Maximum angle made by rotating hinge with energy and gravity

In summary: To clarify, ##E_0## is the energy that the hinged rod has when it is at rest, before the collision. Then, after the collision, you would equate this kinetic energy with the maximum gravitational energy achieved by the rod at its highest point (say, ##\theta_{\mathrm{m}}##).
  • #1
AppleiPad556
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Homework Statement
(see below)
Relevant Equations
##E_{rotational}=\frac{1}{2}I\omega^2##
##\omega=\frac{\Delta\theta}{{\Delta}t}##
##I=\frac{1}{3}ma^2## (for rectangular plate with axis of rotation along a side)
##E_{spring}=\frac{1}{2}kx^2##
Hello!
I had a random question while playing around with a garbage can that I hoped y'all could help me walk through:
Let's say that I have a hinge on a table, rotating with gravity acting perpendicular to it. Energy is provided into the hinge, let's say by a spring, like so:
1640034367536.png

I want to know the maximum angle the hinge will make with the table given a certain amount of energy (ultimately how much energy will cause the hinge to "tip" over 90° and fall over the other direction).

For the energy from the spring, I would set up the spring so that its rest position is in line with the hinge resting position, which means I can start with the maximum kinetic energy:
$$E_{spring}=\frac{1}{2}kx^2$$
Where ##k## is the spring constant and ##x## is the displacement.
However, it's almost certain that the spring will continue to move after impact from the spring, meaning I would have to subtract the energy it has still from this total kinetic energy. Something like:
$$E=\frac{1}{2}kx^2-\frac{1}{2}mv^2$$
We'll let this be ##E##.

Assuming that the hinge has no friction and that all energy is conserved, I begin with the Energy formula for rotational motion:
$$E_{rotational}=\frac{1}{2}I\omega^2$$
Where
  • ##I## is the moment of inertia (in this case, for a rectangular plate with rotation along a side, ##\frac{1}{3}ma^2##, where ##m## is mass and ##a## is length)
  • ##\omega## is the angular velocity.
I rewrite angular velocity in terms of displacement and time, to get:
$$E_{rotational}=\frac{1}{2}(\frac{1}{3}ma^2)(\frac{\Delta\theta}{{\Delta}t})^2$$
Now ##\Delta\theta## and ##{\Delta}t## both have initial values of ##0##, so I can simply look at their final values.
Solving for ##\theta## I get:
$$\theta=\sqrt{\frac{6Et^2}{ma^2}}$$

My question is whether this logic makes sense and whether it seems reasonable to solve for the angle made by the hinge? Am I missing anything that could also play a role in this? How does gravity come into play?
Thanks!
 
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  • #2
If the collision imparts an energy ##E_0 < m_{\mathrm{rod}}ga/2## to the hinged rod then, whilst it's true that ##E_0 = \frac{1}{2} I \omega_0^2##, where ##\omega_0 = \frac{d\theta}{dt} \big{|}_{t=0}## is the initial rotational speed of the rod, the next step should be to equate this kinetic energy with the maximum gravitational energy achieved by the rod at its highest point (say, ##\theta_{\mathrm{m}}##). This means to put ##E_0 \equiv \frac{m_{\mathrm{rod}} g a}{2} \sin{\theta_{\mathrm{m}}}## and then re-arrange for ##\theta_{\mathrm{m}}##.
 
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  • #3
Gravity comes into play through gravitational potential energy of the center of mass of the table. That must be part of the energy conservation equation. I suggest that you solve the back end of the problem first. By this I mean assume that the table starts from the horizontal position with initial rotational energy ##E_0##. Find the angle by which it rotates in terms of ##E_0## and the other given variables.

Once you have that, then worry about how the table gets its initial energy. If it is from a spring, then
1. You need to find the speed of the mass on the spring just before the collision. You have to make some assumptions about the distance the mass covers before it strikes the table.
2. You need to have a model for the collision, e.g. perfectly elastic, perfectly inelastic or in-between.
3. You need to conserve angular momentum about the hinge under the assumption that the collision is instantaneous. This will allow you to find the angular speed of the table after the collision and hence ##E_0##.

As you can see the front part of the problem is much more involved. This is why I suggested doing the back end first which is probably your question. It looks like the spring is just an example you thought about how the table might acquire its initial energy.
 
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  • #4
Thank you @ergospherical and @kuruman for the prompt replies! I'm a little bit more lost though...
kuruman said:
Gravity comes into play through gravitational potential energy of the center of mass of the table.
Perhaps I'm misunderstanding this, but just to clarify, I'm looking at a hinged rectangular plate that is separate from the table: the hinge is simply on the table (or else it will dangle freely!)
kuruman said:
It looks like the spring is just an example you thought about how the table might acquire its initial energy.
You're right on this!
In regards to the remaining considerations that you brought up, could you clarify what you mean by the back and front ends of the problem?
I'd probably assume completely elastic, but out of curiosity how would one solve for a model that is not completely inelastic nor elastic? I only know how to do one or the other.

ergospherical said:
This means to put E0≡mrodga2sin⁡θm and then re-arrange for θm.
Correct me if I'm wrong, but are you saying that I would equate the GPE made by the hinge at its maximum with the kinetic energy that it receives from the spring?
I'm not sure where ##E_0 < m_{\mathrm{rod}}ga/2## and ##E_0 \equiv \frac{m_{\mathrm{rod}} g a}{2} \sin{\theta_{\mathrm{m}}}## came from.

