A solid cylinder of mass 2kg rolls down (pure rolling) an inclined plane from a height of 4m. Its rotational kinetic energy, when its reaches the foot of the plane is (Take g=10m`s^(-2))`A. 20 JB. 40 JC. `80/3`JD. 80 J

A solid cylinder of mass 2kg rolls down (pure rolling) an inclined pla

1.	A solid cylinder of mass 2kg rolls down (pure rolling) an inclined plane from a height of 4m. Its rotational kinetic energy, when its reaches the foot of the plane is (Take g=10m`s^(-2))`A. 20 JB. 40 JC. `80/3`JD. 80 J
Answer» c) At foot of the plane, For rotational motion about CM `2Tr = 1/2Mr^(2)alpha` Where, r = radius of cylinder `rArr 2T = 1/2Ma`…………….(ii) `From Eqs. (i) and (ii), we get `a = 2/3g` and `T=(mg)/6` Now, from equation of motion ,`v^(2) = u^(2) + 2gh`, we get `v^(2) = 2(2/3g) h rArr v= sqrt((4gh)/3)` `mgh = 1/2 komega^(2) + 1/2mu^(2)` `80 = 3/4mu^(2)` `therefore` Rotational KE = 1/2lomega^(2)` `=1/4 mu^(2) = 1/4 xx (80 xx4)/3 = 80/3` J

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A solid cylinder of mass 2kg rolls down (pure rolling) an inclined plane from a height of 4m. Its rotational kinetic energy, when its reaches the foot of the plane is (Take g=10m`s^(-2))`A. 20 JB. 40 JC. `80/3`JD. 80 J

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