- #1

JosefMartens

- 4

- 0

## 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: E

_{kin}= ½* m*v

^{2}+ ½*I*θ

^{2}

Ekin = ½(mr² + I)θ² = ½ * I

_{equivalent}* θ²

Mkin = I

_{equivalent}* θ

ΔW = ΔEpot/ΔX

Mpot = ΔW/Δθ

Mtot = Mpot + Mkin