Perhaps the easiest solution is to use energy to determine the v^2 in zut’s equation. At the top of the arc, velocity is 0 and all energy is potential (mgh). If the mass swings until level with the chain, then potential energy is mgl and the total force would be 3mg at the bottom of the swing. If you swing higher than the chain, all bets are off because you would start getting impact forces as the rope jerked when you fell straight down. Hope this is clear, I know it’s not the best worded answer…
OK, I’m still missing it. The velocity vector is always tangential to the arc of motion. How could a force vector derived from a tangential vector ever affect a normal force?
Seriously, I draw a diagram and see that at any point there is a force due to gravity of mg acting horizontally towards the ground. This force has, as its components, a force extending normal to the tangent (what is exherted on the swing) and a force tangential to the path, which is causing the velocity in the first place.
At the bottom of the swing, the normal force is the weight of the pendulum, and the other force is gone completely (zero acceleration). Why is any other factor even considered? there is no other force acting on the mass other than gravity, and the velocity obtained from swinging is lost by being in the wrong direction to add force to the equation.
Someone please slap me across the head on this!! zut, calculate the damn thing assuming we start from the horizontal. Draw a diagram and explain to me how the hell I am missing something (i have my own diagram to reference, you see), because no matter how long I look at this picture all I see is gravity pulling on the rope… adding motion changes the force vecotr associated with movement, but the larger the velocity gets the more of an angle it is applied until it is completely orthogonal and not a component of pressure on the swing. Are you telling me that horizontal motion results in an orthogonal force???
Eyer, I think some of that energy is used in motion, no? It wouldn’t all go into force pushing down on the swing, which is what made me state my original statement.
Well, let’s make some assumptions here.
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The pendulum behaves itself as a simple pendulum over all angles of interest, not just the small ones.
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We choose the zero of time so that the swing is horizontal; theta(0) = Pi/2
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We further choose the setup so that the maximum height the swing reaches is the horizontal, so omega(0) = 0, where omega is the time derivative of theta.
Given these simplifications, the general equation for the angle theta is
theta(t) = Pi/2 * cos(w*t), where w[sup]2[/sup] = g/L, L being the length of the rope and g being the gravitational acceleration.
The centripetal force is mLomega[sup]2[/sup], and a little algebra gives F[sub]c[/sub] = mg(Pi[sup]2[/sup]/4 - theta[sup]2[/sup]), directed radially inward.
The tangential force is mLalpha, alpha being the second time derivative of theta; this is transparently F[sub]T[/sub] = -mgtheta, directed tangentially (the minus sign accounts for being opposite the direction of motion).
The y component of F[sub]T[/sub] is -mgthetasin(theta) and the x component is -mgthetacos(theta). For F[sub]c[/sub], the y component is mg(Pi[sup]2[/sup]/4 - theta[sup]2[/sup])cos(theta) and the x component is -mg*(Pi[sup]2[/sup]/4 - theta[sup]2[/sup])*sin(theta).
Adding these up, we get:
F[sub]x[/sub] = -mgthetacos(theta) - mg*(Pi[sup]2[/sup]/4 - theta[sup]2[/sup])*sin(theta)
F[sub]y[/sub] = -mgthetasin(theta) + mg*(Pi[sup]2[/sup]/4 - theta[sup]2[/sup])cos(theta) - mg
where the last term in F[sub]y[/sub] is just the gravitational acceleration. At theta = 0, F[sub]x[/sub] vanishes (this is good) and F[sub]y[/sub] = -mg(1-Pi[sup]2[/sup]/4), or about 3/2mg, directed upwards.
Now watch someone who actually remembers classical mechanics tear this to shreds… (And I’m very proud of the fact that I got all this bloody vb code right the FIRST TIME!)
I think you might be confusing two accelerations (or forces, if you prefer):
- The angular acceleration, alpha, which is the change in the angular velocity of the weight. Although I’m calling it “angular,” you could think of it as a linear acceleration, alpha*L, tangent to the arc of the weight’s travel. I think this is the acceleration you’re thinking of, one where the weight is speeding up and slowing down.
