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Considering the formula for angular impulse given by: $$(\mathbf{H}_o)_1+\int_{t_1}^{t_2}\sum{M_o}\,dt=(\mathbf{H}_o)_2$$ Why the sum of torques with respect to 'O' is zero in this system?. Should not the weight of the masses in this system generate a torque that is not equal to zero?

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J. J. Gonzalez
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In most collision problems, we assume that the collision takes place over a very short period of time $\Delta t$. You are correct that there is a net torque of magnitude $M \approx |m_1 r_1 - m_2 r_2| g$ about the pivot during the collision. But because of the above assumption, the angular impulse delivered by this torque during the collision is approximately $M \Delta t$, and so it can be neglected in the equation for angular impulse that you have written above.

Michael Seifert
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  • So, are the intervals of collision 'Δt' measured from any starting point till the impact?, in this case the collision time is measured since the instant the plug is dropped at a height of 600 mm, but this 'Δt' is very small, so it makes the net torque about point 'O' (which can increase or decrease the angular momentum of the system) be discarded, resulting in a conservation of angular momentum. Is this correct? – J. J. Gonzalez Jul 02 '22 at 16:04
  • @JuanJose The interval $\Delta t$ that I'm referring to here is the brief time period in which the collision occurs. In your notation, $t_1$ would be just before the falling mass hits the block, and $t_2$ would be immediately after it has lodged itself in the block. Typically one would solve this problem in two stages: first find the speed of the falling mass right before it hits the block; then use angular momentum conservation to find the rotational speed of the arm after the falling mass has lodged in the block. – Michael Seifert Jul 02 '22 at 18:49