Forces on an object in vertical circular motion

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Homework Statement


Imagine a vertically placed clock. What forces have an effect on the object when it is in the position of 3am?

Homework Equations


No need od equations i guess.

The Attempt at a Solution


So I figured out that when the object is in the position of 12am then weight of the object is the final centripetal force.

Also when it is in the position of 6am then final centripetal force is the force of rope which has to compensate the weight of the object.

But when it is in the position of 3am then weight is perpendicular to centripetal force and I know(or at least I think it's correct) that centripetal force should be the resultant of more forces. But I cant figure out any other forces in there.

Can someone help me a bit please? Thanks
 

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  • #2
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I think the idea of centripetal force may be tripping you up. An object has some forces acting on it. If the object is moving in a curved path then some (or all) of those forces must be acting centripetally, that is, perpendicular to the direction of motion and towards the direction of the curve. You can tell from the velocity and the curvature how large that centripetal component of the applied forces must be, and sometimes that is useful for calculation. However the centripetal force is not something extra or separate from the applied forces. There are applied forces, and some of those forces may be acting centripetally.

In this question they just asked what forces are present, so I don’t think you need to worry about whether they are centripetal or not (yet, probably follow on questions coming). What forces are acting on the object? There are only 2.
 
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  • #3
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So I figured out that when the object is in the position of 12am then weight of the object is the final centripetal force.
Not usually. Saying things like “in the position of 12 o’clock” leaves out some crucial information. For example we’ll have to assume the object is swinging in a circle, and there is no mention of velocity. Assuming we are talking about the object moving in a circle, the weight is certainly acting in the centripetal direction, but does it have to be the only force? Doesn’t it depend on how fast the object is moving and so how much centripetal force is required?
 
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Also when it is in the position of 6am then final centripetal force is the force of rope which has to compensate the weight of the object.
Here, you are correct the rope is the only available centripetal force. You are also right that the rope has to counter the weight. However that would be tru even if the object wasn’t swinging in a circle. At the bottom the rope has to provide force to counter the weight AND enough extra force to provide the centripetal force for the circular motion
 
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  • #5
CWatters
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If an object is moving on a path of radius r at velocity v then there must be a net centripetal force of mv^2/r acting on it towards the centre. Consider two cases...

1) Rigid slow moving clock hands...

At 12 o'clock gravity provides too much centripetal force. The excess tries to compress the hand which resists like a spring.

At the 3 o'clock position gravity tries to bend the hand which again is resisted by the spring force of the material the hand is made from. Leaving tension in the hand to provide the required centripetal force.

2) Bucket on a rope...

At 12 clock tension in the rope and gravity combine in the same direction to provide the necessary centripetal force mv^2/r. If they were to provide more than mv^2/r (eg because v was too small) the radius would reduce and the bucket would fall.

At the 3 o'clock position there are three forces acting. The force of gravity is balanced by the inertial force of the bucket (the bucket is decelerating as it rises) leaving tension in the rope to provide the required centripetal force.

At 6 o'clock tension in the rope and gravity must combine to provide the necessary centripetal force mv^2/r only now they act in opposite directions ,(compared to 12 o'clock) so the tension must be higher.
 
  • #6
HAF
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I think the idea of centripetal force may be tripping you up. An object has some forces acting on it. If the object is moving in a curved path then some (or all) of those forces must be acting centripetally, that is, perpendicular to the direction of motion and towards the direction of the curve. You can tell from the velocity and the curvature how large that centripetal component of the applied forces must be, and sometimes that is useful for calculation. However the centripetal force is not something extra or separate from the applied forces. There are applied forces, and some of those forces may be acting centripetally.

In this question they just asked what forces are present, so I don’t think you need to worry about whether they are centripetal or not (yet, probably follow on questions coming). What forces are acting on the object? There are only 2.
So am I right if I say that there are only weight and force of the rope?

How is then possible that the object won't stop "in the 6am position"? Because it's too fast?
 
  • #7
HAF
57
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If an object is moving on a path of radius r at velocity v then there must be a net centripetal force of mv^2/r acting on it towards the centre. Consider two cases...

1) Rigid slow moving clock hands...

At 12 o'clock gravity provides too much centripetal force. The excess tries to compress the hand which resists like a spring.

At the 3 o'clock position gravity tries to bend the hand which again is resisted by the spring force of the material the hand is made from. Leaving tension in the hand to provide the required centripetal force.

2) Bucket on a rope...

At 12 clock tension in the rope and gravity combine in the same direction to provide the necessary centripetal force mv^2/r. If they were to provide more than mv^2/r (eg because v was too small) the radius would reduce and the bucket would fall.

At the 3 o'clock position there are three forces acting. The force of gravity is balanced by the inertial force of the bucket (the bucket is decelerating as it rises) leaving tension in the rope to provide the required centripetal force.

At 6 o'clock tension in the rope and gravity must combine to provide the necessary centripetal force mv^2/r only now they act in opposite directions ,(compared to 12 o'clock) so the tension must be higher.
So basically if I understood everything correctly.

Tension of the rope is compensating weight in every position?

Does it mean that at 6 o'clock position if we had given the max tension of the rope we can calculate in what speed will the rope tear up right?

The equation would be like this?

F(max. tension) = F(centripetal) + mg
 
  • #8
haruspex
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The equation would be like this?

F(max. tension) = F(centripetal) + mg
No. Reread post #2, which lays it out very clearly and correctly.
The applied forces are mg and the force from the arm. You can if you wish resolve the force from the arm into radial (tension) and tangential components, or whatever.
The resultant of the applied forces leads to the acceleration: ΣF=ma.
In the present case we know something about the acceleration. This allows us to make inferences regarding the applied forces. For this purpose, it is useful to resolve the acceleration into radial and tangential components.
 
  • #9
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So am I right if I say that there are only weight and force of the rope?

How is then possible that the object won't stop "in the 6am position"? Because it's too fast?
Yes, correct, tension in the rope and gravity. And yes again, it doesn’t stop at the bottom because it is moving (has momentum and kinetic energy).
 

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