A solid cylinder of radius R rolls without slipping with a center of mass speed v0 = gR3 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: [figure]
- 0
- 5gR7
- gR15
- 3gR7
Answer
(2) 5gR7
Solution
At the edge the cylinder rotates about the corner with = v/R; Icorner = 32mR2, so KE = 34mv2.
After turning through : 34mv2 = 34mv02 + mgR(1- ) v2 = gR3 + 43gR(1- ).
Contact is lost when normal force = 0: mg = mv2R v2 = gR .
gR = 53gR - 43gR = 57
v = gR = 5gR7
Answer: (B)