MY NOTES A solid sphere is released from the top of a ramp that is at a height h. - 2.30 m. It rolls down the ramp without slipping. The bottom of the ramp is at a height of h, = 1.91 m above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.14 m. (a) Through what horizontal distance d, in meters, does the ball travel before landing? m (b) How many revolutions does the ball make during its fall? rev Supporting Materials Physical ConstantsA uniform rod of mass 1.80 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m, - 4.50 kg is attached to one end and a second mass m, = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles. m11 (a) What is the moment of inertia of the system in kg . m-? kg . m 2 (b) If the rod rotates with an angular speed of 2.40 rad/s, how much kinetic energy, in joules, does the system have? (c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined, in kg . m-? kg . m- (d) If the rod is of negligible mass, what is the kinetic energy, in joules, when the angular speed is 2.40 rad/s?The angular speed at the moment it leave the ramp = 16. 69 radian/ see The angle rowolution during its fall 0 = wxt - - [t = time elapse during fall from = 16 .69 x 2hz 9 = 16.69 X 2x 1. 91 9. 8 - 10.42 radian. Since in one revolution, ball rotate ( complete) 2TT radian so in 10.42 radian, the number of rrevolution will be h= 10 . 42 - 10.42 2x3. 14 n = 1. 65 9 ) revolution I get same answer,. you have asked angular speed = 16.69 radian/see Angular speed is defined as w= Angle rotate/ revebore time . 2 TIX number of revolution time But putting all variable we will get same