Coin sliding on a table slides across an expanding fabric tape

Thread starter yo yo
Start date Apr 21, 2025

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The discussion focuses on calculating the exit velocity of a coin sliding across an expanding fabric tape. Initially, the coin's velocity is subtracted to determine its speed relative to the ribbon upon entry. To find the exit velocity relative to the table, the velocity of the ribbon must be added back. This process ensures accurate measurement of the coin's movement as it interacts with the expanding surface. Understanding these velocity relationships is crucial for precise calculations in this scenario.

Apr 21, 2025

yo yo

Homework Statement: why is the vector v subtracted at the beginning of the movement, and added at the departure from the fabric? I mean, it's clear why we're subtracting, but the fabric has already been transferred... An impulse? then why are we subtracting?
(the original is in Russian so Fтр is the power of friction, u отн is relative speed and u к is final speed)

Relevant Equations: idk

Last edited: Apr 21, 2025

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Apr 21, 2025

haruspex

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It is subtracted at the start to find the velocity relative to the ribbon on entry so that the exit velocity relative to the ribbon can be calculated. But to equate that to the given exit velocity, we have to get it back to being relative to the table. To do that we must add the velocity of the ribbon.

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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