Question:
A thin uniform rod of length L and certain mass is kept on a frictionless horizontal table with a massless string of length L fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O. If a horizontal impulse P is imparted to the rod at a distance x = L/n from the mid-point of the rod (see figure), then the rod and string revolve together around the point O, with the rod remaining aligned with the string. In such a case, the value of n is _____.
A thin uniform rod of length L and certain mass is kept on a frictionless horizontal table with a massless string of length L fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O. If a horizontal impulse P is imparted to the rod at a distance x = L/n from the mid-point of the rod (see figure), then the rod and string revolve together around the point O, with the rod remaining aligned with the string. In such a case, the value of n is _____.
Updated On: Jan 21, 2026
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Correct Answer: 18
Solution and Explanation
Step 1: Define Angular Impulse \( I_A \)
\[ I_A = \text{angular impulse} = \vec{p} \times \vec{r} = \rho \left( \frac{l}{2} + x \right) \]
Step 2: Express Linear Impulse \( I \)
\[ I = \Delta p \]
Since \( p = mv - 0 \), we get:
\[ p = mv \]
Step 3: Write the Expression for \( I_A \)
\[ I_A = mv \times \left( \frac{3l}{2} + x \right) \]
Step 4: Define Angular Momentum \( L \)
\[ L = \text{angular momentum} = I_0 \omega + mv \times r \]
Substituting values:
\[ L = \frac{ml^2}{12} \omega + mv \times \frac{3l}{2} \]
Step 5: Relate Angular Impulse to Change in Angular Momentum
\[ mv \left( \frac{3l}{2} + x \right) = \frac{ml^2}{12} \omega + mv \frac{3l}{2} \]
Step 6: Solve for \( v \)
\[ mvx = \frac{ml^2}{12} \omega \]
Solving for \( v \):
\[ v = \frac{3l\omega}{2} \]
Step 7: Solve for \( x \)
\[ m\omega \cdot \frac{3l}{2} x = \frac{ml^2}{12} \times \omega \]
Solving for \( x \):
\[ x = \frac{l}{18} = \frac{l}{n} \]
Final Answer:
\( n = 18 \)
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