Cylinder on moving wedge A cylinder of mass m is rolling down, without slip- ping, on a wedge of mass M. The wedge is freely sliding on a horizontal table without friction and the axis of the cylinder remains horizontal during the motion. Gravity acts vertically. 1. Using Newton's laws and the horizontal Cartesian coordinates of the axis of the cylinder and the wedge, x and X, find the acceleration of the cylinder and the wedge, X and X. 2. Using Largangian techniques and the relative position of the cylinder's axis on the wedges, find s and X. Expressing s in terms of x compare to your Newtonian solution. 1. Newtonian method Forces on cylinder: F gravity, Fwedge, Fric Forces on wedge: -Fwedge,-Ftriction where F=F cos(a) friction Equations of motion for center of masses F+F+Fw=mF -F-Fw=MR Equation of motion for rotation Fƒ cos(a) = I cos(a)ē R Where R is the radius of the cylinder. M m Constraint: the cylinder is rolling without slipping, so dx = dX + Rcos (a) de, that is x=x+R cos(a)ē Summing the first two equations of motion shows that F₁-mr-MR and projecting to the x direction -mx=MX X α shows that the horizontal center of mass is conserved - due to lack of external horizontal forces. Com- bining the second equation of motion (projected in the x direction) with the rotational one and the constraint, after eliminating x using the first equation of motion, we obtain gm sin(a) cos(a) k² (m+M)+M

Cylinder on moving wedge A cylinder of mass m is rolling down, without slip- ping, on a wedge of mass M. The wedge is freely sliding on a horizontal table without friction and the axis of the cylinder remains horizontal during the motion. Gravity acts vertically. 1. Using Newton's laws and the horizontal Cartesian coordinates of the axis of the cylinder and the wedge, x and X, find the acceleration of the cylinder and the wedge, X and X. 2. Using Largangian techniques and the relative position of the cylinder's axis on the wedges, find s and X. Expressing s in terms of x compare to your Newtonian solution. 1. Newtonian method Forces on cylinder: F gravity, Fwedge, Fric Forces on wedge: -Fwedge,-Ftriction where F=F cos(a) friction Equations of motion for center of masses F+F+Fw=mF -F-Fw=MR Equation of motion for rotation Fƒ cos(a) = I cos(a)ē R Where R is the radius of the cylinder. M m Constraint: the cylinder is rolling without slipping, so dx = dX + Rcos (a) de, that is x=x+R cos(a)ē Summing the first two equations of motion shows that F₁-mr-MR and projecting to the x direction -mx=MX X α shows that the horizontal center of mass is conserved - due to lack of external horizontal forces. Com- bining the second equation of motion (projected in the x direction) with the rotational one and the constraint, after eliminating x using the first equation of motion, we obtain gm sin(a) cos(a) k² (m+M)+M