A carpet of mass M made of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. If R = 1.05 m then calculate the horizontal velocity of the axis (in m/s) of the cylindrical part of the carpet when its radius reduces to R/2. (Take g = 10 m/s2)

Figure shows the forces acting on the carpet in two cases:i. when the radius of the roll is R.ii. when the radius is R/2Mass of the roll of radius R/2 = M/4  (Mass is proportional to the area of cross section)The mass of the part of the carpet spread on the floor = (3/4) M.The mechanical energy of the system will remains constant.The part of the carpet spread on the ground will have no kinetic energy. According to conservation of energy, ΔK+ΔU=0⇒Kf−Ki+ΔU=0              …(i)As the roll of the carpet is not sliding, we can consider that the roll of carpet as a rolling cylinder. We can calculate the kinetic energy of the moving cylindrical roll by the relationK=(1+β)mv22=1+12mv22=3mv24M = mass of the roll at any timeHence the kinetic energy of the roll when its radius is R/2Kf=3(M/4)v24=3Mv216Change in kinetic energy, ΔK=Kf−Ki=3Mv216Change in gravitational potential energy.ΔU=−34MgR+14MgR2=−78MgRFinally substituting ΔU and ΔK in the equation (i), we have⇒3Mv216+−78MgR=0⇒v=143gR=143×10×1.05=7 m/s