Actually you were right lol. I calculated the hanging mass when there is that much of the chain hanging. I got 2.9252kg the divided that by the entire mass and came out with the correct answer for the acceleration. Now all I need is how to find the velocity at that point.
Alrighty well thank you very much for helping. The second part of the question is asking for the velocity at that point. Rather than an answer could anyone help me with what to start with for this point seeing as how the acceleration is varying. Thanks.
Homework Statement
Given: A uniform flexible chain whose mass
is 4.1 kg and length is 5 m. A table whose top
is frictionless.
Initially you are holding the chain at rest
and one-half of the length of the chain is hung
over the edge of the table. When you let loose
of the chain it falls...
Futher insight: it is a multiple choice problem
A. None
B. (4 pi^2mR) / T^2
C. (8pi^3mR^2)/T^2
D. 2uSmgR
E. (4pi^2uSmgR)/T^2
although I have come across a thought that surprised me. Under these conditions you wouldn't be able to use uS unless it was on the breaking point between uS and...
Would be much appreciated if someone could reply to me at least with which method to use, me and a few of my friends are working and unfortunatly we all keep getting different answers.
Homework Statement
An object of mass m sits at a distance R form the axis of rotation of a record that is being played. The record makes a complete revolution in time T and friction holds the object in its place relative to the record. The coefficient of static friction between the object and...