S Conservation of energy implies that T +V= Tinitial+Initial. Initially, the kinetic energy is zero, Tinitial Thus, for a later time, the kinetic energy is given by T=Initial -V= mgs sin o where s is the distance traveled down the ramp. The kinetic energy is simply T=i IG + mR2 is the moment of inertia about the instantaneous center of rotation C, and w is the angular 1+(Ic/mR2) since W=U/R Equilibrium and Stability If all the forces acting on the body are conservative, then the potential energy can be used very effectively to determine the equilibrium positions of a system and the nature of the stability at these positions. Let us assume that all the forces acting on the system can be derived from a potential energy function, v. It is clear that if F=-VV=0 for some position, this will be a point of equilibrium in the sense that if the body is at rest(kinetic energy zero), then there will be no forces(and hence, no acceleration) to change the equilibrium, since the resultant force F is zero. Once equilibrium has been established the stability of he equilibrium point can be determine by examining the shape of the potential function. If the potential function has a minimum at the equilibrium point, then the equilibrium will be stable. This means that if the potential energy is at a minimum, there is no potential energy left that can be traded for kinetic energy. Analogously, if the potential energy is at a marimum, then the equilibrium point is unstable. Example Equilibrium and Stability A cylinder of radius R, for which the center of gravity, G, is at a distance d from the geometric center, C lies on a rough plane inclined at an angle oConservation of energy implies that T +V = Tinitial+Vinitial. Initially, the kinetic energy is zero, Tinitial = 0. Thus, for a later time, the kinetic energy is given by T = Vinitial − V = mgs sin φ , where s is the distance traveled down the ramp. The kinetic energy is simply T = 1 2 IC ω 2 , where IC = IG + mR2 is the moment of inertia about the instantaneous center of rotation C, and ω is the angular velocity. Thus, IC ω 2 = 2mgs sin φ , or, v 2 = 2gs sin φ 1 + (IG/mR2) , since ω = v/R. Equilibrium and Stability If all the forces acting on the body are conservative, then the potential energy can be used very effectively to determine the equilibrium positions of a system and the nature of the stability at these positions. Let us assume that all the forces acting on the system can be derived from a potential energy function, V . It is clear that if F = −∇V = 0 for some position, this will be a point of equilibrium in the sense that if the body is at rest (kinetic energy zero), then there will be no forces (and hence, no acceleration) to change the equilibrium, since the resultant force F is zero. Once equilibrium has been established, the stability of the equilibrium point can be determine by examining the shape of the potential function. If the potential function has a minimum at the equilibrium point, then the equilibrium will be stable. This means that if the potential energy is at a minimum, there is no potential energy left that can be traded for kinetic energy. Analogously, if the potential energy is at a maximum, then the equilibrium point is unstable. Example Equilibrium and Stability A cylinder of radius R, for which the center of gravity, G, is at a distance d from the geometric center, C, lies on a rough plane inclined at an angle φ. 5