- The centrifugal acceleration, or force, which forces the weight to swing in a circle, instead of going in a straight line. This is what I’m referring to with the additional mv[sup]2[/sup]/L term.
Thought experiment: suppose you’re on the space shuttle (no gravity = easier thought experiment). You have a small weight tied to the end of a rope. Whirl the rope around your head at a constant velocity. Is there tension in the string? You bet! That’s the centrifugal force. If you then start whirling the weight faster, that’s angular acceleration. The centrifugal force is the mv[sup]2[/sup]/L term. This term is always in the radial direction, as I hope is intuitive from the above thought experiment. In the same way that whirling a weight around your head puts tension in the string, the velocity of the pendulum adds an additional tension into the rope.
All right. I hope the explanation above is adequate enough to show that I’m taking the gravitational force and adding in centrifugal force, but ignoring angular acceleration (which is zero at the bottom of the arc anyway).
OK, let’s see. I’ll use assumptions from my earlier post.
Assuming we start from vertical, equate kinetic and potential energy to find the velocity of the weight at the bottom of the arc: mgL = (1/2)mv[sup]2[/sup], so v = sqrt(2gL). At the bottom of the arc, total F = mg + mv[sup]2[/sup]/L = mg +m(2gL)/L = 3mg. Pretty significant, eh?
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Eek! I’m thinkin’ you dropped a minus sign somewhere, as a force directed upwards implies you’re pushing on the rope, yes?
the maximum acceleration would be when the swing is horizontal–it’s the acceleration due to gravity: 32 feet per second squared. That’s the only acceleration. It would decrease to zero at the bottom of the swing, then deceleration would take over, again at maximum when the swing is horizontal, until it momentarily becomes zero again as the swing stops and begins its swing in the other direction. When the swing is horizontal, it is in freefall. Its “weight” doesn’t exist. It’s weightless. When the swing reaches the center of its swing, when its supports are vertical, the maximum stress is on the swing–the weight of whatever is in the swing plus the weight of the swing and chain. If at that instant the swing support disappeared, the swing would not move downward but horizontally to the ground. The centrifugal force would carry the swing forward from the center and the weight of the swing is again zero because its acceleration changes from zero to the acceleration due to gravity. There would be some centrifugal force throughout the arc of movement of the swing, but it would never be greater than the force due to gravity. I think the law of conservation of energy would rule that out.
Eek! I’m thinkin’ you dropped a minus sign somewhere, as a force directed upwards implies you’re pushing on the rope, yes?
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Nope, we want a positive net force pulling up on the bob at the bottom of the pendulum so that at the bottom of the arc it accelerates upwards. Remember, the tension in the rope (the centripetal force) pulls upwards, and gravity always pulls down. Hence, the two forces should partially cancel, and as I’ve written them, they do. It’s not clear to me that what I have is correct, mind you, but qualitatively it’s okay. Of course, the big problem is that we can’t hope to use a harmonic oscillator approximation for such large angle displacements, but I haven’t been able to solve the general case and I’m too bloody lazy to look up a solution in one of my references. I HATE problems like this; so much easier to just stick with solving the equations of motion! 
I’m sorry, but I’m still missing it even after all this. here is my reasoning why centripetal acceleration is an extra term with no meaning in a pendulum which is only acted on by gravity.
At t = 0 the swing is completely horizontal. There is no force on the swing (theoretically) and all force acts on the mass to move it down.
At t = x < p/4 we see that the swing takes some of the force given by gravity. This breaks up our force into the two components: one tangential and one normal (I hope I’m using normal right… I mean the force directed out of the circular path if I’m not). At any instant, these forces cannot exceed the force applied which is only mg. What force doesn’t go to the swing goes into creating motion with an accelleration at any instant of
gsin(theta).
So t = 0 the swing feels no weight. t < P/4 the swing feels mgcos(theta) where theta is the angle off of horizontal, and gsin(theta) accellerates the mass further. No additional force is ever applied so the two component forces must always equal mg!
At t = p/4, theta = 90 and swing force is a maximum, mgcos(theta) or simply mg.
Please, please explain what is wrong with this. I cannot see where any other force comes from.
