Two blocks are connected by a massless string that passes over a frictionless peg as shown in figure. One end of the string is attached to a mass m_1=3kg, i.e., a distance R=1.20m from the peg. The other end of the string is connected to a block of mass m_2=6kg resting on a table. From what angle theta, measured from the vertical, must the 3-kg block be released in order to just lift the 6kg block off the table?

Two blocks are connected by a massless string that passes over a frict

1.	Two blocks are connected by a massless string that passes over a frictionless peg as shown in figure. One end of the string is attached to a mass m_1=3kg, i.e., a distance R=1.20m from the peg. The other end of the string is connected to a block of mass m_2=6kg resting on a table. From what angle theta, measured from the vertical, must the 3-kg block be released in order to just lift the 6kg block off the table?
Answer» Solution :This problem involves several concepts. First we will apply conservation of energy to find the SPEED of the BLOCK `m_1` at the bottom of the circular path as a function of `theta` and the radius of the path, R. From Newton's second law we will DETERMINE the tension at the bottom of its path as function of given PARAMETERS. Finally, the block `m_2` will lift off the ground when the upward force (tension) exerted by the cord just exceeds the weight of the block. We take bottom of the circle as REFERENCE level. From conservation of energy, we have `KE_i+U_i=KE_f+U_f` `0+m_1g(R-Rcostheta)=1/2m_1v^2+0` `v^2=2gR(1-costheta)` (i) Applying Newton's second law on block of mass `m_2`, we have `sum F_n=T-m_1g=m_1v^2/R` `T=m_1g+(m_1v^2)/(R)` (ii) As the string is massless, tenstion T is constant throughout. When `m_2` just lifts off the normal reaction becomes zero. For block `m_2`, we have `T=m_2g` (iii) From Eqs. (i), (ii), and (iii), we get `m_2g=m_1g+m_1(2gR(1-costheta))/(R)` `cos theta=(3m_1-m_2)/(2m_1)=(3xx3-6)/(2xx3)=1/2` `theta=60^@`