A solid sphere of uniform density starts from rest and roll without slipping a d

Question

A solid sphere of uniform density starts from rest and roll without slipping a distance of d=2.4 m down a Theta=34?/b> incline. The sphere has a mass M=5.3 kg and a radius R=0.28 m. Of the total kinetic energy of the sphere, what is translational? What is the translational kinetic energy of the sphere when it reaches the bottom of the incline? What is the translational speed of the sphere as it reaches the bottom of the ramp? Now let's change the problem a little. S suppose now that there is no frictional force between the sphere and the inline. Now, what is the translational kinetic energy of the sphere at the bottom of the

Explanation / Answer

mass m = 5.3 kg radius R = 0.28 m distance d = 2.4 m angle = 34 degrees (a). required ration = K / ( K + K ' ) where K = translational K.E = ( 1/ 2) m v ^ 2            K' = rotational K.E = ( 1/ 2) I ^ 2                 = ( 1/ 2) ( 2 / 5 ) m R ^ 2 * ( v / R ) ^ 2 substitue these values weget require ratio   =1 / ( 1 + ( 2/ 5 ) )                                                                 = 5 / 7 (b). Translational K.E when it reach the bottomground   = ( 1/ 2) m V ^ 2 where V = speed of the sphere when it reach theground               = [ 2 g h ]            h= height of the ramp = d sin substitue values weget answer from this we findanswers         

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