Two cord are wrapped around a cylinder and attached to theceiling like in the fi

Question

Two cord are wrapped around a cylinder and attached to theceiling like in the figure below M= mass of the cylinder R= radius of the cylinder h= 3R= the amount of string that is unwrapped g = the acceleration due to gravity If all question are to be answered in terms of the knowns M,R, and g then how do I find the following: 1)linear acceleration of the cylinder 2) tension in each string supporting the cylinder 3)magnitude of the torque rotating the cylinder 4)angular acceleration of the cylinder 5)the final linear speed of the cylinder 6)the final angular speed of the cylinder 7) angle (in radians) through which the cylinder rotates 8) final kinetic energy of the cylinder 9) the final angular momentum of the cylinder

Explanation / Answer

In general we can write an energy equation Pemax = Etotal= Pe(h) + Ke(V) +Ker() Pemax =mgH ; H=maximiu release point where H=height of the celing less 3R Pe(h) =mgh ( this is a h than h=3R specified in theproblem) Ke(V) = 0.5mV2 Ker () = 0.5 I 2 where angular speed is related totranslational velocity V as =V/R and The moment of inertial for the cylinder is I=0.5mR2 No we have mgH= mgh + 0.5mV2 + 0.5 I2 mgH= mgh + 0.5mV2 + 0.50.5mR2 (V/R)2 gH= gh + 0.5V2 + 0.25V2   (nice an simple) V2 = g(H- h)/( 3/4) V2 = (4/3)g(H- h) h is the position of thecylinder above the floor. 1)linear acceleration of thecylinder V= at and h=0.5at2   then h=0.5a(V/a) 2   a= 0.5V2/h   Just before cylinder hitsteh floor use H (height of the celing less 3R) 2) tension in each string supporting thecylinder T=0.5 (g - a)m 3)magnitude of the torque rotating thecylinder = RT 4)angular acceleration of thecylinder = a/R 5)the final linear speed of thecylinder h=0 gH= gh + 0.5V2 + 0.25V2    we have gH= 0.5V2 + 0.25V2   V=2 ( gH/3) 6)the final angular speed of thecylinder =V/R 7) angle (in radians) through which thecylinder rotates = 2 (H/R) 8) final kinetic energy of thecylinder Ke max= Pemax= mgH 9) the final angular momentum of thecylinder gH= gh + 0.5V2 + 0.25V2   (nice an simple) V2 = g(H- h)/( 3/4) V2 = (4/3)g(H- h) h is the position of thecylinder above the floor. 1)linear acceleration of thecylinder V= at and h=0.5at2   then h=0.5a(V/a) 2   a= 0.5V2/h   Just before cylinder hitsteh floor use H (height of the celing less 3R) 2) tension in each string supporting thecylinder T=0.5 (g - a)m 3)magnitude of the torque rotating thecylinder = RT 4)angular acceleration of thecylinder = a/R 5)the final linear speed of thecylinder h=0 gH= gh + 0.5V2 + 0.25V2    we have gH= 0.5V2 + 0.25V2   V=2 ( gH/3) 6)the final angular speed of thecylinder =V/R 7) angle (in radians) through which thecylinder rotates = 2 (H/R) 8) final kinetic energy of thecylinder Ke max= Pemax= mgH 9) the final angular momentum of thecylinder 1)linear acceleration of thecylinder V= at and h=0.5at2   then h=0.5a(V/a) 2   a= 0.5V2/h   Just before cylinder hitsteh floor use H (height of the celing less 3R) 2) tension in each string supporting thecylinder T=0.5 (g - a)m 3)magnitude of the torque rotating thecylinder = RT 4)angular acceleration of thecylinder = a/R 5)the final linear speed of thecylinder h=0 gH= gh + 0.5V2 + 0.25V2    we have gH= 0.5V2 + 0.25V2   V=2 ( gH/3) 6)the final angular speed of thecylinder =V/R 7) angle (in radians) through which thecylinder rotates = 2 (H/R) 8) final kinetic energy of thecylinder Ke max= Pemax= mgH 9) the final angular momentum of thecylinder L=I L=0.5mR2 V/R L=   0.5mR V Just plug and paly Please let me know if you have any questions

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