Peak torque and power for an elevator motor

  • #1

Homework Statement


A loaded elevator acge has a mass of 1500kg and a countermass of 1000kg. The cage is accelerated with 2m/s² to 1,4m/s and lifted to a height of 15m and stops at stand still. The inertia of the motor or the radius of the cablewheel is NOT given. The steady state speed of the motor is 30 revolutions per second.

Homework Equations


a) what is the mechanical energy required to lift the weight to 15m
b) What is the equivalent inertia at the motor axis
c) What is the peak torque and power

The Attempt at a Solution


The mechanical energy needed is the raise in potential energy : mgh with m = 500kg
I have a solution of a likewise excercise in my book, but i don't understand the solution.
I understand that P = dW/dt and i guess the torque M = P/d(theta) yet in the solution he uses dW/d(theta) for M (torque).
Also why is dW = dEpot/dX? And how does he calculate I(equivalent).
I also don't fully understand why he calculates 2 torques: 1 with the potential energy to counter gravity and 1 with the kinetic energy to accelerate the mass?

here is the solution: Ekin = ½* m*v2 + ½*I*θ2
Ekin = ½(mr² + I)θ² = ½ * Iequivalent * θ²
Mkin = Iequivalent * θ
ΔW = ΔEpot/ΔX
Mpot = ΔW/Δθ
Mtot = Mpot + Mkin
 

Answers and Replies

  • #2
haruspex
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he uses dW/d(theta) for M (torque).
Yes. Just as work = force x displacement, you also have work = torque x displacement angle.
ΔW = ΔEpot/ΔX
This looks dimensionally wrong, but without a definition of the variables I cannot be sure. Anyway, it is not used in what you posted.
 
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  • #3
Ah yes that makes sense. Do you have any idea why he calculates 2 torques and how he gets Iequivalent?

But wait he uses θ as angle velocity..?
 
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  • #4
haruspex
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Ah yes that makes sense. Do you have any idea why he calculates 2 torques and how he gets Iequivalent?
Because the cage weight has a constant lever arm about the pulley, the cage mass can be represented as a moment of inertia about the pulley axle. This is being added to that of the pulley to produce an effective MoI of the system.
 
  • #5
Ah so 1 torque is calculated for the work done by lifting the cage 15m in the air and the 2nd is because of the moment of inertia of the system. But the pulley radius or inertia of pulley or motor is not given. So how am i supposed to calculate the kinetic energy?
 
  • #6
haruspex
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Ah so 1 torque is calculated for the work done by lifting the cage 15m in the air and the 2nd is because of the moment of inertia of the system. But the pulley radius or inertia of pulley or motor is not given. So how am i supposed to calculate the kinetic energy?
I think you are supposed to assume that the 1,4m/s corresponds to 30 revs per second - except, that is extremely fast. Should it be 30 radians per second? 30 revs per minute?
 
  • #7
The 30revs is at the motor axis, there could be a reduction in between for example. And yes this 30revs corresponds with 1,4 m/s of the cage. But i don't see how this will get me Ekin.
 
  • #8
haruspex
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But wait he uses θ as angle velocity..?
No, as angular displacement.
The 30revs is at the motor axis, there could be a reduction in between for example. And yes this 30revs corresponds with 1,4 m/s of the cage. But i don't see how this will get me Ekin.
You are probably right, but even if not, without the MoI you cannot find the KE that was put into the pulley.
 

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