Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
94 replies · 6K views
- 13,438
- 8,110
A swing normally hangs straight down. The chains are vertical. "Vertical" is in the direction of gravity. "Horizontal" is perpendicular to this.Callumnc1 said:
As you swing back and forth the angle (measured from the vertical center line) of the chains reaches larger and larger peaks depending on how hard you pump.
If you pump hard enough, the peaks can reach 90 degrees. That is, the chains at the top of the arc are horizontal.
For a child swinger, problems arise when you pass the horizontal. When you reach the top of the arc and swing back down, the chains will have gone slack. You will fall downward on slack chains until they come taut again in an inelastic collision. The collision is jarring and, hence, dangerous. This collision also absorbs energy. Once you reach this situation, you can no longer pump gradually, adding a little energy with each oscillation. Instead, you need to inject all of the excess energy required to surpass the horizontal anew on every half cycle. It becomes hard work. [Plus, you are spending the next quarter cycle trying to get back to a stable orientation when you should have been prepping for the pump action at the bottom -- it keeps you busy dancing on the edge of control]
Last edited:
member 731016
Thank you for your help @jbriggs444 ! I think I understand about what you were meaning now :)jbriggs444 said:
- 42,946
- 10,550
During the part of the process I described, the ant running to one corner, I took angular momentum as conserved. But the result is that the mass centre of the ant+bob is displaced from being vertically below the pivot. So now gravity exerts a torque about the pivot.Callumnc1 said:
Jake357
- 9
- 2
The work done is proportional to the difference in the height from the ground, which means thatCallumnc1 said:
W=mg(h2-h1), which is also the difference in the potential energy of Spiderman.
The difference in height is: l-l cos 60=12-12 cos 60=6 m.
[Solution redacted by the Mentors]
Last edited by a moderator:
member 731016
Third method to solve the work done by spider man:
I think we can solve this using the work integral with respect to theta and have the limits of integration as the initial angle spider man makes with the vertical and the finial angle spider man moves with the vertical.
I think we have to integrate with respect to theta, because for each differential displacement of spider man along his path the differential angle between spider man's weight and his differential displacement vector is not the same.
##W = \int_{\theta_i}^{\theta_f} mgr\cos\theta dr##
##W = \int_{\theta_i}^{\theta_f} mgr^2\cos\theta~d\theta##
In progress...
Many thanks!
I think we can solve this using the work integral with respect to theta and have the limits of integration as the initial angle spider man makes with the vertical and the finial angle spider man moves with the vertical.
I think we have to integrate with respect to theta, because for each differential displacement of spider man along his path the differential angle between spider man's weight and his differential displacement vector is not the same.
##W = \int_{\theta_i}^{\theta_f} mgr\cos\theta dr##
##W = \int_{\theta_i}^{\theta_f} mgr^2\cos\theta~d\theta##
In progress...
Many thanks!
- 42,946
- 10,550
Please define r and θ in that equation.Callumnc1 said:
- 42,946
- 10,550
So r is a constant, but you integrate wrt it?Callumnc1 said:
- 42,946
- 10,550
Right.Callumnc1 said:
Don't just throw integrals together that seem to have the right ingredients. Think what the integral is saying.
In the present case, you are considering spiderman moving as the rope swings, so the independent variable is the angle. As it swings through a small angle ##d\theta##, Spiderman moves a distance (what?) at an angle (what?) to the vertical force (what?) thereby doing work (what?).
member 731016
Thank you for your reply @haruspex ! That is good advice! I have tried to include my reasoning below.haruspex said:
Since we know that the when the spider man moves an infinitesimal displacement ##\vec {ds} = dx\hat i + dy\hat y## when a ##F_g## is applied by the earth
## dW = F_g \cdot \vec {ds} ##
We know the path spider man travels though is a circular arc of length ##S = r\theta## (The arc length is dependent of the theta)
Given that the arc length of the path is ##S = r\theta##, we take the derivative of arc length with respect to theta giving ## ds = r\theta d\theta##
## dW = Fr\cos\theta d\theta ## Where ##d\theta## is the angle between the weight and displacement vector ##\vec ds## tangent to the circular path at each point
Ok so know we need to get ##\cos\theta## in terms of the angle the rope makes with the vertical ##\phi## (I just defined that variable)
However, do you please know how do you do that?
I think maybe from geometry (if I proved it correctly) ##\theta = \phi##
Many thanks!
- 42,946
- 10,550
When you wrote S=rθ, weren’t you taking θ to be that variable?Callumnc1 said:
- 42,946
- 10,550
Callumnc1 said:
Then this is wrong:
Please update those equations using your new definitions.Callumnc1 said:
Edit: just noticed your step from ##S = r\theta## to ## ds = r\theta d\theta## is also wrong.
member 731016
Here are the new equations @haruspex ,
We start with ## dW = Fr\cos\theta d\theta ##
And we must integrate from ##\phi_1## to ##\phi_2## so ## dW = \int_{\phi_1}^{\phi_2} Fr\cos\theta d\theta ##
Therefore, since ##\theta## depends on ##\phi## we must get ##\theta## in terms of ##\phi##. I did this geometrically and proved that ##\theta = \phi##.
Am I correct?
Many thanks!
