Problems 7/18 Page 542, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: The system shown consists of two moving parts, a wheel and an arm.  The arm is shown in green.  The wheel is shown in blue.  The arm and the wheel are each in contact with the ground (shown in purple).  Thus this problem involves three coordinate systems.  Of the three, only one is shown in the figure.  The figure depicts axes attached to the arm (shown in light blue).  It is important to recognize that in attacking the problem it may be useful to think of coordinate systems attached to the ground and to the wheel.  The ground system is a fixed, inertial coordinate system.  The wheel system turns with the wheel.  The arm is a slender rod.  The x axis shown in the figure is attached to the rod and is along its length.  The arm moves so that the x axis is always in a horizontal plane.  The arm is pivoted at a point O, which is fixed in the ground.  The arm rotates about a vertical axis at a constant rate w0 (shown in yellow).  The z axis shown in the figure is vertical and is attached to the arm.  The wheel is mounted on the arm at a distance b from O.  The wheel is of radius r.  The wheel is free to rotate about the arm axis(x).  The wheel is mounted so that its plane is perpendicular to the rod length (x axis).  The wheel is in contact with the ground at a point a distance r below the arm.  The wheel rolls without slipping on the ground

Find: The angular velocity of the wheel relative to the ground.  The angular acceleration of the wheel relative to the ground.

Parameter Values:
b = Length of arm from fixed pivot point to wheel connection point.
r = Radius of wheel.
w0 = Constant rate of rotation of arm about the vertical axis.
 
From the first fundamental formula:
w_wheel/ground = w_wheel/arm + w_arm/ground.
The arm is rotating relative to the ground at a constant rate w0 about a vertical axis.  The z direction is vertical.  Thus:
w_arm/ground = w0 k
The wheel rotates about the arm x axis.  The rate of rotation is not given and will be labeled w.  Thus:
w_wheel/arm = w i
Thus:
w_wheel/ground = w0 k  + w i
To determine the rate of rotation of the wheel about the arm, we must exploit the fact that the wheel is rolling without slipping on the ground.  This means that the point on the periphery of the wheel that is currently in contact with the ground must have zero velocity.  Thus we are interested in developing an expression for the velocity of a point on the periphery of the wheel. To find the velocity of a point we must differentiate a position vector.  That position vector must run from a point fixed in the ground to the point of interest.  The time derivative must be performed relative to the ground.  In this case the point O is fixed in the ground and the vector from point O to any point on the periphery of the wheel can be written:
r = b i + r e_r
In the above formula e_r is a unit vector attached to and turning with the wheel and pointing toward the point of interest on the periphery of the wheel.  The velocity of interest is the time derivative of the above vector:
v = d/dt r |_ground = d/dt { b i + r e_r  } |_ground
The derivative as posed is difficult to evaluate because neither of the unit vectors used is fixed in the ground system.  However our second fundamental formula can come to the rescue.  Noting the derivative of a sum is the sum of the derivatives, the velocity of a point on the periphery of the wheel is given by:
v = d/dt { b i } |_arm + w_arm/ground. X b i + d/dt { r e_r  } |_wheel + w_wheel/ground  X  r e_r
Each of the two time derivatives given above vanishes because the magnitude of the vector is constant and the direction of the vector is constant relative to the coordinate system in which the derivative is being evaluated.  Using the expressions for the angular velocities leads to the following result for the velocity of any point on the periphery of the wheel:
v = w0 k   X b i +  ( w0 k  + w i ) X  r e_r
Before we can perform the above cross products we must pick a particular point on the periphery of the wheel.  We are interested in the point on the wheel in contact with the ground.  The velocity of this point is known to be zero.  This point is directly below the center of the wheel.  Thus the unit vector that points toward this point is in the negative z direction.  Making this substitution and performing the required cross products:
v = 0 = w0 k   X b i +  ( w0 k  + w i ) X  -r k
   = w0 b j + w r j = ( w0 b + w r) j
Knowing that this velocity is zero, enables the determination of the angular speed w.
w = -w0 b / r
Thus the angular speed of the wheel relative to the arm is a constant and the angular velocity of the wheel relative to the ground becomes:
w_wheel/ground = w0 k  -w0 b / r  i
In order to determine the angular acceleration we must take the time derivative of the angular velocity:
a_wheel/ground = d/dt( w_wheel/ground  )|_ground
      = d/dt( w0 k  -w0 b / r  i   )|_ground
This derivative is difficult because the unit vectors used are not fixed in the ground.  One more time the second fundamental formula comes to our rescue.  As the xyz axis system is fixed to the arm, the desired derivative can be expressed as:
a_wheel/ground = d/dt( w0 k  -w0 b / r  i   )|_arm  + w_arm/ground.  X w0 k  -w0 b / r  i 
The derivative term vanishes because the magnitudes of each of the components are constants and the unit vectors used are fixed in the arm system.  Using the expression for the angular velocity of the arm:
a_wheel/ground = w0 k  X w0 k  -w0 b / r  i 
Performing the cross product:
a_wheel/ground = -w0² b / r  j
Wow!