Pulling down on the string connected to a whirling block on a table

AI Thread Summary
The discussion focuses on the equations of motion for two blocks connected by a string, where block A moves in a circle and block B hangs vertically. The tension in the string is initially calculated using the equations T - m_a*g = m_a*a and T = m_b*ω²*R, but there are concerns about the assumptions made regarding tension at the moment of release. It is noted that the tension just before block B is released is different from the tension immediately after release due to the change in dynamics. Additionally, the use of polar coordinates for block A's motion is suggested for better accuracy in expressing acceleration. Clarification on the subscripts for the blocks is also recommended to avoid confusion.
farfromdaijoubu
Messages
8
Reaction score
2
Homework Statement
A horizontal frictionless table has a small hole in its centre. Block A on the table is connected to block B hanging beneath by a massless string which passes through the hole. Initially, B is held stationary and A rotates at constant radius R with constant angular velocity \omega If B is released at t=0, what is its acceleration immediately afterward?
Relevant Equations
[tex]T-m_ag = m_aa[/tex]
[tex]T=m_b\omega^2R[/tex]
At the time of release, the equation of motion of blocks A and B T-m_ag = m_aa and T=m_b\omega^2R respectively, where T is the tension in the string. Solving for the acceleration a then gives a=\frac{m_b\omega^2R - m_ag}{m_a}. Not sure what I did wrong or what incorrect assumptions I made here...

Thanks
pf1.png
 
Physics news on Phys.org
Welcome to PF!

farfromdaijoubu said:
Relevant Equations:: T-m_ag = m_aa
T=m_b\omega^2R
Block ##A## is the block that is initially moving in a circle. Block ##B## is the hanging block. It looks like your subscripts ##a## and ##b## should be switched if ##a## is meant to refer to block ##A## and ##b## to block ##B##.

A couple of things to consider:

(1) When block ##B## is released, the tension ##T## will "instantaneously" change from what it was just before ##B## was released. Just before ##B## is released, the tension is ##m_a \omega^2 R## because ##A## is moving in uniform circular motion before ##B## is released. But this will generally not be the tension just after ##B## is released.

(2) Block ##A## moves on a plane where polar coordinates would be appropriate for this problem. You might want to review how acceleration is expressed in polar coordinates. For example, see equation 4 here.
 
  • Like
Likes farfromdaijoubu and Lnewqban
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top