Total equations:
Instantaneous velocity: gtsin(theta)
Instantaneous acceleration: gsin(theta)
Force felt by swing: mgcos(theta)
If I could solve for theta in terms of time I could even generate an instantaneous position (length travelled) equation but not worth it if this is just going to be wrong anyway. But I cannot see what is wrong with this.
Just wanted to check the “email me when someone replies” box.
OK, sorry about my quick post yesterday, I did not have very much time.
Try this, at the bottom of the arc draw a force diagram for the mass of the pendulum. What you will get is a dot with two vectors, one pointed up, one down (lets call this the positive and negative y direction). The up vector is the tension in the rope, the down vector is gravity mg.
So using Sum of forces in the Y direction = mass times acceleration in the y direction we find:
T(ension) - mg = ma
Since the mass is experiencing circular motion, the acceleration must equal the velocity squared over the radius of motion (or L, the length of the pendulum rope). So solving for the Tension
T= mv^2/L + mg
To find the v^2 term, use energy, namely at the top of the arc kinetic energy is zero and at the bottom, set potential energy to zero. So
1/2 m v^2 = m g h
Where h is the height of the bob at the top of the swing and is equal to L times (1- cosine theta) where theta is the angle the rope makes from vertical. If theta = pi/2 (meaning the rope is horizontal at the top of the swing) h = L and the tension at the bottom of the arc is 3mg, as said before.
So, the formula is (after simplification)
T = mg (3 - 2 cosine(theta))
If you only swing up 45 degrees, the force is 1.59 mg.
What you’re going is approaching this as a statics problem, which it’s not; it’s a dynamics problem. Tell me what you think about these experiments:
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Take a weight (say, a heavy nut or washer or something) and tie it onto the end of the string. Grab the other end of the string and dangle it from your hand. What’s the tension in the string? It’s mg, right? That’s the static part. Now, quickly jerk your hand upwards, making the weight fly upwards. The tension in the string when you jerk it should be more than mg, because you’re accelerating the weight upwards. Agreed? That additional tension is from the dynamic portion of the problem.
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Now take the same weight and whirl it around your head. If you whirl if fast enough, it will go nearly horizontal. What’s the tension in the string? It’s certainly greater than mg (and, as an aside, you can figure out what it is by measuring the angle between the string and the horizontal. The tension in the string is then mg/sin(theta)). The increase in tension in the string is the dynamic part of the problem, and comes from the centripetal acceleration.
Now, what do these two examples have to do with the pendulum? Think of the pendulum dynamically at the bottom of its arc. What does the pendulum do next? It continues on its swing, and starts to move upward. Right? If the weight starts to move upward, it must be accelerating upwards, which means that there must be a force applied to the weight, which must come from the rope, which means there must be additional tension in the rope.
(This site, on amusement park rides might or might not help.)
But if you are calculating the tension in the rope, both the gravitational force and the centripetal force increase tension in the rope, and therefore they add. If you are calculating the forces on the weight, you have gravitational force, tension in the rope, and some acceleration term, which must all add to zero.
OK, not really, but I am becoming even more frustrated with this.
I fully undertand centripedal force in applications involving swinging a bucket of water around your head without it spilling. Indeed in such an event there is more tension on the line than is provided for by the weight of the mass. HOWEVER, the person swinging the thing is the source of extra force… there is gravity and a person applying force to the bucket, so of course there will be more than mg.
Dynamics schmynamics, this force needs to come from somewhere. Where does it come from? Is my diagram really a static representation? I account for every force that is applied to the mass. One, gravity. Two, the normal force the swing must apply to cause circular motion. Three, the tangential force which is what also causes circular motion. Forces two and three are truly just compnents of mg, there is no other force applied in the problem (no one is pushing the swing, the chain isn’t moving, etc).
Can centripetal force even be accoutned for in my diagram? It seems that my swing-component force is the tension… damn it, what am I missing? I cannot see where this extra force comes from… force doesn’t just appear out of nowhere and the only force applied is mg. ACK! ACK!
I tried to do it as an energy problem, but energy isn’t a vector so there’s no way to tell which energy is applied where, and much of the energy is going to be in the velocity of the object… which is tangential to the circular path and hence not a factor in the apparent weight at the bottom.
