A solid sphere of mass 0.604kg rolls without slipping along a horizontal surface with a translational speed of 5.04m/s. It comes to an incline that makes an angle of 32 degrees with the horizontal surface. Ignoring energy losses due to friction, to what vertical height above the horizontal surface does the sphere rise on the incline?

Question

A solid sphere of mass 0.604kg rolls without slipping along a horizontal surface with a translational speed of 5.04m/s.  It comes to an incline that makes an angle of 32 degrees with the horizontal surface.  Ignoring energy losses due to friction, to what vertical height above the horizontal surface does the sphere rise on the incline?

anonymous · Answer

total energy needs to be conserved 
so energy at horzontal surface=energy at height h(to which it rises)
so can u complete it showing the equation?
another thing:if its gonna roll then it should have some angular velocity omega
kinetic energy=1/2*m*v^2
hint: TOTAL K.E=sum of linear+rotational

anonymous · Answer

By the conservation of energy ::
Total mechanical energy at initial=total mechanical energy at final
  1/2mv^2+1/2I(omega)^2+0=0+0+mgh
here omega=v/r;
     I=2/5mr^2;
   find h from this equation....