Thanks!
 
  • #5
AppleiPad556 said:
I'm looking at a hinged rectangular plate
Yes, the change in GPE of the plate.
AppleiPad556 said:
the back and front ends of the problem?
I.e. look at the later part of the motion (just the plate swinging up, given some velocity) first. Then go back and consider what velocity the mass needed to supply the plate's angular momentum, and finally the spring extension needed to supply that velocity to the mass.
Don't forget that the mass rising x from the spring is also overcoming gravity.
AppleiPad556 said:
how would one solve for a model that is not completely inelastic nor elastic?
You choose a coefficient of restitution.
I've only ever seen that done in respect of linear motion (see Newton’s Experimental Law). Might take a bit of thought on how that translates to this case. Maybe just assume the system KE final is coefficient squared times system KE initial.
AppleiPad556 said:
equate the GPE made by the hinge at its maximum with the kinetic energy that it receives from the spring?
Equate the plate's maximum gain in GPE to the KE it was given by the collision.
 
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  • #6
Thank you @haruspex!
haruspex said:
I.e. look at the later part of the motion (just the plate swinging up, given some velocity) first. Then go back and consider what velocity the mass needed to supply the plate's angular momentum, and finally the spring extension needed to supply that velocity to the mass.
Don't forget that the mass rising x from the spring is also overcoming gravity.
Thanks for this clarification! I think I'll look at only the latter part of the motion first, with a certain energy and the angular displacement that results.
haruspex said:
Equate the plate's maximum gain in GPE to the KE it was given by the collision.
OK, this makes sense. Equate GPE and KE and solve for ##\theta##.
I have the kinetic energy given by the spring, and I have the equation for rotational energy. Is the gravitational potential energy what ergospherical said earlier? I know for linear motion that GPE is ##mgh##, but I'm not sure how it'll translate (pun not intended) for rotational motion.
And to confirm, how do I use the rotational energy formula in this, if at all?

Thanks again!
 
  • #7
AppleiPad556 said:
for linear motion that GPE is mgh,
It's not to do with the motion that brought the object to that height, it is merely being at that height. Specifically, the height reached by the plate's mass centre (assuming it is uniform).
AppleiPad556 said:
how do I use the rotational energy formula in this
Equating the energies will let you deduce the initial angular velocity.
 
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  • #8
haruspex said:
It's not to do with the motion that brought the object to that height, it is merely being at that height. Specifically, the height reached by the plate's mass centre (assuming it is uniform).
Sorry, perhaps I should not have worded it like that. My question was more how I would calculate the maximum height of a rotating object to solve for GPE: use a triangle and the angle?
haruspex said:
Equating the energies will let you deduce the initial angular velocity.
So equating rotational energy and the kinetic energy provided on the hinge to solve for angular velocity, and thus angular displacement? If so, I think this is what I had initially. Or equate GPE of the hinge and KE provided by the spring? If it's this, then which formula has angular velocity?

Forgive me if these questions are stupid...I initially thought given conservation of energy, the initial energy provided to the hinge would become rotational energy of the hinge and would be sufficient to calculate the angular displacement...
 
  • #9
AppleiPad556 said:
how I would calculate the maximum height of a rotating object
The initial rotational KE all turns into gain in GPE. Finding the height the mass centre reaches is easy. If you want the angle it reaches you'll need to do a bit of geometry.
AppleiPad556 said:
So equating rotational energy and the kinetic energy provided on the hinge to solve for angular velocity, and thus angular displacement?
If you know the rotational energy given then, as I wrote above, finding the height reached is easy. No need to find the angular velocity. The question is, how to find that initial energy? That might be via finding the initial angular velocity.
AppleiPad556 said:
Or equate GPE of the hinge and KE provided by the spring?
No, because you might want to allow for a collision that's not perfectly elastic, and even if it is elastic some KE might be retained by the mass. You will need to use conservation of angular momentum.
 
  • #10
Thanks @haruspex for the clarifications!
haruspex said:
finding the height reached is easy
If I were to solve for the angle made must I solve for height? Assume I use energy (see below), would the angular displacement originating from angular velocity be sufficient?
haruspex said:
and even if it is elastic some KE might be retained by the mass.
In this case, assuming a perfectly elastic collision, would me considering the velocity the mass after collision, subtracting it from the total kinetic energy (##\frac{1}{2}kx^2-\frac{1}{2}mv^2##) the mass have to result in the final KE transferred be sufficient?
Momentum is a concept that I haven't grasped very strongly yet; I suppose I would need to find the velocity of the mass (perhaps using tracking software) before the collision as mentioned above, and then equate that to the angular momentum (with angular velocity in terms of displacement), and solve for displacement then?

Thanks again!
 
