A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle inclination 300. The centre of mass of cylinder has speed of 4 ms-1. The distance travelled by the cylinder on the inclined surface will be(Take, g = 10 ms-2

When a body rolls, i.e. have rotational motion, the total kinetic energy of the system will beKE=12mv21+K2R2where, m = mass of body, v = velocity and K = radius of gyration.Given,  m=2kg,θ=30∘ and v=4ms−1Let h be the height of the inclined plane, then from law of conservation of energy,KE=PE12mv21+K2R2=mghSubstituting the given values in the above equation, we get12×2×161+12=2×10×h  For cylinder, K2R2=12⇒ 8×32=10h⇒h=1.2mFrom the above diagram,  sin⁡θ=hxx=hsin⁡θ=1.2sin⁡30∘=1.2×2=2.4m