Unusual question regarding torque

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The discussion focuses on calculating the accelerations of two equivalent masses placed at two corners of an equilateral triangle, with the third corner fixed. When the system is released, gravity acts downward, affecting the motion of the masses. The torque and moment of inertia around the fixed point are essential for determining the rotational acceleration. The relationship between torque, moment of inertia, and rotational acceleration is emphasized, using the formula torque = moment of inertia × rotational acceleration. Understanding these concepts is crucial for solving the problem accurately.
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two equivalent mases, m, are placed on two corners of a triangle with 3 sides each of length a (i.e an equilateral traingle). The third corner is fixed (not free to move).
The triangle orginally starts with one side perfectly vertical. Gravity acts downward intially, calculate the accelerations of the masses right after the system is let go .
 
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Find torque and moment of inertia around the fixed point

Use torque = moment of inertioa X rotational acceleration
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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