# Circular Moments

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#1
A hoop of weight 20N can rotate freely about a pin fixed in a wall. A string has one end attached to the pin, runs round the circumference of the hoop to its lowest point, and is then held horizontally at its other end. A gradually increasing horizontal force is now applied to the string so that the hoop begins to rotate about the pin. Find the tension in the string when the hoop has rotated through 40 degrees.

Image: https://imgur.com/a/sZ3B23y

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As you can see in the image, the string exerts normal contact force on the hoop over a continuous arc. Now what I noticed is that if I take moments about the centre of the circle, then the moment of these normal forces is 0 as they all pass through the centre of the circle. Now the only forces left are tension in the string and the force at the hinge which is tangent to the circle facing right (normal component of the force at the hinge will pass through the centre). But the issue with this is that the moment about the weight is also 0.

Another way that I conceived was to split up the circle into 2 parts, and use the centre of mass of a curved arc idea. But that is out of syllabus. So I was wondering if there is any other way?
Last edited by esrever; 1 year ago
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1 year ago
#2
(Original post by esrever)
...
Can you link/scan the whole of the original question.
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1 year ago
#3
(Original post by esrever)
A hoop of weight 20N can rotate freely about a pin fixed in a wall. A string has one end attached to the pin, runs round the circumference of the hoop to its lowest point, and is then held horizontally at its other end. A gradually increasing horizontal force is now applied to the string so that the hoop begins to rotate about the pin. Find the tension in the string when the hoop has rotated through 40 degrees.

Image: https://imgur.com/a/sZ3B23y

---------------------------

As you can see in the image, the string exerts normal contact force on the hoop over a continuous arc. Now what I noticed is that if I take moments about the centre of the circle, then the moment of these normal forces is 0 as they all pass through the centre of the circle. Now the only forces left are tension in the string and the force at the hinge which is tangent to the circle facing right (normal component of the force at the hinge will pass through the centre). But the issue with this is that the moment about the weight is also 0.

Another way that I conceived was to split up the circle into 2 parts, and use the centre of mass of a curved arc idea. But that is out of syllabus. So I was wondering if there is any other way?

I'd imagine you'd have two moments about the pin
1) The hoop weight which acts vertically through the center of the hoop
2) The tension which goes horizontally from the bottom of the hoop.
Equate and solve for the tension?
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#4
(Original post by ghostwalker)
Can you link/scan the whole of the original question.
I'm sorry but I didn't quite get you. I already wrote the question exactly as it was presented to me.
Last edited by esrever; 1 year ago
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#5
(Original post by mqb2766)
I'd imagine you'd have two moments about the pin
1) The hoop weight which acts vertically through the center of the hoop
2) The tension which goes horizontally from the bottom of the hoop.
Equate and solve for the tension?
There are also numerous contact forces acting where the string is in contact with the hoop.
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1 year ago
#6
(Original post by esrever)
There are also numerous contact forces acting where the string is in contact with the hoop.
Its equivalent to having a string tied at the bottom of the hoop with a single point of contact?
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#7
(Original post by mqb2766)
Its equivalent to having a string tied at the bottom of the hoop with a single point of contact?
But there are also sideways forces from the string contact. The net force of contact cannot be taken at the bottom right? Because this would change the moment of the contact forces.
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1 year ago
#8
(Original post by esrever)
But there are also sideways forces from the string contact. The net force of contact cannot be taken at the bottom right? Because this would change the moment of the contact forces.
If you turned the diagram round 90 degrees so the tension goes downwards, this would be more usual for the string (with a mass attached, acting under gravity). The tension would be acting downwards where the moment would be taken at the point of contact with the circle. Same with this case.
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