A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that in can freely oscillate in the vertical plane. A particle of move 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be 

 

4 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago

L_i = L_f
mvL = Iω
mvL = (ML³/3 + mL²)ω


Before collision
After collision
0.1 * 80 * 1 = (0.9 * 1²)/3 + 0.1 * 1²)ω
8 = (3/10 + 1/10)ω 8 = (4/10)ω ω = 20 rad/sec

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Physics System of Particles and Rotational Motion 2025

Physics System of Particles and Rotational Motion 2025

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