# How to size a motor for a balanced lever?

I'm new to mechanical engineering.

I have a 5m long lever with a fulcrum positioned 1m from one end. The whole setup is balanced and I need to calculate the torque of a motor that would move the lever around.

My biggest concern is the inertia of the whole rig, as I need it to do precision movements.

rough sketch: Assuming that:

• the counterweight and load are point masses,
• the load mass is $$m_L$$
• the bar has mass $$m_b$$

Then the counterbalance mass should be equal to $$m_{cw} = \frac{3 m_b + 8 m_L}{2}$$

The moment of inertia for this problem is:

$$I_{total} = I_{counterweight} + I_{bar} + I_{load}$$

where:

• $$I_{cw}$$ is the counterweight mass moment of inertia $$I_{cw} = m_{cw} L_{cw}^2 = \frac{3 m_b + 8 m_L}{2} [kg.m^2]$$
• $$I_{load}$$ is the load mass moment of inertia $$I_{L} = m_{L} L_{L}^2 = 16\cdot m_L [kg.m^2]$$
• $$I_{bar}$$ is the bar mass moment of inertia

$$I_{L} = \frac{m_{b} L_{b}^2}{12} + m_{b}\cdot (\frac{3}{2})^2 = \frac{13}{3} \cdot m_b [kg.m^2]$$

Therefore (if I done all the algebra correct):

$$I_{tot} = \frac{35}{6} m_b + 20\cdot m_L [kg.m^2]$$

• Thanks, mate. This helps a lot. Sep 17 at 14:50
1. Calculate the moment of inertia of your lever. If you can't do the maths then use a 3D CAD package to draw it and use the mass properties tool (or whatever) to show the moment of inertia (and a load of other useful stuff). You'll need to specify the material density. OnShape is free if you don't mind your designs being public. See their mass properties help page.

I need to calculate the torque of a motor that would move the lever around.

1. You need to calculate the torque of a motor that will accelerate and decelerate your load. This will be higher than that required to keep it moving.

If we ignore the weight of your lever you need 1/4 force to lift anything hooked at the short end. like to lift 100kg you need 25kg force.

If you want to calculate further you need to have the data on the weight of the lever, loading, speed, etc.

The relationship is simple:

$$\tau = I\alpha = mr^2 \alpha$$

$$\tau$$ = torque

$$I$$ = moment of inertia

$$\alpha$$ = angular acceleration

$$m$$ = mass

$$r$$ = distance from the mass to the pivot point

Does this help? Obviously not.

For your situation, you know r (4m & 1m), the missing term is the angular acceleration ($$\alpha$$), which is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows:

$$\alpha = \dfrac{\Delta \omega}{\Delta t}$$ where Δω is the change in angular velocity and Δt is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s2. So, it really means the torque required is depending upon how fast you want the mass to rotate.

For a balanced system as shown, the torque required is:

$$\tau = (\sum I_i) \alpha$$