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I have two silly queries:

  1. In case $1$,a rod $AB$ is attached to a string at point $C$. Here the weight is acting at the center of mass $O$. My question is: * Is it even possible to keep a rod hanging like this*? If it were possible,then the rod is in rotational equilibrium,but as we can see the line of action of the weight is not going through $C$. So,there should be net torque with respect to $C$. In light of this logic,how can a rod keep hanging in this position?

  2. In case $2$,we imagine a cylinder whose cross sectional area is large enough standing on ground. Now we all know the center of mass $O$ is at the center of the cylinder. As the body is in rotational equilibrium, the net torque with respect to any point should be $0$. But here, the line of action of weight that is acting at $O$ does not pass through $D$. So,there should be a net torque about $D$. But,still we don't see the cylinder falling down with respect to $D$ even if there is a net torque. Why does this happen?

I will be grateful for clearing my doubts.

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  • $\begingroup$ If you have drawn the only force on the cylinder, then it must be accelerating downward. If it is at rest, then there must be other forces on it. $\endgroup$
    – BowlOfRed
    Commented Aug 13, 2022 at 19:25

1 Answer 1

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  1. There is a net torque about the fixed point C from gravity, so the body will rotate (and oscillate) until ultimately it hangs without moving vertically where gravity provides no torque.
  2. There is a force of constraint on the object from the ground in addition to the force of gravity; the sum of the forces provides translational and rotational equilibrium.
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