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In the diagrma, a particle (A, mass 0.6kg) is moving in a vertical circle. The question is: When it gets to the lowest point, what is the tension in the light rod that is between the center of the circle (B) and the mass A?

At that point the particle is not moving vertically, so why doesn't the tension = the weight (0.6g)? The tension in the rod is providing the centripetal force so it's not like the tension and the centripetal force have to equal 0.6g?

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You can look up D'Alembert forces and see that the vertical acceleration is equivalent to a downwards force. –  ja72 Apr 12 '13 at 19:29
Is there sufficient information to solve this problem? Does the particle just make it over the top, or is the thing doing a gazillion RPM? It matters... –  DJohnM Jun 11 '13 at 22:09

2 Answers 2

Assume that at the moment of interest (mass at the very bottom of its circular path), the mass is travelling at a velocity $V (m/s)$.

Then the total force acting on the mass, at that moment, $F_{Total}$, must be sufficient to keep the mass moving in a circle, radius $r$;$$F_{Total}=\frac {m V^2}{r}=\frac {0.6 V^2}{0.5}=1.2V^2$$and this force, at that moment, must be acting vertically upward.

What are the forces acting that make up this total? The force of gravity, $0.6 g$, acts downward, and the tension in the stick, $F_{Tension}$, acts upward; so $$F_{Total}=F_{Tension}+(-0.6g)$$ $$F_{Tension}=F_{Total}+0.6g$$ $$F_{Tension}=1.2V^2+0.6g$$

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It's more becouse you still need the centripedal force to change the velocity vector of the mass A to keep it int the circle trajectory.

The correct dinamical equation is:

$ m\frac{V^2}{r} = T - mg \quad \quad (1) $

Thats why the tension $T$ can't be equal to the weigh $mg$.


If $T = mg$ then $m\frac{V^2}{r}=0$ due to equation $(1)$ then there ara 2 possible situations (if there is mass):

  • $V=0$, and the object is not moving
  • $r=\infty$ so there is no circle and the object is moving in a straigh line.

Both situations contradict your system, so the assumption that $T = mg$ must be false.

It is a common misconception that the tension should be the centripedal force, but that's not right, as all forces normal to the movement conribute to the centripedal force. If the sistem wasn't vertical, yes, the centripedal force would be only tension. You can actually test this with a pice of robe and some weight in one end: try to spin it with different inclinations and compare the feelinf in your hand in different positions.

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Why is the tension the centripetal though? If the tension isn't the centripetal force then what is? –  Jonathan. Apr 12 '13 at 17:57
The sum of the tension and the gravitational are the centripedal. –  Angel Joaniquet Tukiainen Apr 15 '13 at 7:41

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