A 1.90 kg thin, spherical shell of radius 0.140 m is released from rest at point A in the figure below, its center of gravity a distance of 1.70 m above the ground. The spherical shell rolls without slippingto the bottom of an incline and back up to point B where it is launched vertically into the air. The spherical shell then rises to its maximum height hmay at point C.HINTV = ()1.70 mhmaxBT0.450 m(a) At point B, find the spherical shell's translational speed v, (In m/s).m/s(b) At point B, find the spherical shell's rotational speed w, (in rad/s).rad/s(c) At point C, find the spherical shell's rotational speed w, (in rad/s).rad/s(d) At point C, find the maximum height hmay of the spherical shell's center of gravity (in m).

Given:- The mass of the spherical shell m = 1.90 kg The radius of the spherical shell r = 0.140 m its center of gravity is a distance of h = 1.70 m Find:- a) At point B, find The spherical shell's translation speed vB b) At point B, find the spherical shell's rotational speed ωA c) At point C, find the spherical shell's rotational speed ωC note:- “Since you have posted a question with multiple sub-parts, we will solve the first three subparts for you. To get the remaining sub-part solved please repost the complete question and mention the subpart to be solved"Solution:- a) We know that the law of energy conservation is given as mgh = 12mvB2+12IωB2+mghB We know that VB = ωBR ωB=VBRI = 23mR2 mgh= 12mvB2+12×23mR2vBR2+mghB So Now mgh= 12mvB2+12×23mR2vBR2+mghB mgh= 12mvB2+26mRvB2+mghBgh = 3vB2+2RvB2+6ghB66gh -6ghB= vB23+2R6gh-hB=vB23+2R6gh-hB=vB23+2RvB= 6g[h-hB]3+2R Now put the given value is as in the equation vB= 6g[h-hB]3+2Rput the value in the equation vB= 6×9.8[1.70-0.450]3+2×0.140vB= 6×9.8[1.70-0.450]3+2×0.140vB= 4.73 m/s b) Now We know that the relation between the angular velocity and the linear velocity vB= ωBR ωB= vBRωB= 4.730.140= 33.785ωB= 33.785 rad/sec c) Now we again apply the law of conservation between the points B and C is given as 12mvB2+mghB= mghC12mvB2+mghB= mghC12vB2+ghB= ghC2ghC= vB2+2ghB2ghC= 4.732+2×9.8×0.450hC= 1.59 m So the hmax= hC=1.59 m