Asolid sphere is rolling on a surface. what fraction of its total kinetic energy is in the form of rotational kinetic energy about the center of mass

Total KE = KE (rotational) + KE (translational) Moment of inertia of sphere is I = (2/5)mr^2 So KE (rotational) = (1/2) x I x w^2 = (1/2) x (2/5)mr^2 x w^2 = (1/5) x m x r^2 x w^2 KE (translational) = (1/2) x m x v^2 = (1/2) x m x (rw)^2 = (1/2) x m x r^2 x w^2 Hence KE = (1/5) x m x r^2 x w^2 + (1/2) x m x r^2 x w^2 = m x r^2 x w^2 ((1/5) + (1/2)) KE = (7/10) m x r^2 x w^2 Calculating the fraction of rotational kinetic energy to total kinetic energy, = rotational kinetic energy / total kinetic energy = (1/5) x m x r^2 x w^2 / (7/10) m x r^2 x w^2 = (1/5) / (7/10) = 2 / 7 The answer is 2 / 7

the simple machine has three main features. they are:

loadfulcrumeffort

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever