Hi. Why did you put a constant cross-sectional area for the rubber, which in real, it's not? And the expansion will reduce the area for the rubber, or is it just an approximation?
Thanks Dr for a great explanation. If at rest radius is 737 mm, material density is 7.8 gm/cc (steel), what would be the expanded radius at 3000 RPM for a Modulus of elasticity, E = 210,000 N/mm²
@@DrBenYelverton Thank you, Dr. I used a radius of 0.3685 meter, density of 7800 kg per meter cubed, angular velocity of 314 radians/sec and E of 200,000,000,000 N per sq meter and got expansion of 0.000192414 mm, which is off a factor of 1000. I should get 0.192414 mm. Where am I going wrong?
The change in radius will come out in metres if you use all the standard units, so your 0.000192414 is in m, not mm. Multiply by 1000 to convert to mm and you get the correct answer!
Question: If I place three weights inside this rubber ring, two relatively close to each other, and one on the otherside, how do I calculate its deformation and the appearance of pushing and pulling psudoforces between the weights?
I don't think this would be possible to solve analytically because the added masses would break the symmetry of the system - you'd probably need to run some sort of simulation instead.
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Amazing video man. Thank you for sharing sir. I wish you all the best
Thanks for saying so, I appreciate your support!
one of the best channels ive found, showed my university friends. Keep it up
Thanks, I appreciate that!
Great concept keep working
Thanks for watching!
Hi. Why did you put a constant cross-sectional area for the rubber, which in real, it's not? And the expansion will reduce the area for the rubber, or is it just an approximation?
It's just an approximation!
Thanks Dr for a great explanation.
If at rest radius is 737 mm, material density is 7.8 gm/cc (steel), what would be the expanded radius at 3000 RPM for a Modulus of elasticity, E = 210,000 N/mm²
Thanks! It should be easy to do that calculation using the equation in the video, as long as you do all the appropriate unit conversions.
@@DrBenYelverton Thank you, Dr. I used a radius of 0.3685 meter, density of 7800 kg per meter cubed, angular velocity of 314 radians/sec and E of 200,000,000,000 N per sq meter and got expansion of 0.000192414 mm, which is off a factor of 1000. I should get 0.192414 mm. Where am I going wrong?
The change in radius will come out in metres if you use all the standard units, so your 0.000192414 is in m, not mm. Multiply by 1000 to convert to mm and you get the correct answer!
@@DrBenYelverton Thank you Dr. Got it. Much appreciated.
Question: If I place three weights inside this rubber ring, two relatively close to each other, and one on the otherside, how do I calculate its deformation and the appearance of pushing and pulling psudoforces between the weights?
I don't think this would be possible to solve analytically because the added masses would break the symmetry of the system - you'd probably need to run some sort of simulation instead.
@@DrBenYelverton yeah, that would throw the system off causing vibrations in the particles leading to uncertainty in their exact relative position.
This sort of questions are inevitably asked in the India's largest engineering entrance test IIT JEE. Thank you
Yeah man , i am also preparing
Did u guys pass it