ontact with the step. The normal force, N, can be calculated from the normal momentum equation. That is, the sum of the forces in the direction of OG should equal the centripetal acceleration W sin 8-n=m Setting W=mg, sin=(R-h/R, and N=0, we can determine the minimum velocity, (uG)rebound, that will produce separation. The resulting expression can be combined with equation 4 to obtain an expression for the minimum inital velocity required for separation 9(R-h) UG rebound (1-)2 We note that for h/(R-h)<ko/(2R2),(uG)min <(uG)rebound, and, in this case, there is aa range of velocities for which it is possible for the cylinder to climb the step without rebounding Center of percussion relative to an Instantaneous Center of motion n some situations, it is of interest to determine how a body should be impulsively set in motion such that a certain prescribed point will be(at least momentarily) the instantaneous center of motion. Consider a rigid body of mass m which is initially at rest. An impulse J is applied at t=0 at point P in the body. The application of j to the body initiates both translational and rotational motion. Thus, The modulus of the initial angular velocity dpr/IG, and the modulus of the velocity of the center of mass is UG =J/m. We now wish to find out the point P such that a prescribed point C in the body the instantaneous center of motion. If C is the center of motion, then UG rcG, or, in magnitude UG=wdc. Therefore d ma prcontact with the step. The normal force, N, can be calculated from the normal momentum equation. That is, the sum of the forces in the direction of OG should equal the centripetal acceleration, W sin θ − N = m v ′ G 2 R . Setting W = mg, sin θ = (R − h)/R, and N = 0, we can determine the minimum velocity, (v ′ G)rebound, that will produce separation. The resulting expression can be combined with equation 4 to obtain an expression for the minimum inital velocity required for separation, (vG) 2 rebound = g(R − h) (1 − Rh k 2 O ) 2 . We note that for h/(R − h) < k2 O/(2R2 ), (vG)min < (vG)rebound, and, in this case, there is a a range of velocities for which it is possible for the cylinder to climb the step without rebounding. Center of Percussion Relative to an Instantaneous Center of Motion In some situations, it is of interest to determine how a body should be impulsively set in motion such that a certain prescribed point will be (at least momentarily) the instantaneous center of motion. Consider a rigid body of mass m which is initially at rest. An impulse J is applied at t = 0 at point P in the body. The application of J to the body initiates both translational and rotational motion. Thus, mvG = J , IGω = r ′ × J . The modulus of the initial angular velocity is ω = dP ′J/IG, and the modulus of the velocity of the center of mass is vG = J/m. We now wish to find out the point P such that a prescribed point C in the body is the instantaneous center of motion. If C is the center of motion, then vG = ω × rCG, or, in magnitude vG = ωdC. Therefore, dc = vG ω = J/m dP ′J/IG = IG mdP ′ , or, d ′ P = IG mdC = k 2 G dC . 4