Question

A particle of mass $m_1$ is fastened to one end of a string and one of $m_2$ to the middle point, the other end of the string being fastened to a fixed point on a smooth horizontal table. The particles are then projected, so that the two portions of the string are always in the same straight line and describes horizontal circles. Find the ratio of tensions in the two parts of the string

• A. $\frac{m_1}{m_1+m_2}$
• B. $\frac{m_1+m_2}{m_1}$
• C. $\frac{2m_1+m_2}{2m_1}$
• D. $\frac{2m_1}{m_1+m_2}$

Answer 1

Answer: (C)

$\frac{2m_1+m_2}{2m_1}$

Let the angular speed of system be $\omega$, and length of string be $l$.

Tension in the other end of the string is $T_1=m_1{\omega }^{2}l+m_2{\omega }^{2}l/2$

Tension in portion between $m_1$ and $m_2$ is $T_2=m_1{\omega }^{2}l$

so, $T_1/T_2=(2m_1+m_2)/2m_1$