15. A disc is free to rotate about an axis passing through its centre and perpendicular to its plane. The moment of inertia o the disc about its rotation axis is I. A light ribbon is tightly wrapped over it in multiple layers. The end of the ribbon is pulled out at a constant speed of u. Let the radius of the ribboned disc be R at any time and thickness of the ribbon be d (<< R). Then the force (F) required to pull the ribbon as a function of radius R is:


1   I d u 2 2 πR 4 2   I d u 2 3 πR 4 3   2 I d u 2 πR 4 4   3 I d u 2 πR 4

let the angular velocity of the disc and ribbon be ωu=ωRdudt=d ωRdtSince u is constantd ωRdt=0Rdωdt+ωdRdt=0αR+uRdRdt=0α=-uR2dRdt________1Let radius of disk be D Now see the length of ribbon is constant, therefore if we consider the whole of ribbon then its area of sideways will be constantArea of side view as shown in figure of ribbon bounded on disc+side view area of ribbon opened=constantπR2-πD2+utd=constantd dtπR2-πD2+utd=0d dtπR2+d dt-πD2+d dtutd=02πRdRdt+0+ud=0dR dt=-ud2πRPut in equation 1α=-uR2-ud2πR=u2d2πR3We knowτorque=Iα=Iu2d2πR3Force, F=τorqueR=Iu2d2πR4

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