Problem involving energy/motion

[Curtiss Oakley]

Homework Statement

A ball of mass m=0.300 kg is connected by a strong massless rod of length L = 0.800 m to a pivot and held in place with the rod vertical. A wind exerts constant force F to the right on the ball as shown below. The ball is released from rest. The wind makes it swing up to attain maximum height H above its starting point before it swings down again. (a)Find H as a function of F. Evaluate H for (b)F=1.00N,and(c)F=10.0N;
There is a photo attached of the problem in totality.

Homework Equations

Change in KE=W
KE=1/2mv^2
W=Fdcos(a)

The Attempt at a Solution

At this point in the semester we have gone over Newton’s laws of motion along with multiple energy theorems (kinetic-work, mehcanical, potential,etc.). My professor has hinted at using Newton’s laws (even though this question is in the energy chapters) but I don’t know where to derive a formula that would include both H and F, while somehow ignoring a (angle of rod’s final position and starting position) or the distance covered by the force of the wind. Thanks in advance!

 

[mfb]

With a constant force from the wind total energy will be conserved if you find a suitable potential. The wind is just like "sidewards gravity" and gravity plus wind are like gravity pointing towards a new direction.

 

[Curtiss Oakley]

With a constant force from the wind total energy will be conserved if you find a suitable potential. The wind is just like "sidewards gravity" and gravity plus wind are like gravity pointing towards a new direction.

So am I treating wind as a conservative energy capable of having a potential energy? If so, how do I go about finding that “suitible potential” equation? I abstractly grasp what you’re saying but don’t want to make a step mathematically if I don’t grasp how it relates to the arithmetic.

 

[jbriggs444]

You have a ball with a mass of 0.30 kg. It is subject to a force from gravity. It is subject to a force from the wind. Both forces are constant across the entire area of interest. The net force on the object is the vector sum of the two forces. Since the individual forces are constant, the net force is also constant.

As far as the math is concerned, a constant force is a constant force. Take the sum of wind plus gravity and pretend that the result is gravity. It has a direction, a magnitude and an associated potential.

If it helps your intuition, you might start by thinking of the wind alone as having a "potential". Suppose that you are standing on a surface subject to a constant wind force of F Newtons leftward. If you walk 1 meter to the right, you will have expended F Joules of work fighting the wind. If you strap on roller skates and slide one meter to the left, the wind will have done F Joules of work on you. This can be thought of as potential energy of F joules per meter.

 

[Curtiss Oakley]

You have a ball with a mass of 0.30 kg. It is subject to a force from gravity. It is subject to a force from the wind. Both forces are constant across the entire area of interest. The net force on the object is the vector sum of the two forces. Since the individual forces are constant, the net force is also constant.

As far as the math is concerned, a constant force is a constant force. Take the sum of wind plus gravity and pretend that the result is gravity. It has a direction, a magnitude and an associated potential.

How do I explain the maximum height being different from the equilibrium point without having any knowledge of momentum? I have a solution for the second part through the cancellation of acting forces, but the swinging above equilibrium then back down doesn’t make sense without momentum (something I haven’t been taught yet)?

 

[jbriggs444]

How do I explain the maximum height being different from the equilibrium point without having any knowledge of momentum? I have a solution for the second part through the cancellation of acting forces, but the swinging above equilibrium then back down doesn’t make sense without momentum (something I haven’t been taught yet)?

Use an energy argument. Kinetic energy plus potential energy is constant. Kinetic energy is zero at both ends of the swing.

 

[Curtiss Oakley]

Use an energy argument. Kinetic energy plus potential energy is constant. Kinetic energy is zero at both ends of the swing.

I understand that, but what is the difference in math between the equilibrium point and the max height (they are two different points in the problem)? Kinetic energy is 0 for both situations, and the potential energy has to compensate for the lack of kinetic energy. So is the potential energy at both points the same?

 

[jbriggs444]

I understand that, but what is the difference in math between the equilibrium point and the max height (they are two different points in the problem)? Kinetic energy is 0 for both situations, and the potential energy has to compensate for the lack of kinetic energy. So is the potential energy at both points the same?

Sure. But only if you consider the combined potential from wind plus gravity. And if you are doing that, what is the direction of "gravity"? What does that mean about how you need to measure "height"?

 

[haruspex]

what is the difference in math between the equilibrium point and the max height (they are two different points in the problem)? Kinetic energy is 0 for both

No. The equilibrium point is where it could rest, so the forces balance there and the acceleration is zero. In the actual sequence, the KE need not be zero there.

 

[kuruman]

If you release a pendulum at angular displacement $\theta$ from the direction of gravity, what is the maximum angular displacement on the other side?
What about when you release a pendulum from rest at angular displacement $\theta$ from the direction of "gravity" mentioned by @jbriggs444 in #8?

 

[Curtiss Oakley]

Sure. But only if you consider the combined potential from wind plus gravity. And if you are doing that, what is the direction of "gravity"? What does that mean about how you need to measure "height"?

The direction of gravity won’t change but the rod impacts the direction of its potential energy, correct? My measurement of height is from the intial point being y=0, and the angle of the rod is in relationship to that height.

 

[Curtiss Oakley]

No. The equilibrium point is where it could rest, so the forces balance there and the acceleration is zero. In the actual sequence, the KE need not be zero there.

I get the cancelling of forced, but why does it not need to be zero? It does equal zero as it lacks velocity and acceleration, and with the object being released from rest it would be zero.

 

[Curtiss Oakley]

If you release a pendulum at angular displacement $\theta$ from the direction of gravity, what is the maximum angular displacement on the other side?
What about when you release a pendulum from rest at angular displacement $\theta$ from the direction of "gravity" mentioned by @jbriggs444 in #8?

If I’m using the direction of “gravity” as wind plus gravity then the theta displacement from direction of gravity will be larger than the initial release point angular displacement.

 

[Curtiss Oakley]

There has to be something wrong with this work (document attached) as it lacks the length of the rod, so if anyone is willing to correct my assumptions/tweak something I’m more than happy to hear it out. Thank you all so much for helping though!

 

[kuruman]

There has to be something wrong with this as it lacks the length of the rod, so if anyone is willing to correct my assumptions/tweak something I’m more than happy to hear it out. Thank you all so much for helping though!

The drawing that you attached in #1 says that the length of the rod is $L$.

 

[Curtiss Oakley]

The drawing that you attached in #1 says that the length of the rod is $L$.

I was referring to my answer on the document attached, let me edit my comment to clarify.

 

[kuruman]

I was referring to my answer on the document attached, let me edit my comment to clarify.

I see. Look at the last drawing, bottom left of your drawing. Label angle $\theta.$ When the pendulum is released, at what angle from the vertical is the equilibrium position? At what angle from the equilibrium position is the maximum displacement? Once you have answered these questions, draw a "length" instead of a force diagram. Draw length $L$ along its initial vertical position and then again along its maximum displacement position. Use trigonometry to find $H$.

 

[haruspex]

I get the cancelling of forced, but why does it not need to be zero? It does equal zero as it lacks velocity and acceleration, and with the object being released from rest it would be zero.

As others have remarked, what you have here is, effectively, a pendulum with gravity acting at an angle. When you release a normal pendulum from some displacement it has zero velocity but an acceleration. At the bottom of the swing it is in the equilibrium position; zero acceleration but some velocity. The swing persists because of the constant back and forth between KE and GPE. If it ever comes to having zero acceleration and zero velocity it stops.

 

[Curtiss Oakley]

Thank you all for the fantastic help! I attached my final answer for any person that may want to get direction and ideas with a similar problem.

 

[kuruman]

Your numbers agree with mine.