Problem 6/136  Page 484, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: Semicircular disk of mass M mounted in a light hoop of radius r.  Released from rest with the semicircle filling the top half of the hoop. Rolls without slipping.

Find: Angular velocity when the semicircle reaches the position where it is in the bottom  half of the hoop. Normal force under the hoop when it reaches the aforementioned position.

Parameter values:
M = 2 kg
r = 0.15 m

1. Mechanical System: Disk plus ring from release until it reaches the desired position.

2. Free Body Diagram: Drawn in a typical intermediate position. Note that the only forces acting on the system are its weight and the contact forces at the contact point of the ring and the surface.  The contact forces include both a normal force and a frictional force sufficient to prevent slipping.

3. Equations
a) Work = 0 + 0 -M g Dh
              = M g 8 r / 3 p
Note that the hoop rolls without slipping.  Thus the normal and frictional forces are applied to a point of zero velocity.  Thus these forces do no work on the system.  Gravity does work on the system because the mass center changes in height.  From the table on page 713 the mass center of the disk is a distance 4 r / 3 p from the geometric center of the hoop.  This distance is intially above the geometric center and ends up directly below the geometric center.  Thus the change in height negative and is of magnitude equal to twice this distance.
 
 b)  Kinetic Energy = 1/2 I_G w² + 1/2 M v_G²
D Kinetic Energy = 1/2 (1/2 - 16/9p²) M r² w² + 1/2 M { (1 - 4/3p) r w }² - 0
                           = (3/4 - 4/3p) M r² w²
Note that as the system is released from rest, its initial linear and angular velocities are zero.   Thus its initial kinetic energy is zero.  The mass moment of inertia of the semicircle about its centroid from the table on page 713 has been used.  Additionally the following two point formula has been used to related the velocity of the mass center of the semicircle to the angular velocity of the system.
v_G = v_O + w X r_G/O
v_O = 0 as the contact point is not slipping on the stationary ground.
r_G/O = r { 1 - 4/3p }j = Distance from ground up to center of ring less distance back down to mass center of semicircle.
v_G =  0 + w k X  r { 1 - 4/3p }j
v_G =  -w  r { 1 - 4/3p }i
The square of this magnitude was used in evaluating the final kinetic energy of the system.

4. Equate and Solve:
M g 8 r / 3 p = (3/4 - 4/3p) M r² w²
 w² = g/r 32/(9p-16)
 w = -13.06 rad/s  (Rolling in the clockwise direction)

In order to solve for the normal force we consider the ring just at it reaches the desired position.

1. Mechanical System: The ring when the semicircle is in the bottom position.

2. Free Body Diagram: Drawn in the desired position.  Shows the weight force downward, the normal force upward, and the friction force acting to the left.

3. Equations.
a) S F = -F i + ( N - M g ) j

b) m a_G = M { a_C + w X ( w X r_G/C ) + a X r_G/C }
In the above C is the center of the ring which can be seen to be moving in a straight line along the X axis.  At the instant of interest the mass center is directly below the center of the ring.  Thus:
m a_G = M { aC i + w k X w k X -4r/3p j ) + a k X 4r/3p j }
    = M { ( aC - 4r/3p a ) i  + 4r/3p  w²  j }

4. Equate and Solve
Considering the y component:
N - M g = M 4r/3p  w² 
Using the value obtained above for the square of the angular speed:
N = M g { 1 + 128/3p(9p-16) }
N = 41.3 N  ( More than twice the weight)