Question

A size-5 soccer ball of diameter 21.1 cm and mass 428 g rolls up a hill without slipping, reaching a maximum height of 4.1 m above the base of the hill. We can model this ball as a thin-walled hollow sphere. At what rate was it rotating (in rad/s ) at the base of the hill?

Answer 1

Sol- from Question we have Diameter of Ball = 21.1cm radius = 21.1 cm = 10.55cm = 10.55*10-2 m mass of Ball = 428 gram = 0.428 Kg Maximum height = 4.1m So we moment of Inertial of hollow sphere I I = 3 MR2 about its center Where M=mass R Yadius Since our ball is rolling without slipping from batter up. Let its angular welocity is w and linear velocity V at the battom so its total energy at bottom is € = £ IW2 + 2 muz Where I I62 - Rotational energy I mv2= Kinetic energy Since this energy convert into patential energy at top of hill so we have ngh=Ź I W2 + 1 mv2 since we know I= 3 MR2 v=Rw mgh = 1.2 m R² W2 + 1 m R²2 mgth = 2 MR2 62 [2 + 1] ma - мр- о?у 2 gh x 3 - w2 w2 5R2 5 R2 0 = 29% 73 Now g = 9.8 m/s h= 4.1m R = 10.55 X 102 m = 0.10 55m W = 2x9.884.1x 3 5Xco-10552 W = 65.81rad/s Hence at bottom our ball was! rotating at angular speed w=65.81radle