# A 1"kg" stone is tied to a 0.5"m" string and swung in a vertical circle. What is the tension in the string at the bottom of the loop?

Aug 13, 2017

$T \approx 59 \text{ N}$ (toward center)

#### Explanation:

Diagram:

where $\vec{T}$ is the force of tension and ${\vec{F}}_{G}$ is the force of gravity.

We have the following information:

• $\mapsto m = 1 \text{ kg}$

• $\mapsto r = 0.5 \text{ m}$

• $\mapsto g = 9.81 {\text{ m"//"s}}^{2}$

We can take inventory of the forces acting on the stone:

${F}_{\text{net}} = \sum {F}_{c} = T - {F}_{G} = m {a}_{c}$

We also have that:

${a}_{c} = {v}^{2} / r$

$\implies T - {F}_{G} = \frac{m {v}^{2}}{r}$

Since we know that ${F}_{G} = m g$, we can solve for tension as:

$\textcolor{b l u e}{T = \frac{m {v}^{2}}{r} + m g}$

We aren't given a velocity, so in order to get a numerical answer, we'll have to look to energy conservation.

$\Delta E = \Delta K + \Delta U + {E}_{\text{th}}$

In this case, at the top of the loop, the stone has both gravitational potential energy and kinetic energy.

• ${U}_{g} = m g h$

• $K = \frac{1}{2} m {v}^{2}$

At the bottom of the loop, the stone has only kinetic energy $\left(h = 0\right)$.

Since the height $h$ is equal to the diameter of the circle, we have $h = 2 r$. Therefore, we have:

$\textcolor{b l u e}{\frac{1}{2} m {v}_{\text{top"^2+mg(2r)=1/2mv_"bottom}}^{2}}$

At the top of the loop, we will assume that the velocity is the critical velocity, the minimum velocity necessary to maintain circular motion, where:

${v}_{\text{critical}} = \sqrt{r g}$

$\implies \textcolor{b l u e}{\frac{1}{2} m r g + m g \left(2 r\right) = \frac{1}{2} m {v}_{\text{bottom}}^{2}}$

We can solve for ${v}_{\text{bottom}}^{2}$:

$\implies {v}_{\text{bottom}}^{2} = \frac{2 \left(\frac{1}{2} m r g + 2 m r g\right)}{m}$

$\implies = \frac{m r g + 4 m r g}{m}$

$\implies = \textcolor{b l u e}{5 r g}$

Putting this into our equation for tension:

$T = \frac{m {v}^{2}}{r} + m g$

$\implies = \frac{m \cdot 5 r g}{r} + m g$

$\implies = 5 m g + m g$

$\implies = \textcolor{b l u e}{6 m g}$

$\implies = 6 \left(1 {\text{kg")(9.81"m"//"s}}^{2}\right)$

$\implies = 58.86 \text{ N}$

$\implies \textcolor{b l u e}{T \approx 59 \text{ N}}$ (toward center)