Tension at at the bottom of the pendulum

In summary: Assuming that the ball is suspended from the little black peg, the angle it makes in point A with the vertical is 90°. So applying my calculations, the tension in point B would be...In summary, the tension in point B is 3mg.
  • #1
Eugen
22
1

Homework Statement


A ball of mass m is attached to a string of length L and released from rest at point A. Show that the tension in the string when the ball reaches point B is 3mg, independent of the length l. (there is an image in attachment )
[/B]

Homework Equations


K = mv2/2
U = mgh
Fcp = mv2/r

The Attempt at a Solution


The mechanical energy at point A must equal the mechanical energy at point B. h is the vertical distance between A and B.
So mv2/2 - mgh = 0 and v2 = 2gh
The net force acting on the ball at point B is the centripetal force, which is mv2/L and is equal with T - mg.
mv2/L = T - mg.
So T = mv2/L + mg.
T = 2mgh/L + mg
θ is the angle between the string and the vertical when the ball is at point A.
If I write h = L - Lcosθ I get to the equation T = mg(2 - 2cosθ + 1), which is 3mg only when θ is 90°. Something must be wrong.
 

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  • #2
Eugen said:
A ball of mass m is attached to a string of length L and released from rest at point A. Show that the tension in the string when the ball reaches point B is 3mg, independent of the length l. (there is an image in attachment

i think the total energy at A is transferred to the bob as kinetic energy -therefore
yourwork out is correct.
Is there a problem if theta is 90 degree?
 
  • #3
drvrm said:
i think the total energy at A is transferred to the bob as kinetic energy -therefore
yourwork out is correct.
Is there a problem if theta is 90 degree?
I think the problem wants to show that T = 3mg no matter at what angle the ball begins to move.

L.E: thinking about what drvrm said, I guess it is not possible that T is independent of θ. So, is the problem text wrong? (l is the horizontal line between point A and the string when the ball is in point B)
 
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  • #4
Eugen said:
I think the problem wants to show that T = 3mg no matter at what angle the ball begins to move.

L.E: thinking about what drvrm said, I guess it is not possible that T is independent of θ. So, is the problem text wrong? (l is the horizontal line between point A and the string when the ball is in point B)
There's nothing (well, not much) wrong with the question, but you may be misinterpreting it. Ignore the angle theta. You are only interested in the motion from A to B. The string is depicted in a later position, at angle theta past B, just to make the physical set-up clearer.

The one thing that is not clear is that it asks for the tension "when the ball reaches point B". The tension exactly at B is indeterminate, for the simple reason that it transitions at that point from an arc of radius l to an arc of radius L. For the question to make sense, you have to read it as "as the ball reaches point B", i.e. immediately before B.
 
  • #5
haruspex said:
The tension exactly at B is indeterminate, for the simple reason that it transitions at that point from an arc of radius l to an arc of radius L.

The ball is suspended from a string tied to the roof. There's only one string, so only one radius, L. l is the horizontal distance between point A and the radius in point B. Or so I interpret the image + the problem text. Unfortunately, the image is blurry.
 
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  • #6
Eugen said:
The ball is suspended from a string tied to the roof. There's only one string, so only one radius, L. l is the horizontal distance between point A and the radius in point B. Or so I interpret the image + the problem text. Unfortunately, the image is blurry.
That's not how I read the diagram. There is a black spot above B and to the right of A, which appears to represent some kind of peg. The dotted line showing the trajectory clearly describes an arc centred on the peg as it goes from A to B. Moreover, my interpretation leads to the expected answer.
 
  • #7
haruspex said:
That's not how I read the diagram. There is a black spot above B and to the right of A, which appears to represent some kind of peg. The dotted line showing the trajectory clearly describes an arc centred on the peg as it goes from A to B. Moreover, my interpretation leads to the expected answer.

Assuming that the ball is suspended from the little black peg, the angle it makes in point A with the vertical is 90°. So applying my calculations, the tension in point B would be 3mg, just as specified in the problem text. But, once it arrives in point B, how could it change the radius while still being suspended from that black peg? It could, if there were 2 strings: L beginning in the roof and l beginning in the black peg. But the problem would become quite complicated in this interpretation, and quite distant from the text provided (it's from walker textbook, energy conservation chapter). I think there's a problem with the problem text and diagram.
 
