Motor, a gearbox and a drum with a cable lifting a mass

Question

A lifting device consists of an electric motor, a gearbox and a drum with a cable for lifting a mass.

The speed after the initial acceleration is $v=0.5\,\text{m/s}$.

The gear efficiency $\eta=0.7$

Maximum motor torque is $T=15 \, \text{Nm}$

Motor inertia $J_m=0.005 \, \text{kgm}^2$

Drum diameter $d = 0.5 \, \text{m}$

Motor rotating speed after initial acceleration $n=2600\, \text{rpm}$

The inertia of the drum $J_L$ is considered small and thus neglected.

I want to Calculate how much time it takes for the motor to accelerate the mass $m=510 \,\text{kg}$ from standstill upwards to the given speed. Assume constant acceleration.

I know applying Newton's second law that:

$$F-mg=ma$$ and then we get the acceleration

$$a=\frac{F-mg}{m}$$ and the speed v

$$v=at$$ so $$t=\frac{mv}{F-mg}$$

The motor rotates the geared system $\omega_M$ and the geared system rotates the drum $\omega_L$.

The mass $m$ causes the torque $T_L=mgr$ on the rope drum. So the power of the load is $P_L=T_L\omega_L$. And therefore $\eta P_m=P_L$ and there I get the $\omega_L$ when $$\omega_L=\frac{\eta\, T_m\, \omega_m}{T_L}$$

But how do I get force $F$ so I can calculate the time or is there better way to do this?

Or can I calculate the acceleration $a = \frac{T}{J_{motor}}$

Answer 1

Since this mechanism is all rigidly coupled, it has one degree-of-freedom. One differential equation combined with boundary conditions is enough to describe the position, velocity and acceleration of the system.

Steps for main problem:

1. Compute missing gear ratio as it is a useful parameter: $i=\frac{\omega_M}{\omega_L} = \frac{\frac{2\pi}{60}}{\frac{v}{r}}$

2. Transform interia: In order to do derive the equation of motion you have to transform all inertia values to ONE position in the described 'force chain'. I take the hanging load as my transformation target as it is of linear kind of motion and hence more intuitive than rotary kind. The transformation itself is nicely derived in here and many mechanics textbooks.

   • Transformed motor inertia $m_{M->L} = \frac{J_M \cdot i^2}{r^2}$ in $[kg]$

   • Transformed drum inertia $m_{D->L} = \frac{J_D}{r^2}$ in $[kg]$

   • Load inertia $m_L$ stays the same

3. Transform forces acting on the mechanism: This is mainly the motor torque $T_M$ which I assume is a constant. It is reduced by $\eta$ which is the constant power loss in the geartrain. There is also a static force $G$ neccessary to prevent the load from falling down.

   • Transformed motor torque $F_{M->L} = T_M \cdot i\cdot \eta$ in $[N]$

   • Weight of load $G_L = m_L \cdot g$

4. Build differential equation according Newtons 2nd law: $ m \cdot a = \sum{F}$

   • $(m_{M->L} + m_{D->L} + m_{L}) \cdot a = F_{M->L} - G_L$
   • $(\frac{J_M \cdot i^2}{r^2} + \frac{J_D}{r^2} + m_L) \cdot a = T_M \cdot i\cdot \eta - m_L \cdot g $

5. Since all parameters are known you can solve for $a$ which is a constant value. Integrating a over a time period $\Delta t_a$ gives you the velocity $v$ of the load, which is given above. Finally:

   $\Delta t_a = \frac{v}{a} = \frac{v \cdot (\frac{J_M \cdot i^2}{r^2} + \frac{J_D}{r^2} + m_L)}{T_M \cdot i\cdot \eta - m_L \cdot g}$

   BTW: The nominator of the fraction is the impulse of the system, the denominator the force trying to change the impulse.

About transformation of inertia:

• Assuming a lossless gearbox, there is conservation of power:

  $P_1 = T_1 \omega_1 = P_2 = T_2 \omega_2 $

• Basic kinematic relationship gives a fixed ratio between input and output:

  $\frac{\omega_2}{\omega_1} = \frac{\dot\omega_2}{\dot\omega_1} = i$

• Newtons 2nd law gives:

  $ J_1 \cdot \dot\omega_1 = T_1$ and $ J_2 \cdot \dot\omega_2 = T_2$

• and you can write by dividing the equations above and using the gear ratio:

  $\frac{T_2}{T_1} = \frac{\dot\omega_2}{\dot\omega_1} \cdot \frac{J_2}{J_1}= i \cdot \frac{J_2}{J_1} $

• Use the conservation of power and replace the ration of torques with the equation above:

  $\frac{T_2}{T_1} = \frac{\omega_1}{\omega_2} => i \cdot \frac{J_2}{J_1} = \frac{1}{i} => J_1 = i^2 \cdot J_2$

When you crank a machine at side 2 with a gearbox with a ratio $i=2$:

1. you need half the torque compared to cranking on the output shaft $T_2= \frac{1}{2}\cdot T_1$

2. you need to crank twice as much for a certain rpm at the load $\omega_2 = 2 \cdot \omega_1 $

3. the inertia of the load (e.g. a flywheel) you feel is cut down by factor of four $J_2 = \frac{1}{2^2}\cdot J_1$

One could also say the moment of inertia is reduced, but only for this common case. If you sit on the output shaft, the motor inertia gets amplified. So probably it's better to speak of a transformed mass moment of inertia. In the case of a lossy gearbox, the torque gets decreased by the efficiency, but the general relation $J_1 = i^2 \cdot J_2$ still holds true.

BTW: For drive machines there is a gear ratio leading to an optimum regarding rotational acceleration: $ i_{opt} = \sqrt{\frac{J_2}{J_1}}$.