Equivalent inertia calculation problem

[p0wertripper]

Homework Statement

Need to calculate equivalent mass moment of inertia of a part that moves linearly at the point where torque is applied. Please find two schematic drawings in the attachement (representing the same mechanism in a slightly different way). Basically, there is a rotating crank to which the torque is applied and it transfers it to the linear motion of the linear actuator. I need to find the equivalent inertia of the linear actuator. Obviously, the equivalent inertia will depend on the angle, so it should be given as a relationship depending on the angle. The mass of the push rod of the actuator is M, the radius of the crank wheel is R.

Homework Equations

T=Jα
F=Ma

The Attempt at a Solution

May it be possible to use energy method? We "replace" the linearly moving actutator push rod with some inertia Jeq that is free to rotate around the point where the torque is applied (see drawing). For small displacements and constant rotational speed ω of the equivalent inertia we can assume constant speed of linear motion of the push rod (i.e. we are trying to say that for small displacements around a certain angular position θ, the kinetic energies of the equivalent rotating inertia Jeq and linearly moving actuator mass M will be equivalent).

In this case Jeq*ω^2=M*v^2 => Jeq=(M*v^2)/ω^2. Then we express ω in terms v, R and θ: ω=[itex]\frac{v*sinθ}{R}[/itex] and substitute in the formulae for Jeq. This results in J=[itex]\frac{m*R^2}{(sinθ)^2}[/itex]

This results in the required solution. The solution is definitely correct for when θ=90°, as it is just m*R^2 in this case. However, I am in doubt for whether it is correct for other angles.

 

[KataKoniK]

Thank you for posting your question. I would like to offer some suggestions for your approach to calculating the equivalent mass moment of inertia of the linear actuator.

Firstly, using the energy method as you have suggested is a valid approach. However, it is important to consider the assumptions and limitations of this method. As you have mentioned, it assumes small displacements and constant rotational speed, which may not hold true in all cases. It would be beneficial to also consider other methods, such as using the equations T=Jα and F=Ma, which you have mentioned in your post.

Secondly, when substituting for ω in terms of v, R, and θ, it is important to note that this equation only holds true for the specific case where the linear motion is perpendicular to the radius of the crank wheel. In other cases, the relationship between ω and v may be different, and therefore the resulting formula for Jeq may also be different.

Lastly, it is always a good practice to check your solution for different values of θ to ensure its validity. You can also compare your solution to other known solutions or perform experiments to verify your calculations.

I hope these suggestions are helpful in your calculations. Good luck with your research!