# How fast will this torsion spring open a lid?

• Prockey
In summary, the torsion spring selected for this application can open the lid in seconds, and the hinge friction will subtract from the net torque for opening.
Prockey
TL;DR Summary
Selection of a torsion spring to open a hinged lid
Hi,

I'm designing a small container with a spring loaded hollow rectangular lid. I want the lid to open when a button is pressed, so I have a torsion spring at the hinge. I want to know if the spring I selected is able to open the lid and also how long it would take to open 90°.

Below is a diagram of the hinge, spring and the center of gravity (COG) of the lid. The mass of the lid is 0.005 kg.

This is what I have so far:
Torque needed to open lid:
Torque = weight * distance to COG * Cos(33)
= 0.005 kg * 9.8 m/s^2 * 0.0226 m * 0.8386
= 0.0009 Nm

I'm going to design a custom spring, but for the sake of this question let's assume this torsion spring from McMaster which has a max torque of 0.117 in-lbs (0.0132 Nm).

Question #1: Since this spring is able to provide 0.0132 Nm and I only need 0.0009 Nm, the spring should work. Is this correct?

I also want to know how fast the lid will open in seconds given this spring.
I'm getting the center of gravity MoI from CAD, which I think I need:

I know that: Torque = Moment of inertia * angular acceleration
So: Angular acceleration = Torque/MoI

Using the spring torque and the Z MoI:
Angular acceleration = 0.0132 Nm/9.371e-007kg*m2

Question #2: Is this correct? This 14087 rad/s^2 answer seems way too fast to me.
Question #3: How do I convert this angular acceleration to the number of seconds it will take the lid to open 90°? I would like it to open in less than 1 second.

Last edited:
Welcome, Prokey.

What is holding the lid in place?
Is there friction or pressure difference between the lid and the opening it covers and between the lid and any surface during the trajectory?
Is the lid rotating with the arm?
How do you plan to stop or reduce the oscillation caused by the spring after the arm-lid assemblyreaches maximum acceleration?

Thanks for the reply, Lnewqban !

I added more details to my diagram. There is a simple button/clip mechanism opposite the hinge holding the lid down. There is no pressure difference or friction, the lid is not sealing in any way. The lid is part of a small handheld container and there is a "hard-stop" on the lower part of the housing to stop the lid from over rotating.

The hinge and spring are a small version of this:

The correct moment of inertia is the MOI about the center of gravity (COG) plus M*R^2, where M is the mass and R is the distance from the hinge to the COG. Search parallel axis theorem to learn more.

Design constraints include:
1) Open in less than one second
2) Do not slam it into the end stop too hard

I suggest starting with a spring at zero torque when the COM is directly above the hinge. Then the spring will be twisted 57 degrees when the lid is closed. If I correctly interpret the McMaster-Carr specification, the spring torque at 57 degrees will be 57 / 90 * 0.117 in-lbs * 4.45 N/lb * 0.0254 m/in = 0.0087 N-m.

The torque for initial acceleration is the spring torque minus the torque to raise the lid against gravity, or 0.0087 - 0.0009 = 0.0078 N-m. Hinge friction will also subtract from the net torque for acceleration, but hinge friction is typically (not always) small enough to ignore.

The initial acceleration will decrease as the lid raises. The exact calculation in closed form would be a challenge because the horizontal distance from the hinge to the COM is a trig function. A numerical solution is a typical real world way to get a near exact lid opening time. Or you can do a rough calculation that just might meet your needs.

Rough calculation: The MOI about the hinge is about 3.5E-6 kg-m^2. The initial acceleration is approximately 0.0078 N-m / 3.5E-6 kg-m^2 = 2200 rad/sec^2. Just as D = 0.5 * a * t^2 for linear acceleration, theta = 0.5 * alpha * t^2 for angular acceleration. Theta is in radians, alpha is in rad/sec^2, and 57 degrees is 1.0 radians.

Solving for t, I get about 30 msec to open 57 degrees. Since the acceleration decreases to zero at that point, multiply by two to get 60 msec. Remember, this is a rough calculation. The remaining 33 degrees to the end stop is at constant velocity.

The lid will open with a bang and slam against the end stop. It will most likely break if opened too many times. I used approximate numbers to show how to do the calculation. Now it's your turn to use real numbers. Enjoy.

berkeman and Prockey
Thank you, this was the guidance I needed!

berkeman and jrmichler

## 1. How do you calculate the speed of a torsion spring?

The speed of a torsion spring can be calculated using the formula v = (2π√(k/m)) * A, where v is the speed, k is the spring constant, m is the mass of the object being moved, and A is the amplitude of the spring's rotation.

## 2. How does the size of the torsion spring affect its speed?

The size of the torsion spring directly affects its speed. A larger spring will have a larger amplitude and therefore a faster speed, while a smaller spring will have a smaller amplitude and a slower speed.

## 3. What factors can affect the speed of a torsion spring?

The speed of a torsion spring can be affected by various factors such as the material of the spring, the angle of rotation, the weight and shape of the object being moved, and the amount of tension applied to the spring.

## 4. Can the speed of a torsion spring be adjusted?

Yes, the speed of a torsion spring can be adjusted by changing the tension applied to the spring or by altering its size or material. Adjusting these factors can change the amplitude and therefore the speed of the spring.

## 5. How can I determine the appropriate torsion spring for my lid opening needs?

The appropriate torsion spring for your lid opening needs will depend on the weight and size of the lid, as well as the desired speed of opening. Consulting with a spring manufacturer or using a spring calculator can help determine the best spring for your specific needs.

• Mechanical Engineering
Replies
2
Views
1K
• Mechanical Engineering
Replies
4
Views
757
• Mechanical Engineering
Replies
5
Views
7K
• Mechanical Engineering
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
1K
• Mechanical Engineering
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Mechanics
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
4K