We start with ## dW = Fr\cos\theta d\theta ##
And we must integrate from ##\phi_1## to ##\phi_2## so ## dW = \int_{\phi_1}^{\phi_2} Fr\cos\theta d\theta ##
Therefore, since ##\theta## depends on ##\phi## we must get ##\theta## in terms of ##\phi##. I did this geometrically and proved that ##\theta = \phi##.
Am I correct?
Many thanks!
- 42,946
- 10,550
If ϕ is the angle between the vertical and the rope then S=rφ.Callumnc1 said:
When ϕ=0, what is the angle between ##\vec ds## and the gravitational force?
member 731016
Thank you for your reply @haruspex !haruspex said:
True I did not realize that s =rφ.
I should then take the of the new arc length equation with respect to phi to get ##\frac{ds}{d\phi} = r##
When ##\phi## is zero, then ##\theta## the angle between ##\vec {ds}## and ##F_g## is 90 degrees.
Are you trying to get me to find the relationship between ##\phi## and ##\theta## by considering a few physical situations?
If I consider the second case where ##\phi## is 90 degrees, then the angle between ##\vec {ds}## and ##F_g## is 180 degrees, correct?
Is there a way to find the relationship between ##\phi## and ##\theta## without considering intuitive cases along the circular path?
Many thanks!
- 42,946
- 10,550
Of course: draw the right diagram. The special cases are just an easy way to check your answer.Callumnc1 said:
If you thought you had proved a different relationship, try to see where that proof went wrong.
member 731016
Thank you for your reply @haruspex !haruspex said:
Based on my special cases in post #81,
1. ##\phi = 0## then ##\theta = 90##
2. ##\phi = 90## then ##\theta = 180##
It looks to me like the relationship is ##\theta = \phi+ 90## so theta is always a phase shift of 90 degrees ahead of phi.
I will draw a diagram again to prove this and post it soon!
Many thanks!
- 42,946
- 10,550
You have drawn ##\theta## as the angle the final displacement makes to the vertical. You defined it as the angle the displacement element ##\vec ds## makes to the vertical at an intermediate point.Callumnc1 said:
Draw a diagram with the rope at two angles to the vertical, ##\phi## and ##\phi+d\phi##.
The displacement element is length ##r.d\phi##. Note that it is at right angles to the rope.
member 731016
- 13,438
- 8,110
##ds## is (or behaves like) an infinitesimal. The incremental displacement is infinitesimally close to being at right angles to both the initial and final angles. Its angle differs from the perpendicular by only ##d \phi## at most.Callumnc1 said:
But that is window dressing. Surely you are after the incremental work done. This should be the vector dot product of the force of gravity, ##\vec{mg}## and the incremental displacement, ##\vec{ds}##.
member 731016
Thank you for your reply @jbriggs444 !jbriggs444 said:
Oh ok that makes sense, now. I have added that too my new diagram (which is very not too scale):
Yes I think I am after the incremental work (## dW = \int_{\phi_1}^{\phi_2} Fr\cos\theta~d\phi ##) in terms of phi which I will integrate from ##\phi_1 = 0## to ##\phi_2 = 60## (Where ##\theta## is the angle between the differential displacement and spider man's weight).
Would you please know how to get theta in terms of phi from the diagram? I think (by considering two cases from post #83) that ##\theta = \phi + 90##
Many thanks!
Last edited by a moderator:
member 731016
I think I may have found the answer!!!
First method:
If we start with the big triangle,
Then,
##\phi + d\phi + 90 + \theta_2 = 180##
##\theta_2 = 90 - \phi - d\phi##
Now for the small triangle,
##90 + \theta_2 + \theta_1 = 180##
## \theta_1 = 90 - \theta_2##
## \theta_1 = 90 - (90 - \phi - d\phi)##
## \theta_1 = \phi + d\phi##
Therefore since ##\theta = 90 + \theta_1## (since ##theta_1## is vertically opposite)
Then ##\theta = 90 + \phi + d\phi ≈ 90 + \phi## since ##d\phi## is a differential
Second method:
I believe you could have also found ##\theta_2## from the medium size triangle,
Where in this case
##\theta_2 = 90 - \phi##
## \theta_1 = \phi + d\phi##
##\theta = 90 + \phi## without having to make the approximation in the first method
Is my proof for ##\theta = 90 + \phi## geometrically correct?
Many thanks!
First method:
If we start with the big triangle,
Then,
##\phi + d\phi + 90 + \theta_2 = 180##
##\theta_2 = 90 - \phi - d\phi##
Now for the small triangle,
##90 + \theta_2 + \theta_1 = 180##
## \theta_1 = 90 - \theta_2##
## \theta_1 = 90 - (90 - \phi - d\phi)##
## \theta_1 = \phi + d\phi##
Therefore since ##\theta = 90 + \theta_1## (since ##theta_1## is vertically opposite)
Then ##\theta = 90 + \phi + d\phi ≈ 90 + \phi## since ##d\phi## is a differential
Second method:
I believe you could have also found ##\theta_2## from the medium size triangle,
Where in this case
##\theta_2 = 90 - \phi##
## \theta_1 = \phi + d\phi##
##\theta = 90 + \phi## without having to make the approximation in the first method
Is my proof for ##\theta = 90 + \phi## geometrically correct?
Many thanks!
