Question

The whole set up shown in the figure is rotating with constant angular velocity \( \oplus \) on a horizontal frictionless table then the ratio of tensions \( \frac{T_{1}}{T_{2}} \) is (Given \( \frac{\ell_{2}}{\ell_{1}}=\frac{2}{1} \) ) 

(A) \( \frac{m_{1}}{m_{2}} \) 

(B) \( \frac{\left(m_{1}+2 m_{2}\right)}{2 m_{2}} \) 

(C) \( \frac{m_{2}}{m} \) 

(D) \( \frac{\left(m_{2}+m_{1}\right)}{m_{2}} \)

Answer 1

we find the value of \(T_2\)

\(T_2 = m_2w^2l_2\)

and find \(T_1\) so

\(T_1 = m_2w^2l_2 + m_2w^2l_1\)

The we find the ratio of \(T_1\) and \(T_2\)

\(T = mg \frac{T_1}{T_2}\)

\(= \frac{m_2w^2l_1 + m_2w^2l_2}{m_2w^2l_2}\)

we can get the expression

\(\frac{T_1}{T_2} = \frac{\left[m_1 + m_2\frac{l_2}{l_1}\right]l_2}{m_2l_2}\)

\(\frac{T_1}{T_2} = \frac{(m_1 + 2m_2)}{2m_2}\)

we can say that

\(\frac{T_1}{T_2} = \frac{(m_1 + 2m_2)}{2m_2}\)