B Slack vs tight rope at the start of a pendulum

[kgehrels]

TL;DR Summary

Is there a difference in the force applied to a pivot point if the rope is slack or under tension to start?

Pretty much what the title says. Everything that I can find online addresses the pendulum problem with the assumption that the rope is under tension to start. What if the rope is slack to start? Will there be an increase in force applied to the pivot point, say if a mass is dropped off a ledge vs being pulled back to apply starting tension to the rope, or is there no, or negligible, difference between the two scenarios?

 

[kuruman]

Do the experiment. Tie a small mass at the end of a piece of string. Hold one end of the string above the mass with one hand and mass with the other. Start with a vertical distance between hands less than the length of the string. Lower the mass slowly. What do you observe? What does each hand feel? What do you feel if you drop the mass instead of lowering it gently?

 

[kgehrels]

Wish I'd thought of that. It's such a simple experiment and so easy to execute and replicate.
Here's what I noticed:

• A slack rope cause a slight increase in force over a tight rope.
• However, it was nowhere near as much force as dropping the mass directly beneath the pivot point.

What is going on then?
My theory is that the acceleration due to gravity on the mass between the start of the drop and the point where the rope is under tension applies angular force to the pivot point. However, some of this force is transferred into increased acceleration in the swing while the rest is absorbed by the pivot point. I assume that this increase in swing acceleration can be added to the acceleration of a mass in the formulas that are used to calculate the forces in a pendulum. So:

• As the drop point moves closer to being under the pivot point more force is applied to the pivot point and less is transformed into acceleration along the arc of the swing.
• As the height of the drop before the rope becomes under tension increase so do the forces that are applied to the pivot point and to the acceleration of the swing...gravity and all that

Am I on the right track with this?
How would one go about calculating how much energy turns into acceleration along the swing and how much turns into force on the pivot point?

 

[PeroK]

Wish I'd thought of that. It's such a simple experiment and so easy to execute and replicate.
Here's what I noticed:

• A slack rope cause a slight increase in force over a tight rope.
• However, it was nowhere near as much force as dropping the mass directly beneath the pivot point.

What is going on then?
My theory is that the acceleration due to gravity on the mass between the start of the drop and the point where the rope is under tension applies angular force to the pivot point. However, some of this force is transferred into increased acceleration in the swing while the rest is absorbed by the pivot point. I assume that this increase in swing acceleration can be added to the acceleration of a mass in the formulas that are used to calculate the forces in a pendulum. So:

• As the drop point moves closer to being under the pivot point more force is applied to the pivot point and less is transformed into acceleration along the arc of the swing.
• As the height of the drop before the rope becomes under tension increase so do the forces that are applied to the pivot point and to the acceleration of the swing...gravity and all that

Am I on the right track with this?
How would one go about calculating how much energy turns into acceleration along the swing and how much turns into force on the pivot point?

When the pendulum is swinging, the pivot is only supporting the weight of the mass. When the mass falls, there is a sudden deceleration when the rope goes tight. If the string were inextensible, then the force would be infinite. In reality, the string must extend at least a little, but it will be a significant force. This is why climbing ropes are elastic. If you took a fall on a static rope, that is almost as bad as hitting the ground.

 

[kgehrels]

Agreed. This is much more clear in a vertical drop done below the pivot point.
However, the deceleration does not go to 0 if the drop is done to the side of the pivot point as some of the force is changed into sideways acceleration. How much of the downwards acceleration due to gravity becomes deceleration as the rope becomes tight and how much turns into sideways acceleration along the arc? (For the ease of math let's assume that the rope doesn't stretch).

 

[PeroK]

Acceleration is a vector. Force is rate of change of momentum. Those are both vectors as well. Any sudden (instantaneous) change in direction entails an infinite force.

You can resolve the problem in general by considering the motion in terms of components normal to and tangential to the line of the string when it goes tight. The sudden deceleration is along the line of the string. Then the force depends on the elasticity of the string.

 

[kuruman]

How much of the downwards acceleration due to gravity becomes deceleration as the rope becomes tight and how much turns into sideways acceleration along the arc? (For the ease of math let's assume that the rope doesn't stretch).

It depends on the angle $\theta$ that the string makes with respect to the vertical at the onset of stretching. If the instantaneous value of the tension is $T$ along the string, and before the mass $m$ starts swinging, its acceleration will have components

$a_t=\left(\dfrac{T}{m}+g\right)\sin\theta~~$ in the tangential (sideways) direction.

$a_c=\left(\dfrac{T}{m}-g\right)\sin\theta~~$ in the centripetal (towards the support) direction.

 

[berkeman]

Thread is closed temporarily for Moderation (reopening of a locked thread).

 

[berkeman]

This thread will remain closed. Turns out this is a re-start of a previously locked thread about a dangerous activity. The OP is hiring a consulting engineer to help with these calculations, since human safety is involved in this activity. Thanks to all who were trying to help.