A hollow sphere of radius 0.230 m, with rotational inertia I = 0.0744 kg·m2 abou

Question

A hollow sphere of radius 0.230 m, with rotational inertia I = 0.0744 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 27.0° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 19.0 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 0.840 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

          The angle of inclination is = 270            The initial kinetic energy is K = 19 J ---------------------------------------------------------------------------             The intiial kinetic energy is K = (1/2)I2 + (1/2)mU2                                                  19 J = (1/2)I( U /R)2 + (1/2)mU2                                                  19 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2                                                          19 J = (1/3)m U2 + (1/2)mU2                                                  19 J = (1/3)m U2 + (1/2)mU2                                                      19 J = (5/6)m U2     -----------(a)                                                  mU2 = 6( 19 J) / 5                                                          = 22.8J ------------(1)         Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2                                                                               = 19 J - ( 22.8J /2)                                                                               = 7.6 J                      But the moment of inertia is I = (2/3)mR2                                                           m = 3I / 2R2                                                            = 0.1569 kg              From the equation (1) mU2 = 22.8 J                                                        U = ( 22.8 J / m )1/2                                                            = 12.05 m/s                   This is the speed of the center of mass    (c) if the sphere moves 0.840 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                                                         = 19J +(0 - mgh)                                                          = 19J- mg(0.840m*sin)                                                          = 18.41 J               This is the final total kinetic energy (d)    From the equation (a) the total kinetic energy is K2 = (5/6)m V2                                                                                      V = [ 6(K2) / 5m ]1/2                                                                                           = 11.86 m/s                                                                                           = 11.86 m/s                                                                                                                                      

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