Could you please explain why in vertical circular motion , the tension is zero at the highest point of the motion . What is the logic behind it ?

Also , I get very confused in vertical circular motion in the presence of vertical and horizontal Electric field . Please explain what is happening in it .

Question

Could you please explain why in vertical circular motion , the tension is zero at the highest point of the motion . What is the logic behind it ?

Also , I get very confused in vertical circular motion in the presence of vertical and horizontal Electric field . Please explain what is happening in it .

anonymous · Answer

@ProfBrainstorm   , Could you please help me out ?

anonymous · Answer

who said the tension is zero? 

if u take a stone tied to a string and whirl it very fast.. i bet you, you ll see that string extremely tensed..  :P

P.S. But i know what you mean.. so i want you rephrase your question.. cause then you ll get some insight ;-)

anonymous · Answer

It's zero if the centripetal force at the top is the same as the weight of the object, which means that the speed of the object at the top is in its minimum possible value: $\sqrt{rg}$. That assumption is valid when we think about it going from the bottom to the top.

Suppose we go from the top to the bottom. 

At the bottom in addition to its weight, the centripetal force that makes it follow the circular path is providing an extra tension. At the top, the tension is no longer zero, but the centripetal force making it follow this path. That's because when going up the ball is moving against $g$ while going down is the other way around and notice also that the tangential speeds are not the same at the top and the bottom, as expected.

You can apply some principles of conservation of energy to find the tension:

$b = bottom$
$t = top$

$\large{KE_{b} = KE_{t} + PE_{t}}$

$\large{\frac{1}{2}mv_{b}^2 = \frac{1}{2}mv_{t}^2 + mg(2r)}$

$\large{\frac{mv_{b}^2}{r} = \frac{mv_{t}^2}{r} + 4mg}$ <-- Multiply both side by $\large{\frac{2}{r}}$. Look that $F_{c(bottom)}$ is on the right-hand side.

$\large{\frac{mv_{b}^2}{r} + mg = \frac{mv_{t}^2}{r} + 5mg}$ <-- Add $w$ to both sides in order to get $T_b$ on the right.

$\large{T_{b} = T_{t} + 5mg}$ <-- Note that $\large{\frac{mv_{t}^2}{r} = T_{t}}$.

$\large{T_{t} = T_{b} - 5mg}$

anonymous · Answer

@Joq  Sigh .. I am looking at what you have done and understood steps individually ..but i am not able to connect everything ..:(
@Mashy  I am very confused ..

anonymous · Answer

@joq , in your second paragraph ..could you explain it more in detail and from scratch ..?
I am not able to conclude anything from it :(

anonymous · Answer

Hi,

Mmm I wasn’t clear enough in my explanation and I think I forgot something at the end of the equation. I willI try to be much clearer and correct that.

Suppose you have a ball tied up to a rope and is being hung down with no motion (neither pendular nor by pulling upward/downward from the rope). The only tension acting along the rope is the weight of the ball, as expected since the ball will be at the bottom.

You can have less tension if you accelerate your hand in the direction of the weight (have you ever heard about the elevator problem? It's the same thing). When the ball is in circular, vertical motion and it’s at the top, the weight is ‘pushing’ the rope downward. And pushing along a rope is the same as subtracting tension from it. If the tangential velocity at the top isn’t enough, the ball will fall and won’t complete a circular path.
 
At the top you have to find a way to cancel the weight.  That will make the tension to be zero, but it allows the ball to follow a circular path. So we need to find a centripetal force that allows the ball to do so:

We now that  $T_t = F_{c(top)} – w$; And that $T_t = 0$, so

$F_{c(top)} – w =0$

$F_{c(top)}=w$

$\frac{mv^2}{r}= mg$

$\frac{v^2}{r} = g$

$v= \sqrt{rg}$

We’ve found the minimum velocity that allows the ball to follow a circular path. If the velocity is less than that, the ball will fall before reaching the top. 

But, what will the tension be at the bottom when coming from the top at $v_t$? At the beginning we had a $T_b=w$ at the bottom, but now there will be a centripetal force adding tension to the rope and $g$ will increase the velocity too ($F_c$ is toward the center of rotation but by Newton’s Third Law there is outward force along the rope). So the tension at the bottom is:

$T_b= F_{c(bottom)} + w$

We need a way to relate $F_{c(bottom)}$ with $F_{c(top)}$, in order to relate $T_b$ with $T_t$. Here we can apply the principle of conservation of energy. Wwhat I’ve written in the comment above would be okey if it weren’t by the mistake I made (I came up with $T_t$ from nothing).

In a comment below I will fixed that, giving the correct formula relating $T_t$ with $T_b$.

anonymous · Answer

First we know that: $T_t = F_{c(top)} - mg \hspace{35pt}(1)$ $T_b= F_{c(bottom)} + mg$ Then: $KE_t + PE_t = KE_b$ $\frac{1}{2}mv_t^2 + mg(2r) = \frac{1}{2}mv_b^2$ $\frac{mv_t^2}{r} + 4mg = \frac{ mv_b^2}{r}$ <-- Multiply by $\frac{2}{r}$. $\frac{mv_t^2}{r} + 5mg = \frac{ mv_b^2}{r} +mg$ <-- Add $mg$ to both sides and write in terms of $F_c$. $F_{c(top)} + 5mg = F_{c(bottom)}+ mg \hspace{35pt}(2)$ We know $T_t = F_{c(top)} –mg$ <-- Then solving for $F_{c(top)}$ yields: $F_{c(top)} = T_t + mg$ <-- Plug this into equation $(2)$: $T_t + 6mg = F_{c(bottom)}+ mg$ $T_t = F_{c(bottom)} - 5mg$ $T_b = F_{c(top)} + 5mg \hspace{35pt} (3)$ Take a look at the last two equations. $5mg$ is an apparent weight relative to the rope that arises because the ball is not stationary (as in the elevator problem)but going upward/downward all the time. As long as the the velocity at the top is $\sqrt{rg}$ we’ll know that $T_t$ will be zero and at the bottom it will be $F_{c(top)} + 5mg$. The neat thing is that we can make $T_t > 0$ by making $v_t > \sqrt{rg}$. But as you can see see, $F_{c(top)}$ will increase, making $T_b$ even larger. i.e, the equations are useful even when the tension at the top is not zero. By increasing the velocity at any point (in this particular case, at the top or at the bottom), the tension up and down is increased as well. If you want $T_t$ and $T_b$ in the same equation, you can combine equations $(1)$ and $(3)$: $T_t = F_{c(top)} - mg \hspace{35pt}(1)$ $\hspace{35pt}$ $T_b = F_{c(top)} + 5mg \hspace{35pt}(3)$ $F_{c(top)} = T_t + mg \hspace{35pt}(1)$ $\hspace{35pt}$ $T_b = F_{c(top)} + 5mg \hspace{35pt}(3)$ $$T_b = T_t + 6mg$$ and $$T_t = T_b - 6mg$$

anonymous · Answer

I hope this seems much clearer. I imagine that this doesn't help you enough in trying to understand vertical motion in the presence of electric fields. I would to help you with that, but honestly I've not learned the part of electromagnetism yet (:

anonymous · Answer

for the second part of your question, do you have a specific example in mind, that would make it easier to know what you are thinking about