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Issac Physics solid spring, challenge 3
hi, this is the question
https://isaacphysics.org/questions/solid_spring?board=be0dac43-6ba3-44d5-9a77-091234eef766&stage=a_level
it's really tough and I got Eb^3pi/3(b-a) which was wrong. does anyone want to have a look pls?
I would upload my method in the discussion part, thanks a lot!
https://isaacphysics.org/questions/solid_spring?board=be0dac43-6ba3-44d5-9a77-091234eef766&stage=a_level
it's really tough and I got Eb^3pi/3(b-a) which was wrong. does anyone want to have a look pls?
I would upload my method in the discussion part, thanks a lot!
Reply 1
hi,
this is my method. I tried to integrate the solid but somehow the answer was wrong.
this is my method. I tried to integrate the solid but somehow the answer was wrong.
Original post by 1831
hi,
this is my method. I tried to integrate the solid but somehow the answer was wrong.
this is my method. I tried to integrate the solid but somehow the answer was wrong.
I dont really understand what youve tried to do, but again the hints are reasonable. Consider a thin slice of the frustum and apply hint 4 to it.. You should get that the
infinitesimal extension is proportional to infinitesmal height / area
Then integrate to get a relationship between F and e (total extension) which gives the spring constant.
Reply 3
Original post by mqb2766
I dont really understand what youve tried to do, but again the hints are reasonable. Consider a thin slice of the frustum and apply hint 4 to it.. You should get that the
infinitesimal extension is proportional to infinitesmal height / area
Then integrate to get a relationship between F and e (total extension) which gives the spring constant.
infinitesimal extension is proportional to infinitesmal height / area
Then integrate to get a relationship between F and e (total extension) which gives the spring constant.
Yea I got E*pi(b^3-a^3)/3L and it’s wrong
.
Original post by 1831
Yea I got E*pi(b^3-a^3)/3L and it’s wrong
.
I agree with that (thats its wrong). You need to formulate the problem properly for a thin disc, then to integrate over the frustum. Can you get the right formula for the infinitesial extension of a thin disc in terms of its (infinitesimal) height?
(edited 3 years ago)
Reply 5
Original post by mqb2766
I agree with that (thats its wrong). You need to formulate the problem properly for a thin disc, then to integrate over the frustum. Can you get the right formula for the infinitesial extension of a thin disc in terms of its (infinitesimal) height?
hi,
if this is what you mean
and also the first solution I used was to kind of trying to create a coordination of the frustum... I think I treated the radius as y and the length as x, so I did dx/dy (dr/dl) and I try to integrate it. which m is the gradient which equals to (b-a)/L. and I tried to get dl/dA from (dr/dA)/(dr/dl), and I integrated the dl/dA from l=0 to l=L. and I got a wrong answer
But I'm still keeping trying it
I did a similar question about the resistance of a frustum before and I used the same method (learnt it from YouTube) and got the right answer so I applied the same method to this one but it seems like it's not quite right.... thanks so much!Reply 6
Original post by mqb2766
I agree with that (thats its wrong). You need to formulate the problem properly for a thin disc, then to integrate over the frustum. Can you get the right formula for the infinitesial extension of a thin disc in terms of its (infinitesimal) height?
I got this but emmmm.…………would u like to tell me where’s the loophole pls 🤣🤣🤣🤣🤣 thxxxx I think my brain is a bit lack of sleep today 🥲🥲🥲
Original post by 1831
I got this but emmmm.…………would u like to tell me where’s the loophole pls 🤣🤣🤣🤣🤣 thxxxx I think my brain is a bit lack of sleep today 🥲🥲🥲
There are a few things which you probably need to be clearer about.
Firstly, the force is applied throughout the frustum, so for the thin disc you have F going upwards and downwards, just as in the complete frustum. From the point of view of the later integration problem, its a constant.
When the force is applied to the thin disc (of height dx), this will produce an extension de, so
E = F*dx / (pi*r^2*de)
where r is the radius which is a function of x, so r(x) from a..b. E and F are constants. Rearrange for
de = ... dx
and note that you want to integrate along the length or height of the frustum to get the total extension. So integrating both sides, where on the right you either put r in terms of x, or change integration variables and integrate wrt r (probably a bit easier).
Setting up the problem is arguably the "hard" / important bit of the problem. You seem to confuse dx and x, treat k as a constant (it won't be as de is a function of x), integrate E to get E and a few other errors/loopholes.
(edited 3 years ago)
Reply 8
Original post by mqb2766
There are a few things which you probably need to be clearer about.
Firstly, the force is applied throughout the frustum, so for the thin disc you have F going upwards and downwards, just as in the complete frustum. From the point of view of the later integration problem, its a constant.
When the force is applied to the thin disc (of height dx), this will produce an extension de, so
E = F*dx / (pi*r^2*de)
where r is the radius which is a function of x, so r(x) from a..b. E and F are constants. Rearrange for
de = ... dx
and note that you want to integrate along the length or height of the frustum to get the total extension. So integrating both sides, where on the right you either put r in terms of x, or change integration variables and integrate wrt r (probably a bit easier).
Setting up the problem is arguably the "hard" / important bit of the problem. You seem to confuse dx and x, treat k as a constant (it won't be as de is a function of x), integrate E to get E and a few other errors/loopholes.
Firstly, the force is applied throughout the frustum, so for the thin disc you have F going upwards and downwards, just as in the complete frustum. From the point of view of the later integration problem, its a constant.
When the force is applied to the thin disc (of height dx), this will produce an extension de, so
E = F*dx / (pi*r^2*de)
where r is the radius which is a function of x, so r(x) from a..b. E and F are constants. Rearrange for
de = ... dx
and note that you want to integrate along the length or height of the frustum to get the total extension. So integrating both sides, where on the right you either put r in terms of x, or change integration variables and integrate wrt r (probably a bit easier).
Setting up the problem is arguably the "hard" / important bit of the problem. You seem to confuse dx and x, treat k as a constant (it won't be as de is a function of x), integrate E to get E and a few other errors/loopholes.
thank you so much!!! omg I got it right hahah! I used the 'harder' method cuz I didn't know how to change variables. thank you so much!!! (would you like to give some advice about the climbing wall pls...)
Reply 9
How did u chnage the variables to integrate wrt. What relationship between r and x do u have?
Thanks in advance
Thanks in advance
Reply 10
Original post by TheKnightmare24
How did u chnage the variables to integrate wrt. What relationship between r and x do u have?
Thanks in advance
Thanks in advance
Hi, as this is more 1-year-old thread, it is advisable not to post any questions in the thread. If you have any issue, please start a new thread and link to this thread (it will be closed soon).
First, it would be better that you show what you have worked to demonstrate your understanding, instead of just relying on other hints which may better benefit them instead of you.
If you have set up the integral “correctly”, I believe you would see two variables: extension in the vertical direction, say y and radius of the infinitesimal small disc, say r.
The relationship between y and r can be recognised by identifying the two-end points: y increases from 0 to L while r decreases from b to a.
If you still have a problem, start a new thread and describe what you have done.
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