Strain Energy to Gravitational Potential Energy

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esrever
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A uniform rod, AB, of length 2m and mass 1kg, has a particle of mass 1kg attached to it at B. It can rotate freely about a horizontal axis through A. The end B is attached by means of an elastic string of natural length 1m to a point a distance 2m above A. The system is held at rest with AB horizontal and is then released. If the rod just reaches the vertical position, find the modulus of elasticity of the string.

I got the modulus of elasticity \lambda =  17.6. I first found the initial strain energy, which turned out to be 1.67\lambda. Kinetic and gravitation potential energies are initially 0J.

At the top when the rod is vertical, the kinetic energy and the strain energy are 0J. Let G be the centre of mass of the rod. It can be found that AG is 1.5m. Also the total mass of the rod and particle is 2kg. So at top, gravitational potential energy is 2*9.8*1.5 = 29.4. Equating 1.67\lambda = 29.4 gives \lambda =  17.6. But the answer apparently is 10.4.

Where did I go wrong? Thank you for your help .
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Eimmanuel
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(Original post by esrever)
A uniform rod, AB, of length 2m and mass 1kg, has a particle of mass 1kg attached to it at B. It can rotate freely about a horizontal axis through A. The end B is attached by means of an elastic string of natural length 1m to a point a distance 2m above A. The system is held at rest with AB horizontal and is then released. If the rod just reaches the vertical position, find the modulus of elasticity of the string.

I got the modulus of elasticity \lambda = 17.6. I first found the initial strain energy, which turned out to be 1.67\lambda. Kinetic and gravitation potential energies are initially 0J.

At the top when the rod is vertical, the kinetic energy and the strain energy are 0J. Let G be the centre of mass of the rod. It can be found that AG is 1.5m. Also the total mass of the rod and particle is 2kg. So at top, gravitational potential energy is 2*9.8*1.5 = 29.4. Equating 1.67\lambda = 29.4 gives \lambda = 17.6. But the answer apparently is 10.4.

Where did I go wrong? Thank you for your help .

I can understand why you have a problem in attempting the question. I find that the problem is quite badly worded.

What is missing in your working is that the string is also stretched when the rod is in the vertical position. So according to the conservation of energy,

Change in elastic potential energy = Change in gravitational potential energy
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esrever
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(Original post by Eimmanuel)
I can understand why you have a problem in attempting the question. I find that the problem is quite badly worded.

What is missing in your working is that the string is also stretched when the rod is in the vertical position. So according to the conservation of energy,

Change in elastic potential energy = Change in gravitational potential energy
Oh I just realised that. Thank you for pointing it out .
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esrever
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(Original post by Eimmanuel)
I can understand why you have a problem in attempting the question. I find that the problem is quite badly worded.

What is missing in your working is that the string is also stretched when the rod is in the vertical position. So according to the conservation of energy,

Change in elastic potential energy = Change in gravitational potential energy
When horizontal, x = 2\sqrt{2} - 1 and GPE = 0. Total energy is \dfrac{\lambda (2\sqrt{2} - 1)^2}{2(1)} = 1.67\lambda. When vertical, x = 2 - 1 = 1 so strain energy is \dfrac{\lambda (1)^2}{2(1)} = 0.5\lambda. Also GPE of rod is 1g(1) = g and GPE of particle is 1g(2) = 2g. Total GPE is 3g.

So 1.67\lambda = 0.5\lambda + 3g. Hence 1.17\lambda = 3g and \lambda = 25.1. But the answer still doesn't match. Is there something else that I missed?
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Eimmanuel
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(Original post by esrever)
When horizontal, x = 2\sqrt{2} - 1 and GPE = 0. Total energy is \dfrac{\lambda (2\sqrt{2} - 1)^2}{2(1)} = 1.67\lambda. When vertical, x = 2 - 1 = 1 so strain energy is \dfrac{\lambda (1)^2}{2(1)} = 0.5\lambda. Also GPE of rod is 1g(1) = g and GPE of particle is 1g(2) = 2g. Total GPE is 3g.

So 1.67\lambda = 0.5\lambda + 3g. Hence 1.17\lambda = 3g and \lambda = 25.1. But the answer still doesn't match. Is there something else that I missed?
Your strain energy or elastic potential energy is incorrect when the rod is vertical. You may want to draw a picture to show what is your thinking. As I said the question is pretty badly worded, I find that it is quite hard to explain without a picture.
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esrever
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(Original post by Eimmanuel)
Your strain energy or elastic potential energy is incorrect when the rod is vertical. You may want to draw a picture to show what is your thinking. As I said the question is pretty badly worded, I find that it is quite hard to explain without a picture.
Thank you. I will try again with the picture.
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