  • #11
AppleiPad556 said:
would the angular displacement originating from angular velocity be sufficient?
You would also need the angular acceleration, and that changes as the plate rises. The only sensible way is to find the height from the KE. You can then find the angle from the height if you wish.
AppleiPad556 said:
In this case, assuming a perfectly elastic collision, would me considering the velocity the mass after collision, subtracting it from the total kinetic energy (12kx2−12mv2) the mass have to result in the final KE transferred be sufficient?
I cannot decipher that.
Assuming it is elastic you need to use both conservation of energy and conservation of angular momentum to find the two velocities after the collision. Two interlinked unknowns generally implies you need two equations, even if you only care about one unknown.
The best is to take angular momentum about the hinge point. That way you don't have to consider the reaction force at the hinge.
If the mass m arrives at the collision moving upwards at speed u, and the plate is length L, what is the angular momentum of the mass about the hinge point?
 
  • #12
Thank you all for the replies.
I'm going to try to summarize what I'm understanding from this thread in hopes that I'll better string everything together:

I'm hoping to determine the angular displacement of a hinge acting against gravity given a certain amount of energy upon a collision (with, say, a spring). I'd like to take measurements of the angle itself and compare it with theoretical values (hence why I'm asking about my theory here)
My initial thought was to use ##1/2I\omega^2##, but what I'm understanding is that it will not work because that is the rotational kinetic energy. Instead, if I were to take an energy approach, I should consider equating the GPE of the hinge with the kinetic energy received upon collision.
  • Firstly, where did @ergospherical's ##E_0 \equiv \frac{m_{\mathrm{rod}} g a}{2} \sin{\theta_{\mathrm{m}}}## come from? Is this the equation for angular GPE, if that exists?
  • It was also suggested to determine GPE through height (which can then solve for the angle. I should be able to do that using tracking software (although it does add an additional variable to the mix, alas). Which point of the hinge should I be measuring for GPE? Is it from the centre of mass or from the end of the hinge (where the collision occurs)?
However, the above also assumes a perfectly elastic collision, so @haruspex suggested to use conservation of angular momentum to accommodate. I'm not very familiar with the concept, but from my understanding if I am able to find the momentum given before the collision (by the spring) and equate that to the angular momentum of the hinge to solve for ##\theta##? So something like the below?
$$mv=I\frac{\Delta\theta}{{\Delta}t}$$
How does gravity play into conservation of momentum, if at all?
I think regardless of whether I use momentum or energy I will assume an elastic collision (closely resembles preliminary tests); I'm still at high school level physics and so the coefficient of restitution is foreign to me, but I will explore it to see if I can tie it in.

haruspex said:
The best is to take angular momentum about the hinge point. That way you don't have to consider the reaction force at the hinge.
Could you clarify this? Is the hinge point the point of collision or the axis of rotation (or neither)?

haruspex said:
If the mass m arrives at the collision moving upwards at speed u, and the plate is length L, what is the angular momentum of the mass about the hinge point?
I know the momentum of the mass is ##mu##, but what is the angular momentum of the mass if it is moving linearly?

Thanks again!
 
  • #13
AppleiPad556 said:
Is the hinge point the point of collision or the axis of rotation
Your illustration shows a plate or rod connected to a table by a hinge. You seem to call the whole plate a "hinge", but that's a misuse of the term. It's just the joint.
AppleiPad556 said:
what is the angular momentum of the mass if it is moving linearly?
It's similar to torque. Just as the torque of a force about a point is the force multiplied by the perpendicular distance to the point, the angular momentum of a body about a point is its linear momentum multiplied by the perpendicular distance to the point.
 
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Related to Maximum angle made by rotating hinge with energy and gravity

1. What is the maximum angle that can be achieved by rotating a hinge with energy and gravity?

The maximum angle that can be achieved by rotating a hinge with energy and gravity depends on several factors, including the strength of the hinge, the amount of energy applied, and the force of gravity. In general, the maximum angle will be determined by the point at which the hinge can no longer support the weight of the object being rotated.

2. How does energy affect the maximum angle of rotation for a hinge?

Energy plays a crucial role in determining the maximum angle of rotation for a hinge. The more energy that is applied to the hinge, the greater the potential for rotation. However, if the energy exceeds the strength of the hinge, it may break or fail to support the weight of the object being rotated.

3. Can the maximum angle of rotation for a hinge be increased?

Yes, the maximum angle of rotation for a hinge can be increased by increasing the strength of the hinge or by applying more energy to it. However, there may be practical limits to how much the angle can be increased based on the design and materials of the hinge.

4. How does gravity affect the maximum angle of rotation for a hinge?

Gravity is a crucial factor in determining the maximum angle of rotation for a hinge. The force of gravity pulls down on the object being rotated, and the hinge must be strong enough to support this weight. If the force of gravity is too great, it may prevent the object from rotating to its full potential.

5. Are there any safety concerns when working with maximum angles of rotation for hinges?

Yes, there are potential safety concerns when working with maximum angles of rotation for hinges. If the hinge is not strong enough to support the weight of the object being rotated, it may break and cause injury. It is important to carefully consider the design and materials of the hinge and to use caution when applying energy to it.

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