0
$\begingroup$

I was reading vertical circular motion, and found that the minimum velocity required to loop the loop is √5gl where l is the length of the string/radius of the vertical circle. However when I also read this

If the velocity given at lowest position v = √4gl,, the bob will be able to reach the highest point but the tension in the thread becomes zero before reaching the top position and afterward it will no longer be in circular motion. As soon as the thread will slack at an angle θ = cos^-1(-2/3) particle becomes free to move and it will follow the projectile path.

And after that , I saw some more illustrations:-

enter image description here

Now I am not able to understand where do I consider the minimum initial velocity as √5gl or √4gl for reaching the maximum position... Basically could anyone just explain me this:-

"even if normal is zero , the bead will not lose contact with the track, normal can act radially outward"

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

In cases like thread take it to be √5gl because thread remains taut only when tension in it is positive, so to complete circular motion we need to ensure that its not zero.(otherwise the thread will be tangled) In cases like rod, take it to be √4gl because it always remains taut. When the bead is attached to the ring, it cannot escape or normal cannot be zero so it is √4gl, and same is the case when you slide it between two frictionless surfaces. If instead it was on a sphere (not attached), it could escape or Normal can be zero, so √5gl. Let us also put the constraint for tension in the string not to be zero.
$T = mgcosθ + mv²/r$
let us take the tension in the string to be zero at topmost point
$v² = rg$
Using, Work Energy Theorem;
$mg2r = 1/2m(v'²-v²)$;
$v' = √5gr$;
Instead if we take a rod, for minimum velocity, velocity at top equals to zero.
$mg2r = 1/2mv'$²;
$v' = √4gr$

Improve this answer
$\endgroup$
1
  • $\begingroup$ If string slacks but not breaks at topmost position, the particle undergoes projectile motion and then little later it again pulls on the string and moves in a circle, right? For a small time/arc circular and parabolic trajectories should be indistinguishable $\endgroup$
    – Aurelius
    Commented Dec 10, 2023 at 8:13
0
$\begingroup$

The minimum speeds of $\sqrt {4gl}$ and $\sqrt {5gl}$ apply in two different situations.

If the bead is constrained to move in a vertical circle (because it is in a pipe or on a wire or attached to the centre by a rigid rod, for example) then the only condition for completing a whole circle is that its speed at the top of the circle is greater than or equal to zero. If its speed at the bottom is $v$ then we can equate KE + PE at top and bottom of the circle and we have:

$KE_{bottom} = KE_{top} + PE_{top} - PE_{bottom} \\ \displaystyle \Rightarrow \frac 1 2 mv^2 \ge 0 + 2mgl \\ \Rightarrow v \ge \sqrt{4gl}$

However, if the bead is attached to the centre of the circle by a string then its speed at the top of the circle must be sufficient to keep the string taut. If the tension in the string is $T$ and its speed at the top of the circle is $v_{top}$ then

$\displaystyle mg + T = \frac {mv_{top}^2} l$

and the condition that the string is taut i.e. $T \ge 0$ gives us

$mv_{top}^2 \ge mgl \\ \displaystyle \Rightarrow KE_{top} \ge \frac 1 2 mgl \\ \displaystyle \Rightarrow \frac 1 2 mv^2 \ge \frac 1 2 mgl + 2mgl \\ \Rightarrow v \ge \sqrt {5gl}$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Why does $\sqrt{4gl}$ not hold in both cases? If string slacks but not breaks at topmost position, the particle undergoes projectile motion and then little later it again pulls on the string and moves in a circle, right? For a small time/arc circular and parabolic trajectories should be indistinguishable. Though IDK maybe parabolic trajectory means string never becomes taut again $\endgroup$
    – Aurelius
    Commented Dec 10, 2023 at 8:12
  • $\begingroup$ @Aurelius When the bead is attached by a string it must be travelling with a speed of at least $\sqrt{gl}$ at the top of the circle otherwise the string goes slack before it reaches the top of the circle. Once the string goes slack the bead starts travelling ballistically (in a parabola) and does not follow the circular path so it never gets to the top of the circle. However, the string may become taut again at some point on the other side of the circle. $\endgroup$
    – gandalf61
    Commented Dec 10, 2023 at 10:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.