Question

A hollow sphere of radius 0.150 m, with rotational inertia I = 0.0460 kg·m² about a line through its center of mass, rolls without slipping up a surface inclined at 23.2° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 44.0 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 1.10 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

a)

Total kinetic energy of the sphere = 44 J

TKE = (0.5) I w² + (0.5) m v² = 44

TKE = (0.5) (2/3) (m r²) (v/r)² + (0.5) m v² = 44

TKE = (0.5) (2/3) m v² + (0.5) m v² = 44

TKE = (0.5) m v² (2/3 + 1)= 44

KE = (0.5) m v² = 44 (3/5)

hence

RKE = Rotational Kinetic energy = (2/5) 44 = 17.6 J

Rotational Kinetic energy is 0.40 times the total kinetic energy

b)

I = moment of inertia of sphere = 0.046 kgm²

m = mass of sphere

r = radius of sphere = 0.150 m

moment of inertia of sphere is given as

I = (2/3) (m r²)

0.046 = (2/3) m (0.150)²

m = 3.07 kg

we have

(0.5) m v² = 44 (3/5)

(0.5) (3.07) v² = 44 (3/5)

v = 4.15 m/s

c)

L = length of incline traveled = 1.10 m

= Angle of incline = 23.2

h = height of the incline

Height of incline is given as

h = L Sin

h = (1.10) Sin23.2

h = 0.433 m

using conservation of energy

initial total kinetic energy = potential energy + final kinetic energy

44 = mgh + TKE_f

44 = (3.07) (9.8) (0.433) + TKE_f

TKE_f = 30.97 J

d)

TKE_f = 30.97 J

(0.5) (m v²) (2/3 + 1)= 30.97

(0.5) (3.07) v² (5/3)= 30.97

v = 3.5 m/s