Question

A-lb Rod OA with length 0.4m is rotating about point O at constant angular velocity. It will push the plate B to move up and down. At this instant, theta = 30 and omega = 0.5 rad/s. Determine the velocity and acceleration of point C on the plate.

Answer 1

Preliminaries For this problem it is convenient to apply the following balance laws: Euler's 1st law to the center of mass of the rolling sphere .Euler's 2nd law about the center of mass of the rolling sphere In order to use the aforementioned balance laws, we will need the following kinematic quantities in an inertial reference frame: The acceleration of the center of mass of the rolling sphere The rate of change of angular momentum of the rolling sphere about the center of mass of the rolling sphere

Let F be a fixed reference frame. Then, choose the following coordinate system fixed in reference frame F Origin at O Ex Along od, when θ = 0 Into Page Ez × Ex Next, let A be a reference frame fixed to the direction OC. Then, choose the following coordinate system fixed in reference frame A Origin at O Along OC Into Page -Ez) The geometry of the bases Ex. Ey, Ezl and fer, ep, ez is shown in Fig. 5-14 where φ be the angle between the direction Ex and the direction er. Using the er Figure 5-14 tion 5-8. Relationship Between Bases Ex,Ey,Ez} and {er, ефіег} for Ques- geometry in Fig. 5-15, we have the following relationship between Ex.Ey, Ez and fer, e, ez) (5.333) Ey sinder + cos феф Next, the angular velocity of reference frame A in reference frame F is given as (5.334)

Furthermore, denoting the reference frame of the rolling sphere by R and ob- serving that describes the rotation of the rolling sphere relative to the fixed vertical direction, we have that (5.335) The position of the center of mass of the rolling sphere is then given as (5.336) Differentiating f in reference frame F using the transport theorem, we have that (5.337) Now we have that (f) 0 (5.338) Consequently, (5.339) Differentiating in reference frame F using the transport theorem, we have that (5.340) dt dt Now we have that (R +r) (5.341) Consequently (5.342) Now we see that Eq. (5.342) is an expression for à in terms of the derivatives of ф. However, in this problem we are interested in obtaining the differential equation of motion in terms of θ. Therefore, we need to eliminate φ in favor of Eliminating ф is accomplished as follows. First, we know that the sphere rolls without slip along the fixed sphere. Denoting Q as the instantaneous point of contact between the two spheres, we have that (5.343)

Then, applying the relative velocity property for two points on a rigid body, we have that (5.344) Now we have that ro-Rer which implies that F- rQre. Then, substituting from Eq. (5.335) into Eq. (5.344), we obtain (5.345) Finally, observing from Eq. (5.343) that vo-0, we obtain as (5.346) Then, setting the result of Eq. (5.346) equal to the result of Eq. (5.339), we obtain (5.347) Solving Eq. (5.347) for φ, we obtain (5.348) Differentiating ф in Eq. (5.348), we obtain (5.349) The acceleration of the center of mass of the rolling sphere is then given as (5.350) R+r R + r Simplifying Eq. (5.350), we obtain (5.351) Next, we need to compute the rate of change of the angular momentum of the sphere relative to the center of mass of the sphere in reference frame F. We (5.352) where is the moment of inertia tensor relative to the center of mass and 。yR is the angular velocity of the sphere in reference frame F. Now since fer,e e:) R + r have that is a principle-axis basis, the moment of inertia tensor İR can be written as (5.353) Then, substituting the expression for w as given in Eq. (5.335) and the ex pression for the moment of inertia tensor from Eq. (5.353) into Eq. (5.352), we obtain H as (5.354)