Question 12: Mass on Suspended Rod
Note: \( T_1 \) (Tension in the cord) is the same in both cases. When Torque is taken about the spring’s connection point, the only other torque is due to rod weight \( mg \), which remains the same.
1. Force Balance Equations:
Case 1 (Initial): $$ T + T_1 = mg $$ Case 2 (Spring extension doubles $\to 2T$, Load $m_0$ added): $$ 2T + T_1 = (m + m_0)g $$
Case 1 (Initial): $$ T + T_1 = mg $$ Case 2 (Spring extension doubles $\to 2T$, Load $m_0$ added): $$ 2T + T_1 = (m + m_0)g $$
2. Torque Equation (About Spring Connection):
$$ T_1 \times l = mg \frac{l}{2} $$ $$ T_1 = \frac{mg}{2} $$
$$ T_1 \times l = mg \frac{l}{2} $$ $$ T_1 = \frac{mg}{2} $$
3. Solving:
From Case 1 equation: \( T = mg – T_1 = mg – mg/2 = mg/2 \).
Substitute \( T \) and \( T_1 \) into Case 2 equation: $$ 2(mg/2) + (mg/2) = (m + m_0)g $$ $$ mg + \frac{mg}{2} = (m + m_0)g $$ $$ \frac{3}{2}mg = mg + m_0 g $$ $$ \frac{1}{2}m = m_0 $$ $$ m_0 = \frac{m}{2} $$
From Case 1 equation: \( T = mg – T_1 = mg – mg/2 = mg/2 \).
Substitute \( T \) and \( T_1 \) into Case 2 equation: $$ 2(mg/2) + (mg/2) = (m + m_0)g $$ $$ mg + \frac{mg}{2} = (m + m_0)g $$ $$ \frac{3}{2}mg = mg + m_0 g $$ $$ \frac{1}{2}m = m_0 $$ $$ m_0 = \frac{m}{2} $$