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Impulse $\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}$

$$ \mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s} $$

Angular impuls $(\vec{M})$

$$ \begin{aligned} & \vec{M}_c=\int \tau d t \\\\ & =\int F \frac{L}{2} d t \\\\ & =\frac{L}{2} \int F d t=\frac{L}{2} \times J \\\\ & =\frac{0.3}{2} \times 0.2 \\\\ & =0.03 \end{aligned} $$

$\begin{aligned} & I_{\mathrm{cm}}=\frac{\mathrm{ML}^2}{12}=\frac{2 \times(0.3)^2}{12}=\frac{0.09}{6} \\\\ & \mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\right) \\\\ & 0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}\right) \\\\ & \omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}\end{aligned}$

$\theta=\omega \mathrm{t}$

$\mathrm{t}=\frac{\theta}{\omega}=\frac{\pi}{2 \times 2}=\frac{\pi}{4} \mathrm{sec}$

$X=4$19:T545, JEE Main 2024 (Online) 1st February Evening Shift Physics - Rotational Motion Question 71 Hindi Explanation

JEE Main 2024 (Online) 1st February Evening Shift Physics - Rotational Motion Question 71 Hindi Explanation

आवेग $\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}$

$$ \mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s} $$

कोणीय आवेग $(\vec{M})$

$$ \begin{aligned} &\vec{M}_c=\int \tau d t \\\\ &=\int F \frac{L}{2} d t \\\\ &=\frac{L}{2} \int F d t=\frac{L}{2} \times J \\\\ &=\frac{0.3}{2} \times 0.2 \\

Rotational Motion

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