Why Is Tension at the Lowest Point of a Vertical Circle Significantly Higher?

[Karol]

Homework Statement

A stone tied to a rope rotates in a vertical circle. prove that the tension in the rope at the lowest point is 6 times the stone's weight bigger than at the highest point.

Homework Equations

Potential energy: $E_P=mgh$
Kinetic energy: $E_K=\frac{1}{2}mV^2$
Radial force: $F_R=m\frac{V^2}{R}$

The Attempt at a Solution

V₀ is the velocity at the top and V₂ is at the bottom and R is the radius.
$$\frac{1}{2}mV_0^2=\frac{1}{2}mV_2^2-2Rmg \rightarrow V_2^2=V_0^2+4gR$$
The ratio of radial forces at the bottom and at the top:
$$\frac{F_B}{F_T}=\frac{\frac{V_B^2}{R}}{\frac{V_T^2}{R}}=\frac{V_B^2}{V_T^2}=\frac{V_0^2+4gR}{V_0^2}=1+\frac{4gR}{V_0^2}$$
First it includes V₀ and R, it's not fixed, and secondly it doesn't even come close to the form.
Of course i have to deduce, at the upper point, the weight of the stone from the radial force and add it at the lowest point, but my solution doesn't even come close.

 

[haruspex]

You are asked to consider the difference in the two forces, not the ratio.
Also, don't forget the force of gravity on the stone. How will that affect the two tensions?

 

[dauto]

You calculated the ratio between the radial forces. You want the ratio between the tensions. Not the same thing.

 

[haruspex]

You want the ratio between the tensions.

No, Karol does not want the ratio of the tensions. The question refers to the ratio between the stone's weight and the difference between the tensions.

 

[HalfLostAndSearching]

I would approach this problem by first understanding the forces acting on the stone in the vertical circular motion. At the top of the circle, the only force acting on the stone is its weight, which is equal to its mass times the acceleration due to gravity (F=mg). At the bottom of the circle, in addition to the weight, there is also a centripetal force acting on the stone, which is provided by the tension in the rope. This centripetal force must be equal to the mass times the centripetal acceleration (F=ma_c).

Using this information, we can set up an equation for the tension at the top and bottom of the circle:
At the top: T=m(g+a_c)
At the bottom: T=m(g-a_c)

We can see that the tension at the bottom is larger than at the top by an amount equal to 2ma_c. Since we know that the centripetal acceleration is given by a_c=\frac{V^2}{R}, we can rewrite this as:
At the top: T=m(g+\frac{V_0^2}{R})
At the bottom: T=m(g-\frac{V_2^2}{R})

Subtracting these two equations, we get:
T_{bottom}-T_{top}=m\left(g-\frac{V_2^2}{R}\right)-m\left(g+\frac{V_0^2}{R}\right)
=m\left(g-g-\frac{V_2^2}{R}-g+\frac{V_0^2}{R}\right)
=m\left(\frac{V_0^2-V_2^2}{R}\right)
=m\left(\frac{V_0^2-(V_0^2+4gR)}{R}\right)
=m\left(\frac{-4gR}{R}\right)
=-4mg

We can see that the difference in tension between the bottom and top is equal to -4mg, which means that the tension at the bottom is 4 times larger than at the top. However, we also know that the weight of the stone is equal to mg, so the tension at the bottom must be equal to the weight of the stone plus an additional force equal to 4mg. This means that the tension at the bottom is 5 times larger than the weight of the stone (4mg