Ball and chain projected up a rough inclined plane
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- เผยแพร่เมื่อ 8 ก.ย. 2024
- A long chain is coiled up at the bottom of a rough inclined plane with a ball attached to one end. The chain has linear density λ, the ball has mass m, the plane makes an angle θ with the horizontal, and the coefficient of friction between the ball and chain and the plane is μ. The ball is projected up the slope with initial velocity u. In this video we find the maximum distance up the slope reached by the ball. A difficult Physics problem in which energy is not conserved!
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About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
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Note: I forgot to draw the friction arrow on the diagram, but don’t worry, this doesn’t affect any of the working as I did include the frictional term in the equation of motion. For completeness there should of course be an arrow labelled μN pointing down the plane!
Brilliant problem as usual! Seems like a lot of the chain problems give rises to Bernoulli DEs. I wish in DE courses could mention them, otherwise these strange forms of DEs often seem to come out of nowhere
Thanks! That's true, I suppose it's because if the chain is uncoiling, we always get that quadratic velocity term from the product rule when differentiating the momentum, and because we can always write the acceleration as ½dv²/dx.
cool, what about a truck holding of a coiled chain that's falling and depleting over time with the case of a truck holds at some velocity or the second case where the truck initially at some velocity (so the depletion of the chains out from the trunk going to increase the truck velocity somehow).
and the last thing the idea about this is probably a ball and a chain act like a pendulum, but at first the chains are loose until it hits the level where the chain acts as a *non* massless string that gives tension.
Thanks for the suggestions!
Fantastic stuff, thanks Ben❤
Hello ! This is a very cool problem ! But i didn't understand at 7:21, how do you project to get a cos ? Is it that when you project on the perpendicular axis you get a cos and when you project on the parallel axis you get a sin ? Because I've never learned this way of projection.
Imagine a right angled triangle whose hypotenuse is the W arrow on the diagram, with one side in the direction of the dashed red line and the other side parallel to the plane. You can then see that the side in the direction of the dashed line is adjacent with respect to θ, and it follows from trigonometry that cosθ = adj/hyp = (perpendicular component)/W. For the parallel component, you're then interested in the opposite side of the triangle instead of the adjacent, so sinθ is the relevant trig function.
Glorious!
When integrating 2l/(m+lx)dx, shouldn't the answer be 2l*ln(m+lx), Dr? Cos l (lambda) is a constant
If you differentiate 2ln(m+λx) you get [2/(m+λx)]×λ, where the extra λ comes from applying the chain rule. For the integral you can just use the same argument in reverse, so the λ on the numerator disappears. That's how I usually do this sort of integral, but to see it more explicitly you can also try integrating using the substitution y = m+λx so that dy = λdx.