Question

Two masses, m1 = 27.0 kg and m2 = 50.0 kg, are connected by a rope that hangs over a pulley (as in fig. 10-60). the pulley is a uniform cylinder of radius 0.40 m and mass 4.8 kg. initially m1 is on the ground and m2 rests 2.5 m above the ground. if the system is released, use conservation of energy to determine the speed of m2 just before it strikes the ground. assume the pulley bearing is frictionless.

Answer 1

3.8 m/s Assuming that the mass of the rope is negligible, we have three objects that will be in motion, m1, m2, and the pulley. The sum of the gravitational potential energy plus the kinetic energy in the system will be constant. And since at the beginning, all of the energy will be gravitational potential, and at the end, all of it will be kinetic, let's start by calculating the gravitational potential energy we can extract from the system. We can ignore the pulley for this situation and merely consider m1 and m2. The energy we can extract from m2 falling is the mass of m2 multiplied by the distance it can fall times the gravitational acceleration. So: E = 50.0 kg * 2.5 m * 9.8 m/s^2 = 1225 kg*m^2/s^2 = 1225 J Now when m2 falls, it will be lifting m1, giving m1 some gravitational potential energy, let's calculate that energy and subtract it from the amount of energy available for the kinetic energy of the system. So: E = 27.0 kg * 2.5m * 9.8 m/s^2 = 661.5 kg*m^2/s^2 = 661.5 J E = 1225 J - 661.5 J = 563.5 J So the kinetic energy that the system will have at the moment m2 strikes the ground is 563.5 Joules. That kinetic energy will be distributed between m1, m2 and the pulley. The pulley will be modeled as a flywheel. Energy in a flywheel: E = (1/2)Iω^2 where E = Energy I = moment of inertia ω = angular velocity Moment of inertia for solid cylinder: I = mr^2/2 where I = moment of inertia m = mass r = radius So let's calculate I: I = mr^2/2 I = 4.8 kg * (0.4 m)^2/2 I = 4.8 kg * 0.16 m^2/2 I = 0.384 kg*m^2 And substitute I into the kinetic energy formula E = (1/2)Iω^2 E = (1/2) 0.384 kg*m^2 ω^2 E = 0.192 kg*m^2 ω^2 Now let's modify the equation to get rid of the ω^2 term and make it a function of rim velocity in m/s. The circumference of the pulley is 2*pi*0.40 and there are 2pi radians for a full revolution. And happily, that means 1 radian of rotation is 0.4 meters. So by dividing velocity in m/s by 0.4, we have radians/sec. So the equation becomes: E = 0.192 kg*m^2 ω^2 E = 0.192 kg*m^2 (v/0.4m)^2 E = 0.192 kg*m^2 v^2/(0.16 m^2) E = 1.2 kg* v^2 Now the energy of a mass moving is: E = 1/2 m v^2 So the total energy of the two masses and the pulley is: E = 1/2mv^2 + 1/2mv^2 + 1.2 kg v^2 E = 1/2*27 kg v^2 + 1/2 50 kg v^2 + 1.2 kg v^2 E = 13.5 kg v^2 + 25 kg v^2 + 1.2 kg v^2 E = (13.5 kg + 25 kg + 1.2 kg) v^2 E = 39.7 kg v^2 Now let's solve for v, substitute the available energy and calculate. E = 39.7 kg v^2 E/ 39.7 kg = v^2 sqrt(E/ 39.7 kg) = v sqrt(563.5 J/ 39.7 kg) = v sqrt(14.19395466 m^2/s^2) = v 3.767486518 m/s = v Finally, round to 2 significant figures since that's the lowest precision datum available, giving a speed of 3.8 m/s just prior to m2 striking the ground.