# 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 ?

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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?

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?

##### 1 Answer

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 *upper wire*, *lower wire*, **weight**, *not mass*, of the ball, and **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

Starting with the upper wire, you have

#T_"1 v" = T_1 * cos(60^@) -># thevertical componentof#T_1#

#T_"1 h" = T_1 * sin(60^@) -># thehorizontal componentof#T_1#

Do the same for the lower wire

#T_"2 v" = T_2 * cos(60^@) -># thevertical componentof#T_2#

#T_"2 h" = T_2 * sin(60^@) -># thehorizontal componentof#T_2#

Now let's examine what forces act on the ball. **Vertically**, the ball is being acted upon by its weight,

#T_"1 v" = T_"2 v" + G" " " "color(purple)((1))#

**Horizontally**, the ball is being acted upon by

#T_"1 h" + T_"2 h" = F_c" " " "color(purple)((2))#

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

#color(blue)(F_c = (m * v^2)/r)" "# , where

The *centripetal force* here will be the **sum** of the two horizontal components of

Looking at the diagram, the radius

#r = L * sin(60^@)#

The weight of the ball will be equal to

#G = m * g#

Rearrange equations

#T_"1 v" - T_"2 v" = m * g#

#T_1 * cos(60^@) - T_2 * cos(60^@) = m * g#

#color(blue)(T_1 - T_2) = m * g * overbrace(1/cos(60^@))^(color(brown)(=2)) = color(blue)(2 * m * g)#

and

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

#color(blue)(T_1 + T_2) = (m * v^2)/L * overbrace(1/(sin(60^@))^2)^(color(brown)(=4/3)) = color(blue)(4/3 * (m * v^2)/L)#

To find the tension in the lower string,

#T_1 - T_2 = 2 * m * g#

#T_2 = "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 = color(green)("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_"centripetal" = (T_1 + T_2) * sin(60^@)#

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 = 4/3 * (m * v^2)/L#

Rearrange to solve for

#v^2 = 3/4 * L/m * (T_1 + T_2)#

Therefore,

#v = sqrt(3/4 * L/m * (T_1 + T_2))#

#v = sqrt(3/4 * "1.06 m"/(1.91 color(red)(cancel(color(black)("kg")))) * 62.5color(red)(cancel(color(black)("kg"))) "m s"^(-2))#

#v = color(green)("5.10 ms"^(-1))#