# A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 with the vertical. Air Resistance is negligible.

• – Calculate the velocity of the rock when the string perishes via the vertical standing.
• – Find the net tension in the line when it completes an angle of $45$ with the vertical position.
• – If the string passes through a vertical position, what will be the tension in the string?

This problem aims to find the tension in a string as it passes through vertical space. The concept required to solve this problem is related to tension and acceleration and the law of thermodynamics which includes kinetic energy and potential energy.

Let’s first look into the phrase tension, which is a force acting along the size of a medium, particularly a force held by an elastic substance, like a rope or a string. It is an action-reaction couple of forces functioning at each end of the elastic medium.

The tension $(T)$ on an elastic substance having mass $(m)$ is equivalent to the mass time gravitational force $(g)$ summed to the mass times acceleration $(a)$ expression. This whole definition can be summed up as:

$T=mg+ma$

The first law of thermodynamics, which is most commonly known as the law of conservation of energy, states that the energy possessed by a substance can neither be created nor can be eliminated but can transform from one form to another. The common expression for this is:

$K.E=P.E$

Before starting the solution, we will list out the given variables,

The length of string $l=0.8 m$,

Mass of the rock $m=0.12kg$,

And the angle with the vertical position $\theta=45^{\circ}$.

Part a:

For this part, we will have to find the rate at which the rock attached to the string passes through the vertical position. According to the first rule of thermodynamics:

$\text{Potential energy}=\text{Kinetic energy}$

Potential energy = $mgl\times (1-\cos\theta)$.

Kinetic energy = $\dfrac{1}{2}mv^2$.

Plugging them in:

$m\times g\times l\times (1-\cos\theta)=\dfrac{1}{2}\times m\times v^2$

Rearranging the above equation for calculating $v$:

$v^2=2\times g\times l\times (1-\cos\theta)$

$=2\times 9.8\times 10.8\times (1-\cos(45))=4.597$

We get,

$v=\sqrt{4.597}=2.1 ms^{-1}$

Part b:

Finding the tension in the line when it completes an angle of $45$ using the first condition of equilibrium:

$\sum{Forces}=0$

$T-m\times g\times \cos\theta=0$

$T=m\times g\times \cos\theta$

$T=0.12\times 9.81\times \cos(45)$

$T=0.83 N$

Part c:

Finally, the tension in the string as it crosses through the vertical position can be found by the first condition of equilibrium:

$T – m\times g = \dfrac{(m\times v^2)}{r}$

$T=mg(1+2(1-\cos\theta))$

$T=0.981\times 0.12 (1+ 2(1-\cos(45)))$

$T=1.867 N\space or\space 1.9 N$

## Numerical Result

Part a: $v=2.1 ms^{-1}$

Part b: $T=0.83 N$

Part c: $T=1.867 N\space or\space 1.9 N$

## Example

A $10 kg$ rock is hanging from a rope. Find the tension in the rope, if the rock:

• – experiences zero acceleration?
• – has an upward acceleration of $5 m/s^2$.
• – has a downward acceleration of $-5 m/s^2$.

To calculate tension:

$T=mg+ma$

Substituting the values:

$T=(10 kg)\times (9.8 m/s^2)+(10 kg)\times (0)$

$T=108 N$

With an upward acceleration of $5 m/s^2$, we get:

$T=(10 kg)\times (9.8 m/s^2)+(10 kg)\times (5 m/s^2)$

$T=148 N$

With a downward acceleration of $-5 m/s^2$, we get:

$T=(10 kg)\times (9.8 m/s^2)+(10 kg)\times (-5 m/s^2)$

$T=48N$