Saved Bookmarks
| 1. |
A solid cylinder is released from rest. If string does not slip relative to cylinder then find relative acceleration of A w.r.t B just after it is released. [ans :2 root2g/3] |
|
Answer» re is ANSWER !!!I HOPE you help !!!DEAR Student ,Here in this case ,let the radius of the cylinder is r and tension be T .the MOMENT of inertia of RIGID cylinder is I = ½ mr² Torque = T r Again , torque = Iα So, T r = Iar⇒T = 12mr2ar2 ⇒T = 12maNow from the conservation of energy we can write that ,∆E=0⇒mgh−12mv2−12Iω2=0⇒mgh−12mv2−12(12mr2)(vr)2=0⇒gh−12v2−14v2=0⇒gh=34v2⇒v=4gh3‾‾‾‾√Regards |
|