A solid cylinder of radius R rolls without slipping with a center of mass speed v_0 = gR 3 on a horizontal surface with a vertical edge, as shown in the figure. Here, g is the acceleration due to the gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:
B. 5gR 7
Let the mass of the cylinder be m and its radius be R. When the cylinder is rolling on the horizontal surface, its kinetic energy is: K_i = 1 2 mv_0^2 + 1 2 I_ cm _0^2 Since it rolls without slipping, _0 = v_0 R and I_ cm = 1 2 mR^2. K_i = 1 2 mv_0^2 + 1 2 ( 1 2 mR^2 ) ( v_0 R )^2 = 3 4 mv_0^2 Given v_0 = gR 3 , we have v_0^2 = gR 3 . K_i = 3 4 m ( gR 3 ) = 1 4 mgR Let the corner be the reference level for potential energy. The initial total mechanical energy of the cylinder just as it reaches the corner is: E_i = K_i + U_i = 1 4 mgR + mgR = 5 4 mgR As the cylinder rotates around the corner, let be the angle the line connecting the corner to the center of mass makes with the vertical. Let v be the speed of the center of mass at this angle. The total mechanical energy at angle is: E_f = 3 4 mv^2 + mgR By conservation of mechanical energy, E_i = E_f: 5 4 mgR = 3 4 mv^2 + mgR The forces acting on the cylinder along the radial direction (towards the corner) are the component of gravity mg and the normal force N. The equation of motion is: mg - N = mv^2 R The cylinder loses contact with the corner when the normal force becomes zero (N = 0): mg = mv^2 R mgR = mv^2 Substituting this into the energy conservation equation: 5 4 mgR = 3 4 mv^2 + mv^2 5 4 mgR = 7 4 mv^2 Solving for v: v^2 = 5 7 gR v = 5gR 7 Answer: 5gR 7
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