A uniform rolling body of radius R is projected up an inclined plane with an initial speed v . It rolls without slipping and reaches a maximum vertical height h = 3v^2 4g before momentarily coming to rest. If the radius of gyration of the body about its center of mass is k , the value of 10 ( k^2 R^2 ) is ______.

Step-by-step solution

By the principle of conservation of mechanical energy, the total initial kinetic energy of the rolling body is converted into potential energy at the maximum height. Loss in K.E. = Gain in P.E. 1 2 mv^2 (1 + k^2 R^2 ) = mgh Given that h = 3v^2 4g , we substitute this into the equation: 1 2 mv^2 (1 + k^2 R^2 ) = mg ( 3v^2 4g ) 1 2 (1 + k^2 R^2 ) = 3 4 1 + k^2 R^2 = 3 2 k^2 R^2 = 1 2 Therefore, the value of 10 ( k^2 R^2 ) is 10 1 2 = 5 . Answer: 5