OK, first let’s get some of the terminology straight:
There is NO such thing as centrifugal force! There is such a thing as centripetal acceleration, but there is no force! When you are on a merry go round that is going really fast, there is not a force trying to pull you off, it is your inertia (objects in motion tend to remain in motion (same speed and direction) unless acted upon by an external force) that is trying to keep you going in a straight line. So why are you not going in a straight line? There must be a force pulling inwards, toward the center of the motion. This force will result in a centripetal acceleration toward the center of the merry go round (or the chain in the OP, or for Earth’s orbit, the sun). What causes this acceleration is different in each case: for the merry go round it is the person hanging on, for the Earth it is gravity (G m1 m2 / r^2), for the swing it is the Tension in the rope.
Now think about this… the mass at the end of the rope is undergoing uniform circular motion at the bottom of the arc (there are no tangential accelerations), so we know that it must be accelerating toward the center of the circle. Since the mass is accelerating upwards, the tension in the rope MUST be greater than mg (which wants to accelerate the mass in the opposite direction). The magnitude of the tension can be found above in previous posts.
Where is this force is coming from? It is a pseudo-normal force that is being applied to keep the swing moving in circular motion. It is the external force that is changing direction of the motion in accordance with Newton’s first law. It is NOT a component of mg, but a counteracting force. It has nothing to do with mg, except for the fact that it counter acts it. If the rope were cut and the force removed from the diagram, mg would be the same and you start accelerating toward a very hard landing.
Think about this: a bungie jumper jumps off a cliff and accelerates at g for 100 feet. At this point, the slack has been used up and he starts to decelerate to a stop over a period of 10 feet. Does the tension in the cable = mg. No, the tension has decelerated the person at 10 times the rate that gravity accelerated them. The tension is much greater, it has nothing to do with gravity.
I don’t know what else to say, but I hope this helps!
Right. My bad. For what it’s worth, I think I agree with Eyer’s earlier post.
The magnitudes of the forces must be equal; if they were unequal the mass would definitely go in one direction or the other-- the swinging action would be bumpy.
Consider it as an energy problem. At the bottom the pendulum is at rest. We have a force mg acting downward, and a (lets say) theoretical 0 potential energy (of course there must always be some since a pundulum can’t swing if its bottom curve meets the ground!). We raise the pendulum arm up, adding to it mgh joules of energy (or whatever units you want). We let go. When it reaches the bottom, we have essentially not changed from the original state except for the motion, which is always tangential (if you are spinning a mass around your head and let go, the mass always travels along the tangent to the path). This motion is a result of a potential–>kinetic conversion.
Now, lets tackle this as follows. We have an energy of mgh at the horizontal. At the bottom of the swing h = 0 and all energy must have gone somewhere. Now, we can calculate velocity as the indefinite integral of instantaneous acceleration. I gave instantaneous acceleration before as mgsin(theta). This gives us a velocity equation of -mgcos(theta) (which I incorrectly gave above, thinking in terms of time instead of angles) plus some constant which accounts for initial velocity. Initial velocity is zero, so we are quite simply left with -mgcos(theta). Since at the bottom cos(theta) is one, the velocity at the bottom is simply -mg.
The kinetic energy here is 0.5(mg)[sup]2[/sup]. NOW, we know that there can be no loss of energy in our ideal situation, so the total energy at the bottom must be
0.5m(mg)[sup]2[/sup] + C = mgh where C is the mysterious force/energy that isn’t used up in motion as I suggest it is and accounts for the extra force applied.
Solving this equation for C yields
C = mgh - 0.5m(mg)[sup]2[/sup]
= mgh - 0.5m[sup]3[/sup]g[sup]2[/sup] (of course the units look screwy but that’s because of the integration: all variables here are only values, and the units have to be assumed)
Here we note that C can only, in real world problems, have values greater than or equal to zero since C is the energy not used by velocity.
First, when is C > 0?
