Chain dropped onto scales: finding the effective mass

Рет қаралды 827

7 ай бұрын

A chain of mass M and length l hangs vertically with its lower end a height h above a set of scales. In this video we find the mass reading on the scales as a function of time after the chain is released from rest. There are two contributions to the reading: the actual mass of the part of the chain currently on the scales, and the force the chain exerts on the scales as it is brought to rest.
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.
My website: benyelverton.com/
#physics #mathematics #dynamics #mechanics #chain #momentum #impulse #calculus #force #newtonslaws #suvat #acceleration #infinitesimal #differentiation #scales #mass #gravity #physicsproblems #maths #math #science #education

Пікірлер: 20

@MissPiggyM976 7 ай бұрын

Well done, thanks!

@DrBenYelverton 7 ай бұрын

Thank you!

@elibrahimi1169 7 ай бұрын

shouldn't you use the product rule for dp/dt since v is not a constant ?

@DrBenYelverton 7 ай бұрын

Good question, I can see why it looks like that mathematically. Let's write it as Δp/Δt, which is Δ(dm×v)/Δt. Note that p is not mv, it's dm×v because the mass is already infinitesimal. Then using the product rule Δ(dm×v)/Δt = dmΔv/Δt + vΔ(dm)/Δt. The second term is in fact zero because dm isn't changing, it's just the fixed mass of our chain element. Finally, Δv has magnitude v, because the velocity of the element reduces from v to zero in time Δt. Taking the limit Δt -> dt gives F = vdm/dt.

@elibrahimi1169 7 ай бұрын

@@DrBenYelverton oh so it's like cross term differentials being negligeable next to ordinary differentials , thank you sir Ben , i get it now

@DrBenYelverton 7 ай бұрын

I suppose the end result is the same but this is slightly different because the second term is actually identically zero - because dm is a fixed value and doesn't change - rather than just negligibly small.

@elibrahimi1169 7 ай бұрын

@@DrBenYelverton i see, ty

@hmu2816 7 ай бұрын

Thanks for the amazing content. It's educational and insightful. I'd appreciate if you could elaborate on the dx/dt construct. When it comes to impact, I was thinking of dt as the interval of time between the moment of impact and the object coming to rest. But we don't explicitly have the dt information in the problem. It seems that the construct implies that the 'last' dt prior to the impact is equivalent to the dt required to bring the object to rest. Could you please explain the intuition behind the "dx/dt" and why this is exactly the same as the velocity of the falling object prior to impact and not some 'other' velocity? Thanks!

@DrBenYelverton 7 ай бұрын

Thanks for your kind words! First, from a purely mathematical perspective, note that we derived x = ½gt² earlier, so dx/dt has to be gt. But gt is the same as v, so if the derivation makes sense up to that point, you're kind of forced to accept that dx/dt is just the velocity of the chain at the moment before impact! However, for a more intuitive way of thinking about it, note that the chain has total instantaneous momentum Mv, and therefore momentum per unit length Mv/l. Now, any part of the chain that hits the scales has its momentum reduced to zero immediately, so the total change in momentum per unit time is (momentum per unit length) × (length of chain hitting scales per unit time). That second term is simply the velocity of the chain! This way you can derive the same expression for F that we got in the video, without ever having to worry about infinitesimal quantities.

@hmu2816 7 ай бұрын

Awesome. Got it. Thanks!

@DrBenYelverton 7 ай бұрын

One point to add for completeness - I said that the total momentum was Mv, but this is only true when the entire chain is still in freefall. However, the momentum per unit length is still Mv/l even after the chain starts hitting the scales, because the mass of chain in freefall is proportional to the length of chain in freefall (i.e. they decrease by the same proportion so that their ratio is constant and equal to M/l).

@hmu2816 7 ай бұрын

Thanks for the follow up. Very clear.