A hollow sphere of radius 0.120 m, with rotational inertia I = 0.0300 kg·m 2 abo

Question

A hollow sphere of radius 0.120 m, with rotational inertia I = 0.0300 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 40.0° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 37.0 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 1.20 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Explanation / Answer

a)

I = moment of inartia = (2/3) mr2 = 0.03

Rotational KE = (0.5) I w2 = (0.5) ((2/3) mr2 ) (v/r)2 = (0.5) (2/3) m v2

transational KE = (0.5) m v2

ratio of RKE and TKE = RKE/TKE = (0.5) (2/3) m v2 / (0.5) m v2 = 2/3

RKE = (2/3) TKE

total kinetic energy = RKE + TKE

37 = RKE + (3/2) RKE

RKE = 14.8 J

b)

(0.5) I w2 = 14.8

(0.5) (0.03) w2 = 14.8

w = 31.41 rad/s

V = r w = 0.12 x 31.41 = 3.77 m/s

c)

height gained , h = L Sin40 = 1.20 Sin40 = 0.77 m

(2/3) mr2 = 0.03

(2/3) m(0.12)2 = 0.03

m = 3.125 kg

using conservation of energy

Total energy at bottom = potential energy + kinetic energy + rotational KE at top

37 = mgh + (0.5) m v2 + (0.5) I w2

37 = (3.125) (9.8) (0.77) + (0.5) (3.125) v2 + (0.5) (2/3) (3.125) v2

v = 2.27 m/s

kinetic energy = KE = (0.5) m v2 = (0.5) (3.125) (2.27)2 = 8.1 J

d)

v = 2.27 m/s

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