Question

The elevator system given in the figure moves up and down with a velocity of v=1 m/sec. As the elevator starts to move, it reaches to its steady state angular velocity in 1 seconds. It is known that all the friction effects within the system are neglected. The other information related with the system are given as below: Mass moment of inertia about motor shaft M-M : JM=1.25 kgm2 Mass moment of inertia about the elevator shaft K-K : JK=1 kgm2 Mass of the elevator cabin : M=400 kg Counter mass : m=100 kg Radius of the pulley : R=0.25 m Gear ratio : i=13 Gravitational acceleration : g=10 m/s2 a. Calculate the angular velocity of motor shaft and elevator shaft. b. Find the equivalent mass moment of the system that is lumped at motor shaft. c. Find the total moment acting on the motor shaft (in Newton-meters) in steady state.

Answer 1

The elevator system moves up and down with a velocity of v = 1 m/sec. As the elevator starts to move, it reaches its steady-state angular velocity in 1 second. It is known that all the friction effects within the system are neglected. the other information related to the system are given as below:

Mass moment of inertia about motor shaft M-M = J_M = 1.25 kgm²

Mass moment of inertia about the elevator shaft K-K = J_K = 1 kgm²

Mass of the elevator cabin = M = 400 kg

Counter mass = m = 100 kg

Radius of the pulley = R = 0.25 m

Gear ratio = i = 13

Gravitational acceleration = g = 10 m/s²

(a)

The velocity of the elevator is \(\mathrm{v}=1 \mathrm{~m} / \mathrm{s}\)

The radius of the pulley \(\mathrm{R}\) is \(0.25 \mathrm{~m}\).

The angular velocity of the elevator shaft is,

\(w_{e}=\frac{v}{R}=\frac{1}{0.25}=4 \mathrm{rad} / \mathrm{s}\)

The gear ratio is 13

So the angular velocity of the motor shaft is

\(w_{m}=i * w_{c}=13 * 4=52 \mathrm{rad} / \mathrm{s}\)

(b)

the equivalent mass moment at the motor shaft is

\(J_{t}=J_{m}+\frac{j_{k}}{N^{2}}\)

\(J_{t}=1.25+\frac{1}{13^{2}}=1.2559 \mathrm{Kgm}^{2}\)

(c)

Total moment acting on the motor shaft is given as

Sum of the moment due to the elevator (mvr), moment due to elevator shaft \(\left(J_{K} w_{e}\right)\), and moment due to motor shaft \(\left(J_{M} w_{m}\right)\)

\(L_{T}=(M-m) v R+J_{K} w_{e}+J_{M} w_{m}=300 * 1 * 0.25+1 * 4+1.25 * 52\)

\(L_{T}=144 N m\)