A subsequent question stemming from here A massless motor is attached to a massless pulley(the blue circle) with a fixed rotational inertia and they are all attached to a mass X (the green platform). A cable (in orange) with spring constant K is attached to the pulley from the black rigid ceiling, the cable could be at the top of the pulley at t=0, or it could be at the side as shown in the picture.

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Assume it is a motor supplying the torque to the pulley. What torque should the motor (assume it has the same outer diameter with the pulley) output to move the mass up? Should the torque produce a force that is the same as the weight of the platform and the cable would experience a tension two times the weight of the platform? If this is so, it could be said that it is the tension from the cable that is pulling up the mass with an acceleration of g, right? But if the motor rpm is constant, then why would there be acceleration?

  • $\begingroup$ You link didn't work $\endgroup$ – Fred Feb 21 at 4:49

Assume the cable is simply looped around the pulley in counter-clock-wise direction, the initial position should be as shown on the sketch (by the side of pulley). Further assume the pulley and the motor are fixed in space, and the platform is attached to another end of the cable. Then the tension in the cable equals mass of the platform times gravitational constant (F = mg), and the torque required to lift the platform must exceed the calculated torque, which equals to the force in the cable times radius of the pulley (T = Fr).

  • $\begingroup$ Thank you, but the cable is only attached to the pulley and the ceiling though? And is this tension that equals to the weight of the platform going to be constant throughout the lifting process? $\endgroup$ – user39178 Feb 20 at 18:42
  • $\begingroup$ @user39178 No. The tension at constant speed will be due to its weight. During upward acceleration and downward deceleration you'll have to add in F = ma. $\endgroup$ – Transistor Feb 20 at 19:31
  • $\begingroup$ but if the input voltage is constant, then back emf is constant and output torque and motor speed is also constant. That means there is no acceleration and the platform is moving up at constant speed, and the cable only carries the weight of the platform, right. And the only time it accelerates at constant voltage input is when the motor is starting up, before it enter steady state? $\endgroup$ – user39178 Feb 20 at 19:43
  • $\begingroup$ @39178 If the cable is infinite stiff, the work done is equal to mgh (h = height of the lift). On the other hand, if the cable is stretchable with a linear spring constant k, then the cable will stretch a distance x = F/k, and the total energy will increase to m*g*(h+x). However, in either case, the force remains constant, as well as the speed of lifting (a = 0). By the way, you need to find a way to connect the platform, pulley, and the cable to form a loop for force flow; otherwise, the lifting mechanism just not there. $\endgroup$ – r13 Feb 20 at 21:10
  • $\begingroup$ Could you elaborate what acceleration you are referring to? Is this acceleration caused by the acceleration of the pulley due to the motor trying to lower the torque it is creating to match the load torque? And why is the total energy increase being (h+x)? the x that happens due to the cable extending should be a negative value, right? $\endgroup$ – user39178 Feb 23 at 1:08

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