A solid cylinder of mass 2 ~kg and radius 50 ~cm rolls up an inclined plane of angle inclination 30^ . The centre of mass of cylinder has speed of 4 ~m / s . The distance travelled by the cylinder on the inclined surface will be : (Take g=10 ~m / s ^2 )

Step-by-step solution

When a body rolls i.e. have rotational motion, the total kinetic energy of the system will be KE = 1 2 mv ^2 (1+ k ^2 R ^2 ) where, m = mass of body, v = velocity and k = radius of gyration Given, m =2 ~kg , =30^ , v =4 ~ms ⁻¹ Let h be the height of the inclined plane, then from law of conservation of energy, aligned KE & = PE 1 2 mv ^2 (1+ k ^2 R ^2 ) & = mgh aligned Substituting the given values in the above equation, we get aligned & 1 2 2 16 (1+ 1 2 )=2 10 h [ For cylinder k ^2 R ^2 = 1 2 ] & 8 3 2 =10 ~h h =1.