What forces are acting on the ant as it swings on a rotating rod?

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The discussion centers on the forces acting on an ant clinging to a rod that rotates vertically. When the rod is released from a horizontal position, the ant experiences both gravitational force and friction as it swings to the lowest point. At the equilibrium position, the net force in the horizontal direction is zero, indicating that only friction and the ant's weight are acting on it. The ant also has a vertical acceleration due to its circular motion, necessitating consideration of forces in both x and y directions. Understanding these dynamics is crucial for calculating the contact force exerted by the rod on the ant.
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a thin rod is fixed on one end and rotate in a vertical motion. Its period of oscillation is pi s .
It is now held in the horizontal position and let go. An ant happen to cling into the rod at distance 1 m from the pivot. Find the contact forcce the rod exert on the ants when it swing to the lowest position. Express the mag. of the force in terms of the ant's body weight.

When it reaches the equilibrium position, the aceleration is 0. hence the net force acting on ant in the horizontal direction should be 0. Only friction and the weight of the ant act on it . Is this way of doing correct , if not , please explain to me how?
 
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Wen,
two questions before I try to answer your question.
1. Is the rod rotating like the spoke of a bicycle wheel while the wheel is held upright?
2. Have you done simple harmonic motion?
 
yes to both .
 
What forces are the ant experiencing at the bottom and in which directions are they?
 
the friction and its weight because the aceleration of both the rod and ant is 0
 
Forget about the rod, just think about the ant. The projected motion of the ant on the x-axis gives a zero acceleration for the ant in the x-direction, but we also need to consider the projection of the ant on the y-axis in order to describe it's motion fully, that is it also has an acceleration in the y-direction as it revolves around the pivot. In what direction will the friction force (called the contact force in your problem) be?
 
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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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