JEE Advanced 2024Paper 1
- Exam
- JEE Advanced
- Year
- 2024
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Rotational Motion
JEE Advanced 2024 · Paper 1 · Physics · Rotational Motion
A thin uniform rod of length L and certain mass is kept on a...
A thin uniform rod of length and certain mass is kept on a frictionless horizontal table with a massless string of length fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point . If a horizontal impulse is imparted to the rod at a distance from the mid-point of the rod (see figure), then the rod and string revolve together around the point , with the rod remaining aligned with the string. In such a case, the value of is _______ .
Correct answer
0
Step-by-step solution
Linear impulse momentum
...(i)
Angular impulse angular momentum
\begin{aligned} & \int \mathrm{r} \times \mathrm{Fdt}=\Delta \mathrm{L} \\ & \begin{aligned} & \mathrm{r} \times \int \mathrm{Fdt}=\mathrm{I}(\omega-0), \text{ and } \mathrm{I} \text{ is moment of inertia about axis of rotation. } \\ & \begin{aligned} &\left(\mathrm{L}+\frac{\mathrm{L}}{2}+\mathrm{x}\right) \times \mathrm{P}=\left(\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2\right) \omega \\ &=\left(\frac{\mathrm{mL}^2}{12}+\mathrm{m}\left(\mathrm{L}+\frac{\mathrm{L}}{2}\right)^2\right) \omega\end{aligned} \\ &\left(\frac{3 \mathrm{~L}}{2}+\mathrm{x}\right) ^2\left(\frac{1}{12}+\left(\frac{3}{2}\right)^2\right) \omega\end{aligned}\end{aligned}\left(\frac{3 L}{2}+x\right) P=\mathrm{mL}^2\left(\frac{7}{3}\right) \omega...(ii)
Divide eq.-(i) & (ii)$