# In the figure, a 1.91 kg ball is connected by means of two massless strings, each of length L = 1.06 m, to a vertical, rotating rod. The strings are tied to the rod with separation d = 1.06 m and are taut. The tension in the upper string is 50.0 N ?

## What are (a) the tension in the lower string, (b) the magnitude of the net force Fnet on the ball, and (c) the speed of the ball? Jan 31, 2016

Here's what I got.

#### Explanation:

The key to this problem lies with breaking up the tension in the two wires into their respective vertical and horizontal components.

This will allow you to get a better image of what forces are responsible for the ball's circular motion.

So, a rough sketch of the force breakdown looks like this Here ${T}_{1}$ represents the tension in the upper wire, ${T}_{2}$ the tension in the lower wire, $G$ the weight, not mass, of the ball, and ${F}_{c}$ the centrifugal, not centripetal, force acting on the ball.

Now, because the two wires and the distance between them form an equilateral triangle, the angles shown in $\textcolor{red}{\text{red}}$ are equal to ${60}^{\circ}$.

Starting with the upper wire, you have

${T}_{\text{1 v}} = {T}_{1} \cdot \cos \left({60}^{\circ}\right) \to$ the vertical component of ${T}_{1}$

${T}_{\text{1 h}} = {T}_{1} \cdot \sin \left({60}^{\circ}\right) \to$ the horizontal component of ${T}_{1}$

Do the same for the lower wire

${T}_{\text{2 v}} = {T}_{2} \cdot \cos \left({60}^{\circ}\right) \to$ the vertical component of ${T}_{2}$

${T}_{\text{2 h}} = {T}_{2} \cdot \sin \left({60}^{\circ}\right) \to$ the horizontal component of ${T}_{2}$

Now let's examine what forces act on the ball. Vertically, the ball is being acted upon by its weight, $G$, ${T}_{\text{2 v}}$, and ${T}_{\text{1 v}}$. This means that you can write

${T}_{\text{1 v" = T_"2 v" + G" " " }} \textcolor{p u r p \le}{\left(1\right)}$

Horizontally, the ball is being acted upon by ${T}_{\text{1 h}}$ and ${T}_{\text{2 h}}$ and ${F}_{c}$. You can thus say that

${T}_{\text{1 h" + T_"2 h" = F_c" " " }} \textcolor{p u r p \le}{\left(2\right)}$

Now, the centrifugal force acting on the ball takes the form

$\textcolor{b l u e}{{F}_{c} = \frac{m \cdot {v}^{2}}{r}} \text{ }$, where

$m$ - the mass of the ball
$v$ - its velocity
$r$ - the radius of the circle described by the ball's motion

The centripetal force here will be the sum of the two horizontal components of ${T}_{1}$ and ${T}_{2}$.

Looking at the diagram, the radius $r$ can be written as

$r = L \cdot \sin \left({60}^{\circ}\right)$

The weight of the ball will be equal to

$G = m \cdot g$

Rearrange equations $\textcolor{p u r p \le}{\left(1\right)}$ and $\textcolor{p u r p \le}{\left(2\right)}$ to get

${T}_{\text{1 v" - T_"2 v}} = m \cdot g$

${T}_{1} \cdot \cos \left({60}^{\circ}\right) - {T}_{2} \cdot \cos \left({60}^{\circ}\right) = m \cdot g$

$\textcolor{b l u e}{{T}_{1} - {T}_{2}} = m \cdot g \cdot {\overbrace{\frac{1}{\cos} \left({60}^{\circ}\right)}}^{\textcolor{b r o w n}{= 2}} = \textcolor{b l u e}{2 \cdot m \cdot g}$

and

T_1 * sin(60^@) + T_2 * sin(60^@) = (m * v^2)/(L * sin(60^@)

$\textcolor{b l u e}{{T}_{1} + {T}_{2}} = \frac{m \cdot {v}^{2}}{L} \cdot {\overbrace{\frac{1}{\sin \left({60}^{\circ}\right)} ^ 2}}^{\textcolor{b r o w n}{= \frac{4}{3}}} = \textcolor{b l u e}{\frac{4}{3} \cdot \frac{m \cdot {v}^{2}}{L}}$

To find the tension in the lower string, ${T}_{2}$, use

${T}_{1} - {T}_{2} = 2 \cdot m \cdot g$

${T}_{2} = {\text{50.0 N" - 2 * "1.91 kg" * 9.81"m s}}^{- 2}$

T_2 = "50.0 N" - 37.47 color(white)(a)overbrace("kg m s"^(-2))^(color(brown)("=N"))

${T}_{2} = \textcolor{g r e e n}{\text{12.5 N}}$

Now, for part (b) you are asked to find the magnitude of the net force that is acting on the ball. Think about the ball's movement.

It's not moving up or down, it is only moving in a circle in the horizontal plane.

Now, all objects that are moving in a circular path are experiencing an acceleration pointed towards the center of that path. This of course implies that objects moving in a circle are being acted upon by an inward force that causes that acceleration.

In other words, objects moving in a circle experience an inward force called the centripetal force that is responsible for keeping them moving along their orbit.

Since we've established that your object is only moving in a circular motion along the horizontal plane, the centripetal force will be the net force acting upon it.

In this case, we know that the centripetal force is equal to the sum of the horizontal components of the two tensions

${F}_{\text{centripetal}} = \left({T}_{1} + {T}_{2}\right) \cdot \sin \left({60}^{\circ}\right)$

This will be equal to

F_"centripetal" = ("50.0 N" + "12.5 N") * sqrt(3)/2 = color(green)("54.1 N")

Finally, the speed of the ball will be given by

${T}_{1} + {T}_{2} = \frac{4}{3} \cdot \frac{m \cdot {v}^{2}}{L}$

Rearrange to solve for $V$

${v}^{2} = \frac{3}{4} \cdot \frac{L}{m} \cdot \left({T}_{1} + {T}_{2}\right)$

Therefore,

$v = \sqrt{\frac{3}{4} \cdot \frac{L}{m} \cdot \left({T}_{1} + {T}_{2}\right)}$

$v = \sqrt{\frac{3}{4} \cdot {\text{1.06 m"/(1.91 color(red)(cancel(color(black)("kg")))) * 62.5color(red)(cancel(color(black)("kg"))) "m s}}^{- 2}}$

$v = \textcolor{g r e e n}{{\text{5.10 ms}}^{- 1}}$