Question

A hollow sphere of radius 0.190 m, with rotational inertia I = 0.0218 kg·m² about a line through its center of mass, rolls without slipping up a surface inclined at 10.8° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 7.90 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 0.600 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Answer 1

Radius of the hollow sphere, r = 0.190 m

Rotational inertia of the sphere, I = 0.0218 kg*m^2

(a) Total kinetic energy (K.E) of the sphere = 7.90 J

Again,

Total K.E. = Translational K.E. + Rotational K.E.

=> 7.90 = (1/2)mv^2 + (1/2) * (2/3)mr^2ω^2 = (1/2) mv^2 + (1/3) mv^2 = (5/6)*mv^2

=> 7.90 = (5/6)* mv^2

Rotational K.E = (1/3) mv^2 = (1/3) * 7.90 * (6/5) = 3.16 J (Answer)

(b) r = 0.19 m

M.I. = 0.0218 = (2/3)*m*(0.19)^2

=>m = (0.0218 * 3) / (2 * 0.19^2) = 0.906 kg

Total K.E. = (5/6) mv^2 = 7.90

v^2 = (7.90*6) / (5*0.906)

v = 3.23 m/s (Answer)

(c) Decrease in total K.E. at the height h = 0.60*sin 10.8 is equal to the increase in potential energy.

mgh = 0.906*9.8* 0.60*sin 10.8 = 1.0 J

Total K.E. at height h is (7.90 - 1.0 ) = 6.90 J (Answer)

(d) At a height h , when the velocity of center of mass is v1 , again as shown in (a) total K.E. is ,

(5/6) m v1^2 = 6.90 J

v1^2 = (6*6.90) / (5 * 0.906)

=> v1 = 3.02 m/s (Answer)