_A =  v_C  +  v_A/C  

                  =   0 +   (w  x  r_A/C)

            v_A =  wr_A/C

where

             v_A    = 48 mph = 70.4 ft/sec

             r_A/C = 11 inches = 11/12 ft  = 0.9167 ft

Now that you know the angular velocity of the tire, w,

and the velocity of one of the points on it, v_A,

you can find the velocity of any other point on the tire.

You want to find v_B , v_D, and v_E:

             v_B =  v_A  +   (w  x  r_B/A)

             v_D =  v_A  +  v_D/A   = v_A + (w  x r_D/A)

             v_E =  v_A  +  v_E/A   = v_A + (w  x  r_E/A)