only other force is the weight, w, which is not an impulsive force. Hence, the impulse generated over a ver short time interval, t2-t1, will be negligible. Thus, Ho =Ho, which implies that, =+mB(Rbm=(1-局 IG +mR2 where ko=VIo/m is the radius of gyration of the cylinder about O Rolling about O After the initial impact, the cy linder rolls about point O. During this process, since there are no dissipation mechanisms, energy will be conserved. Therefore, we can use the conservation of energy principle and require that the sum of the potential and kinetic energies remains constant R h工 The change in potential energy, mgh, is obtained by a decrease in the kinetic energy, thus Tow=mgh+lou R2 ugh The residual velocity after the cylinder has climbed the step, uG, can be expressed in terms of the initial relocity, UG, using 4, as, =(1-mh)2n If we want to determine the minimum initial velocity, (uG)min, that would allow the cylinder to climb the step, we set the residual velocity, uG, to zero and obtain ghaR So far, we have assumed that the cylinder pivots around point O without loosing contact with the step. It is clear that if the velocity uG is very large, then, as soon as the turning starts, a force will be require to produce the centripetal acceleration. For small velocities, the weight should be sufficient, but velocities, it is possible that the normal reaction force, N, will become zero, and the cylinder will looseonly other force is the weight, W, which is not an impulsive force. Hence, the impulse generated over a very short time interval, t2 → t1, will be negligible. Thus, HO = H′ O, which implies that, v ′ G = IG + mR(R − h) IG + mR2 vG = (1 − Rh k 2 O ) vG , (4) where kO = p IO/m is the radius of gyration of the cylinder about O. Rolling about O After the initial impact, the cylinder rolls about point O. During this process, since there are no dissipation mechanisms, energy will be conserved. Therefore, we can use the conservation of energy principle and require that the sum of the potential and kinetic energies remains constant. The change in potential energy, mgh, is obtained by a decrease in the kinetic energy, thus 1 2 IOω ′2 = mgh + 1 2 IOω ′′2 , or, v ′′2 G = v ′2 G − 2ghR2 k 2 O . The residual velocity after the cylinder has climbed the step, v ′′ G, can be expressed in terms of the initial velocity, vG, using 4, as, v ′′2 G = (1 − Rh k 2 O ) 2 v 2 G − 2ghR2 k 2 O . If we want to determine the minimum initial velocity, (vG)min, that would allow the cylinder to climb the step, we set the residual velocity, v ′′ G, to zero and obtain, (vG) 2 min = 2gh R 2 k 2 O (1 − Rh k 2 O ) 2 . Rebounding So far, we have assumed that the cylinder pivots around point O without loosing contact with the step. It is clear that if the velocity v ′ G is very large, then, as soon as the turning starts, a force will be required to produce the centripetal acceleration. For small velocities, the weight should be sufficient, but, for larger velocities, it is possible that the normal reaction force, N, will become zero, and the cylinder will loose 3