# 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.

• what is the speed of the rock when the string passes through the vertical position?
• what is the tension in the string when it makes an angle of $45$ with the vertical?
• what is the tension in the string as it passes through the vertical?

The purpose of this question is to find the speed of the rock and the tension in the string as the rock is fastened to a string to form a pendulum.

A pendulum is an object which is hung from a fixed location and can swing back and forth due to the impact of gravity. Pendulums are utilized to control clock movement since the time frame for each complete revolution, known as the period, is constant. When a pendulum is dislocated laterally from its equilibrium or resting position, it experiences a restoring force from gravity, which accelerates it back toward the equilibrium position. In other words, when it is released, the restoring force influencing its mass causes it to oscillate around the equilibrium state, swinging back and forth.

A pendulum bob moves in a circle. As a result, it is influenced by a centripetal or a center-seeking force. The tension in the string makes the bob pursue the circular path of the pendulum. The force due to gravity and the string’s tension combine to make the total force on the bob which acts on the bottom of the swing of the pendulum.

Work out the speed of the string as follows:

$mgl(1-\cos\theta)=\dfrac{1}{2}mv^2$

Or $v=\sqrt{2gl(1-\cos\theta)}$

Substitute the given values as:

$v=\sqrt{2\times 9.8\times 0.80\times (1-\cos45^\circ)}$

$v=2.14\,m/s$

Now, work out the tension in the string making an angle of $45^\circ$ with the vertical:

$T-mg\cos\theta=0$

$T=mg\cos\theta$

$T=0.12 \times 9.8 \times \cos45^\circ=0.83\,N$

Finally, the tension in the string when it passes through the vertical is:

$T-mg=\dfrac{mv^2}{r}$

$T=mg+\dfrac{mv^2}{r}$

Here $r$ is the radius of the circular path and equals the string’s length. So substituting the values:

$T=(0.12)(9.8)+\dfrac{(0.12)(9.8)^2}{(0.80)}$

$T=1.86\,N$

## Example

The oscillation period of a simple pendulum is $0.3\,s$ with $g=9.8\,m/s^2$. Find the length of its string.

### Solution

The period of the simple pendulum is given by:

$T=2\pi\sqrt{\dfrac{l}{g}}$

Where $l$ is the length and $g$ is the gravity. Now, squaring both sides:

$T^2=\dfrac{4\pi^2l}{g}$

Solve the above equation for $l$:

Or $l=\dfrac{gT^2}{4\pi^2}$

$l=\dfrac{9.8\times (0.3)^2}{4\pi^2}$

$l=\dfrac{0.882}{4\pi^2}$

$l=0.02\,m$