Fallingrope physics

# Fallingrope physics - we have F N – mg dm/dt v rel y = 0...

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A Falling Rope A uniform rope of mass M and length L is held with its lower end just touching the surface of a scale. The rope is released and begins to fall. Find the force of the scale on the rope just as the midpoint on the rope first touches the scale. Solution: First of all let’s draw a sketch of the situation including the initial and final configuration and the coordinate system of reference. There are two external forces on the portion of the rope on the scale: the force of gravity (mg) and the normal force (F N ) exerted by the scale. For a system with continuously variable mass Newton’s second law become: F net ext + dM/dt v rel = M d v /dt In this case the forces are along the y direction, so we can write the previous equation in component form: F net ext y + dm/dt v rel y = m dv y /dt Where m is the mass of the rope on the scale. Since the velocity of the system remains zero (for the part that is on the balance), then dv y /dt = 0, and substituting for F net ext y = F N – mg

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Unformatted text preview: we have: F N – mg + dm/dt v rel y = 0 In order to find F N we need to have an expression for dm/dt. Let’s first find the relation between dm and dl , where dm is the mass of a short rope segment of length dl that falls on the scale during time dt . Since the rope is uniform, the relation between dm and dl is dm/dl = M / L and multiplying both side for dt dm/dt = M/ L dl/dt = -M/L v rel y Note that d/dt is the impact speed of the segment, so v rel y =-dl/dt (v rel y is negative because up is the positive y direction and the rope is falling). Substituting in the equation for the force we have: F N = mg + M/L v 2 rel y To find v 2 rel y you need to note that, until it touches the scale, each point along the rope falls with constant acceleration g, therefore using the usual equation for a motion with constant acceleration we have: v 2 rel y = v 2 rel 0 +2a y Δ y = 0 + 2(-g) (-L/2) = gL and substituting in F N = Mg/2 + MgL/L = 3/2 Mg where m=M/2...
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Fallingrope physics - we have F N – mg dm/dt v rel y = 0...

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