A small block with mass m rests on a frictionless horizontal tabletop a distance
ID: 1906606 • Letter: A
Question
A small block with mass m rests on a frictionless horizontal tabletop a distance r from a hole in the center of the table. A string tied to the small block passes down through the hole, and a larger block with mass M is suspended from the free end of the string. The small block is set into uniform circular motion with radius r and speed v. v was found to be v=sqrt(m2gr/m1)
if block m2 is pulled down, so small block is rotating at r2, what is its new velocity?
i think the system is coserved and no external forces are acting on system. can I use conservaition of angulat momentum here? I.net = I.1 + I.2 and I?1=I?2 -->
?2 = I?1/I2 --> plug in v1 -->
?2 = ( (m1r^2)sqrt(m2gr/m1) ) / (m2r^2)
is this right or am I totally confused? please help
could you figure out the net work done by pulling the big block down too?
Explanation / Answer
F(inward) = Mg F(outward) = mv^2/r For no motion, Mg = mv^2/r v = sqrt(Mgr/m)