# Peak torque and power for an elevator motor

• JosefMartens
In summary, the homework statement states that a loaded elevator accelerates with 2m/s² to 1,4m/s and is lifted to a height of 15m. The cage has a mass of 500kg and an equivalent inertia at the motor axis of I. The peak torque and power is calculated to be Mpot = Iequivalent * θ and Mtot = Mpot + Mkin, respectively.
JosefMartens

## 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

JosefMartens said:
he uses dW/d(theta) for M (torque).
Yes. Just as work = force x displacement, you also have work = torque x displacement angle.
JosefMartens said:
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.

JosefMartens
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..?

Last edited:
JosefMartens said:
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.

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?

JosefMartens said:
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?

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.

JosefMartens said:
But wait he uses θ as angle velocity..?
No, as angular displacement.
JosefMartens said:
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.

## 1. What is peak torque and power for an elevator motor?

Peak torque and power for an elevator motor refers to the maximum amount of force and energy that the motor can produce at its highest level of operation. This is typically measured in Newton-meters (Nm) for torque and kilowatts (kW) for power.

## 2. Why is peak torque and power important for an elevator motor?

Peak torque and power are important for an elevator motor because they determine the maximum load and speed that the motor can handle. This is crucial for ensuring the safe and efficient operation of the elevator, as well as determining the size and capacity of the elevator.

## 3. How is peak torque and power calculated for an elevator motor?

Peak torque and power for an elevator motor can be calculated by multiplying the rated torque and power by the motor's torque and speed multiplier. This value is typically provided by the motor manufacturer and can vary depending on the motor's design and construction.

## 4. What factors can affect the peak torque and power of an elevator motor?

The peak torque and power of an elevator motor can be affected by several factors, including the motor's design and construction, the load and speed of the elevator, and any external factors such as friction or resistance. Regular maintenance and proper lubrication can also impact the motor's performance.

## 5. How can I ensure that the elevator motor is operating at its peak torque and power?

To ensure that the elevator motor is operating at its peak torque and power, it is important to regularly monitor and maintain the motor. This includes checking for any signs of wear or damage, ensuring proper lubrication, and addressing any issues that may affect the motor's performance. It is also important to follow the manufacturer's guidelines for load and speed limits to prevent overloading the motor.

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