0 < mgh - 0.5m[sup]3[/sup]g[sup]2[/sup]
0 < 2mgh - m[sup]3[/sup]g[sup]2[/sup]
2mgh < -m[sup]3[/sup]g[sup]2[/sup]
2h < -m[sup]2[/sup]g
Thus, for C to be positive, h must be negative (m and g aren’t negative so the right side can never be positive)
If h is negative, then mgh is negative. Thus the original equation 0.5mv[sup]2[/sup] + C = mgh gives us a contradiction. We made the requirement that C be positive, and kinetic energy is always positive, and two positive numbers do not add to be negative.
There is no extra energy, there is no extra force.
You seem to be mixing up force and energy. If the person swinging the bucket swings it at a constant speed, he is most certainly NOT adding energy into the system (well, except for replacing that lost to friction and so forth). The person is the “source” of the force only in that he’s holding the other end of the rope.
How about this: Erect a six-foot-high post in your back yard. Put a swivel hook on the top. Tie a rope to the swivel and a weight to the other end. Now give the weight a good push and let go. What happens? The weight whirls around the post in exactly the same way that a bucket would whirl around your head. I hope it is intuitive that the tension in the rope is greater than mg.
Where does this extra force come from? If you jerk a yo-yo upwards, there’ll be extra tension in the string because you accelerated the yo-yo upwards. In the same way, the extra tension in the string comes from accelerating the weight.
Now, think of the pendulum again. Tell me where you miss this course of reasoning:
- Pretend the pendulum is swinging.
- When the rope is vertical, the pendulum is at it’s lowest point. Right?
- At this point, the velocity of the pendulum is completely horizontal, right?
- The next instant of time, the pendulum is slightly higher than it was before, right?
- To get to that point, the pendulum must have some vertical componant of velocity, right?
- But wait a minute! When the pendulum is at its lowest point, it has no vertical velocity (#3, above)!
- So the vertical velocity has changed. Change in velocity is acceleration, so the weight must be accelerating. Upwards, in fact.
- Since F=ma, the acceleration must cause a force.
- The only thing the weight is connected to is the rope; therefore, the rope must bear the additional force.
Potential energy when weight is raised up = mgh
(h is measured from the bottom of the arc)
Kinetic energy when weight is raised up = 0
Kinetic energy at bottom of arc = (1/2)mv[sup]2[/sup]
Potential energy at bottom of arc = 0
mgh + 0 = 0 + (1/2)mv[sup]2[/sup]
The only unknown here is v, so
v = sqrt(2gh)
There is no additional C in the equation. There is no additional energy to consider. However, force is not energy. I’m not sure, but I think you’re trying to say that since there is no additonal energy, there cannot be any greater tension in the rope than mg. If that is so, then consider the static case, where the pendulum is hanging unmoving. In this case, there is no energy, either kinetic or potential. Is there then no tension in the rope? What if I sat on the weight? Would the tension not increase? (sitting on the weight adds no energy to the system either)
There is no C (i.e. WORK) in the energy equation for this system, the reason being is that the Tension in the string is perpendicular to the motion. Work is defined as a force acting over a distance F times d (actually dot product, but whatever). If you take the dot product of the tension with the motion of the mass, it will always be 0 because the vectors are perpendicular.
With this system (i.e. no friction, ideal) the only work done is by gravity, and the work done is equal to mgh.
But this still has nothing to do with the fact that in circular motion, the mass is accelerating inwards. This corresponds to the force of tension in the string which is greater than mg at the bottom of the arc.
Anyway, here are some links that may help (neat site!)…
http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html#circ
http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cf
I don’t know guys, I just can’t get this. 
I look at my force diagram and I see circular motion, replete with acceleration all over the place! And yet I can never manipulate the numbers to get any more than mg acting “up” on the mass (towards the center of the circular path).
I think we should just leave it as erl doesn’t get centripetal acceleration (and hence force) and be done with it. If you all say it would experience 3mg of tension on the rope, then I’ll take your word for it. I just can’t see it.
I’m thinking that your diagram is simply missing terms. As Eyer8 pointed out, all circular motion has associated centripetal acceleration. This acceleration has nothing (directly) to do with the gravitational acceleration, g. Would it help if I pointed out that, if you set up a pendulum on the Space Shuttle and gave it a push to start it, there would still be tension in the rope (equal to mv[sup]2[/sup]/L, in fact)?
Another web site demonstrating centripetal acceleration with pictures! Oooh!