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]
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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‾‾‾‾√
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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‾‾‾‾√
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asharanya7913:
u can copy and paste this and edit the mistake
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