A hollow sphere of radius 0.35 m, with rotational inertia I = 0.060 kg m2 about

Question

A hollow sphere of radius 0.35 m, with rotational inertia I = 0.060 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 10° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 95 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?

Now, the sphere moves 1.0 m up the incline from its initial position.

c) What is its total kinetic energy now?
d) What is the speed of its center of mass now?

PLEASE HELP :)

Explanation / Answer

  The moment of inertia is I = 0.060 kg .m2    The angle of inclination is = 100            The initial kinetic energy is K = 95 J             The intiial kinetic energy is K = (1/2)I2 + (1/2)mu2                                                  95 J = (1/2)I( u /R)2 + (1/2)mu2                                                  95 J = (1/2)(2/3)mR2* u2 / R2 + (1/2)mu2                                                          95 J = (1/3)m u2 + (1/2)mu2                                                   95 J = (5/6)m u2     -----------(1)                                                  mu2 = 6* 95 J / 5                                                          = 114 J ------------(2)   b ) the rotational kinetic energy is (1/2)I2 = K - (1/2)mu2                                                                               = 95 J - ( 114 J / 2)                                                                               = 38 J                      But the moment of inertia is I = (2/3)mR2                                                           m = 3I / 2R2                                                            = 3 * 0.060 kg m^2 / 2 ( 0.35m)^2                                                            = 0.73 kg              From the equation (2) mu2 = 102 J                                                        u = ( 114 J / m )                                                            = 10.6 m/s                   This is the speed of the center of mass    (c) if the sphere moves 1.0 m up then                    From the conservation of energy at intiial point and the final point                                      K.E_11 + P.E_1 = K.E_2 + P.E_2                                                   K2 = K1 + P.E1 - P.E2                                                         = 95J +(0 - mgh)                                                          = 95J - mg(1.0m*sin)                                                          = 95 J - 0.73 kg * 9.8 m/s^2 ( 1.0 m sin 10)                                                         = 93.7 J                                                                        (d)    From the equation (1) the total kinetic energy is K.E_2 = (5/6)m v2                                                                                       v = [ 6*K.E_2* / 5m ]                                                                                           = 6 * 93.7 J / 5 * 0.73 )                                                                                           = 12.4 m/s                                                                                           = 6 * 93.7 J / 5 * 0.73 )                                                                                           = 12.4 m/s                                              

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