10. A hollow sphere of radius 0.150 m, with rotational inertia I = 0.0636 kg·m2

Question

10. A hollow sphere of radius 0.150 m, with rotational inertia I = 0.0636 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 12.2° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 48.0 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?

Explanation / Answer

at the radius of the hallow sphere is R = 0.15 m The moment of inertia is I = 0.0636 kg .m2 The angle of inclination is ? = 12.2 degree The total kinetic energy is K = 48 J --------------------------------------------------------------------------- The intiial kinetic energy is K = (1/2)I?2 + (1/2)mU2 48 J = (1/2)I( U /R)2 + (1/2)mU2 48 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2 48 J = (1/3)m U2 + (1/2)mU2 48 J = (1/3)m U2 + (1/2)mU2 48 J = (5/6)m U2 -----------(a) mU2 = 6* 48 J / 5 = 57.6 J ------------(1) Then the rotational kinetic energy is (1/2)I?2 = K - (1/2)mU2 = 48 J - ( 57.6 J /2) = 19.2 J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R^2 = 4.24 kg From the equation (1) mU2 = 48 J U = ( 57.6 J /4.24 kg )1/2 = 3.68 m/s This is the speed of the center of mass (c) if the sphere moves 1.10 m up then From the conservation of energy at intiial point and the final point K1 + P.E1 = K2 + P.E2 K2 = K1 + P.E1 - P.E2 = 48J +(0 - mgh) = 48 J - mg(1.1*sin12.2) = 38.34 J This is the final total kinetic energy (d) From the equation (a) the total kinetic energy is K = (5/6)m V2 V = [ 6*38.34 / 5*4.24 ]1/2 = 3.294 m/s

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