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  • Two identical particles, each of mass '  ' are attached to two ends of a mass

Two identical particles, each of mass ' 1 \mathrm{~kg} ' are attached to two ends of a massless rod is shown in the figure. A particle of mass ' 2 \mathrm{~kg} ' travelling with a speed of ' 1 \mathrm{~m} / \mathrm{s} ' hits one of the mass in the direction perpendicular to the rod. If the length of the rod is given ' 1 \mathrm{~m} ' then find the angular velocity of the system. (Assume 2 \mathrm{~kg} mass comes to rest after the collision]

Option: 1

2 \mathrm{rad} / \mathrm{s}


Option: 2

1 \mathrm{rad} / \mathrm{s}


Option: 3

4 \mathrm{rad} / \mathrm{s}


Option: 4

3 \mathrm{rad} / \mathrm{s}


Answers (1)

best_answer

Angular impulse = change in angular momentum [Applying it about the centre of the mass]

\begin{aligned} & 2 \times 1 \times \frac{1}{2}=\left[I_{c \cdot m}\right] \omega \\ & I_{c \cdot m}=1 \times\left(\frac{1}{2}\right)^2+1 \times\left(\frac{1}{2}\right)^2=\frac{1}{2} \mathrm{~kg}-m^2 \\ & 1=\frac{1}{2} \times \omega \\ & \omega=2 \mathrm{rad} / \mathrm{s} \end{aligned}

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Anam Khan

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