Little confusion regarding centripetal force in vertical circle

[exuberant.me]

consider a pendulum. The mass 'm' is hung and now we are interested in finding the velocity so that it completes one circle. Clearly we can do it easily by conserving energy.

Now my problem is with the top most point.
Clearly the tension is minimum at this point so that string becomes slack.
The forces acting on the mass is mg downwards, Tension downwards
and since its in circular motion so the centripetal force acts towards the center.
"How then does the particle not fall", is what my confusion is.
since all the forces are downwards.
However, i know this is completely untrue. The centripetal force equals mg + T acting on the mass but what about the "direction" it should act in outward direction but as i have read and know that centripetal force acts towards the center. Clear it please.
Thanks in advance!

 

[Redbelly98]

Minimum tension does not mean the string becomes slack. That would require zero tension (or perhaps coming up with a negative value of tension, which is impossible, when solving a problem). Minimum tension does not have to be zero, and the string does not become slack as long as there is some tension.

You're correct that the net force must be downward when the object is at the top of the circle.

 

[exuberant.me]

oooo yes i got it
that means its just the component of mg that keeps the string tight and at the top most point this component of mg is 0 right...!
Thank u!

 

[tiny-tim]

hi exuberant.me!

Clearly the tension is minimum at this point so that string becomes slack.
The forces acting on the mass is mg downwards, Tension downwards
and since its in circular motion so the centripetal force acts towards the center.
"How then does the particle not fall", is what my confusion is.
since all the forces are downwards.

yes, the only force is downward, and therefore the acceleration is downward

(vector) acceleration is rate of change of (vector) velocity, so the velocity downward-component is increasing (from zero) …

but the velocity sideways-component stays the same, so the velocity becomes slightly sloping downward, not 100% downward

however, the above reasoning is irrelevant, since (if there is exactly enough energy to just reach the vertical) the string actually becomes slack before it reaches the vertical ​

 

[exuberant.me]

But where does this outward velocity come from??
Is it imaginary or what is it that keeps the string straight and not slack?

 

[tiny-tim]

sorry … i wasn't clear

if the mass reaches the top with non-zero speed, then it will keep going even if the tension is zero (which however you could only achieve by shortening the string)

in fact the mass won't reach the top (if the string stays the same length) because it will have started following a parabola (with slack string) earlier

 

[Redbelly98]

oooo yes i got it
that means its just the component of mg that keeps the string tight and at the top most point this component of mg is 0 right...!
Thank u!

Not sure I understand your question here. The force of gravity is always mg, directed downward. And mg can't be zero, ever, so long as m is not zero.

 

[nasu]

The forces acting on the mass is mg downwards, Tension downwards
and since its in circular motion so the centripetal force acts towards the center.
"How then does the particle not fall", is what my confusion is.
since all the forces are downwards.

The sum of tension an weight IS the centripetal force.
There is no third force.
"Centripetal" is called any force or force resultant which produces the centripetal acceleration.

 

[InfinitySucks]

Velocity at top of the circle must be at least

v_t = √(gR)

R is the radius of the circle.

To make a complete circle without falling off the top, velocity at the bottom of the circle must be at least

v_b = √(5gR).

Another way to look at it is

v_b = √(5) * v_t

All these you can find from centripetal force and conservation of energy.