- #1
bartekac
- 15
- 0
Hi,
So I am currently working on a rather simple problem of a projectile being launched by a spring at a certain angle. Ignoring friction, from conservation of energy we know that the velocity of the launched projectile would be ##v = \sqrt{\frac{kx^2}{m}}## (with ##m## being the mass of the projectile, ##k## the spring constant and ##x## the initial displacement of the spring). Then I applied simple kinematics to get that $$\text{Range} = \frac{v\cos{\alpha}}{g}\left(\sqrt{(v\sin{\alpha})^2+2gY} + v\sin{\alpha}\right )$$ (##Y## is the initial height and ##\alpha## is the angle of the launch relaive to the horizontal).
What I was slightly confused about was to what extent this model would be applicable to a real world scenario in terms of just the spring extension process. That is, if we still assume that the losses due to external friction are negligible can we be sure that the extending spring itself is going to reach the computed velocity ##v## even if there is no projectile mass launched (i.e. the only mass the spring has to accelerate is the mass of itself)?
For instance, let's say that we have a spring with a very large ##k = 10^6 ~\text{N/m}## and we compress it by ##10 ~\text{cm}##. If we take the effective mass that the spring has to push to be about ##m = 0.1 ~\text{kg}## the final velocity should be about ##316~\text{m/s}## according to the theoretical model
The question I'm asking here is whether, if we neglect external friction, can we be sure that the endpoint of the spring is going to approximately reach the computed theoretical velocity for a given mass ##m## and a given value of ##x##?
Even though it strictly follows from conservation of energy that it would be the case, I'm not sure if that's what would actually happen in a real world scenario. Would the internal non-conservative work done in the system due to internal friction be highly significant and thus make the final velocity calculation invalid?
I would appreciate any sort of feedback on this question :)
So I am currently working on a rather simple problem of a projectile being launched by a spring at a certain angle. Ignoring friction, from conservation of energy we know that the velocity of the launched projectile would be ##v = \sqrt{\frac{kx^2}{m}}## (with ##m## being the mass of the projectile, ##k## the spring constant and ##x## the initial displacement of the spring). Then I applied simple kinematics to get that $$\text{Range} = \frac{v\cos{\alpha}}{g}\left(\sqrt{(v\sin{\alpha})^2+2gY} + v\sin{\alpha}\right )$$ (##Y## is the initial height and ##\alpha## is the angle of the launch relaive to the horizontal).
What I was slightly confused about was to what extent this model would be applicable to a real world scenario in terms of just the spring extension process. That is, if we still assume that the losses due to external friction are negligible can we be sure that the extending spring itself is going to reach the computed velocity ##v## even if there is no projectile mass launched (i.e. the only mass the spring has to accelerate is the mass of itself)?
For instance, let's say that we have a spring with a very large ##k = 10^6 ~\text{N/m}## and we compress it by ##10 ~\text{cm}##. If we take the effective mass that the spring has to push to be about ##m = 0.1 ~\text{kg}## the final velocity should be about ##316~\text{m/s}## according to the theoretical model
The question I'm asking here is whether, if we neglect external friction, can we be sure that the endpoint of the spring is going to approximately reach the computed theoretical velocity for a given mass ##m## and a given value of ##x##?
Even though it strictly follows from conservation of energy that it would be the case, I'm not sure if that's what would actually happen in a real world scenario. Would the internal non-conservative work done in the system due to internal friction be highly significant and thus make the final velocity calculation invalid?
I would appreciate any sort of feedback on this question :)