  • #8
Eugen said:
Assuming that the ball is suspended from the little black peg, the angle it makes in point A with the vertical is 90°. So applying my calculations, the tension in point B would be 3mg, just as specified in the problem text. But, once it arrives in point B, how could it change the radius while still being suspended from that black peg?
It is not, according to the diagram, suspended from the peg. It is suspended from a point L-l above the peg. But the peg interferes with the string while the mass is left of the vertical position.
I agree that it seems pointless to have such an elaborate set-up. The problem as stated would have worked just as well suspending it from the peg. Are there more parts to the question?
 
  • #9
haruspex said:
It is not, according to the diagram, suspended from the peg. It is suspended from a point L-l above the peg. But the peg interferes with the string while the mass is left of the vertical position.
I agree that it seems pointless to have such an elaborate set-up. The problem as stated would have worked just as well suspending it from the peg. Are there more parts to the question?

Point b of the problem text: "Give a detailed physical explanation for the fact that the tension at the point B is independent of length l."

It can't be suspended above the peg if it is to hang (at least initially) from a string of length l. (that's what I understand if it is to change radius to L at point B).
Anyway, I don't think we should waste our time doing diagram hermeneutics.
 
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  • #10
Eugen said:
Point b of the problem text: "Give a detailed physical explanation for the fact that the tension at the point B is independent of length l."

It can't be suspended above the peg if it is to hang (at least initially) from a string of length l. (that's what I understand if it is to change radius to L at point B).
Anyway, I don't think we should waste our time doing diagram hermeneutics.
Not sure what you are saying there, but here is my interpretation of the diagram. The string is length L, and attached to some point, call it O. The string hangs vertically from there to a peg at point P, then is pulled under and around the peg to go horizontally to point A. The length of PA is lowercase l.
When released, the mass will swing in an arc of radius l until it is vertically below P, then will continue in an arc of radius L.
That is entirely consistent with the text, with the diagram, and with the result to be proved (provided we read 'reaches B' as immediately before reaching B).
 
  • #11
haruspex said:
Not sure what you are saying there, but here is my interpretation of the diagram. The string is length L, and attached to some point, call it O. The string hangs vertically from there to a peg at point P, then is pulled under and around the peg to go horizontally to point A. The length of PA is lowercase l.
When released, the mass will swing in an arc of radius l until it is vertically below P, then will continue in an arc of radius L.
That is entirely consistent with the text, with the diagram, and with the result to be proved (provided we read 'reaches B' as immediately before reaching B).

You are right, this explanation is consistent with all the problem data. But besides knowing some physics, one has to be a bit of Sherlock Holmes to solve it... :smile:
 

Related to Tension at at the bottom of the pendulum

1. What causes tension at the bottom of a pendulum?

The tension at the bottom of a pendulum is caused by the force of gravity acting on the mass of the object. As the pendulum swings, the weight of the object pulls down, creating tension in the string or rod that is holding the object.

2. How does tension affect the motion of a pendulum?

Tension plays a crucial role in the motion of a pendulum. It is responsible for keeping the pendulum oscillating back and forth, as it acts as the restoring force that pulls the object back towards its starting point. Without tension, the pendulum would not be able to swing back and forth.

3. What happens to the tension at the bottom of a pendulum as it swings?

The tension at the bottom of a pendulum changes as the pendulum swings. At the bottom of its swing, the tension is at its maximum as the weight of the object is pulling directly down. As the pendulum swings towards the top of its arc, the tension decreases as the weight of the object is pulling less directly down.

4. Can the tension at the bottom of a pendulum be measured?

Yes, the tension at the bottom of a pendulum can be measured using a spring scale or a force sensor. By attaching the scale or sensor to the string or rod holding the object, the tension can be measured at various points in the pendulum's swing and can be used to analyze the motion of the pendulum.

5. How does the length of the pendulum affect the tension at the bottom?

The length of the pendulum does not directly affect the tension at the bottom. However, it does affect the period (or time) of the pendulum's swing, which in turn can affect the tension. A shorter pendulum will have a shorter period and therefore a higher frequency of oscillation, resulting in a higher tension at the bottom. A longer pendulum will have a longer period and a lower frequency, resulting in a lower tension at the bottom.

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