A hollow sphere of radius 0.140 m, with rotational inertia l = 0.0690 kgvm2 about a line through its center of mass, rolls without slipping up a surface inclined at 356 to the horizontal. At a certain initial position, the sphere's total kinetic energy is 98.0 J. (a) How much of this initial kinetic energy is rotational? (bl What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 1.40 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass? l l (a) Number i Units J (bl Number 4.72 Units m/s v (c) Number Units J M I (d) Number Units m/s v C The speed of the centre of mass of the sphere at initial position is 4.71 mls I's the sphere is moved up the ericline , the change in the height will be , h = 140 x Sim ( 35 . 6 ' ) = 0 8149 m - According to the law of conservation of the total mechanical energy Ki + Vi = kf + Uf Let , the initial value of the potential energy be zero; then the final potential energy will be , Of = mghFor the given values of the Kinetic of potential energy g 8 0 ] + 0 = kf + (5 . 28 kg) ( 9 . 8 m/s2 ) . (0 . 8149 mo) Kf = 55 . 83 J Therefore, the total Kinetic energy ky = 55.83 ] . The velocity at this point is 2 x 55 . 83 .. . V = 2 kf 5 . 28 V 111 . 66 21.14 5. 28 = 4.59 m/s V = 54.59 mm's