A circular pendulum of length 1.1 mgoes around at an angle of 28 degreesto the vertical. Predict the speed of the mass at the end of the string. Also predict the period, the time it takes to go around once. Remember that the radius of the circle is the length of the string times the sine of the angle that the string makes to the vertical.

A circular pendulum of length 1.1 m goes around at an angle of 28 degrees to the vertical.

According to the diagram below the tension force F_Thas two components in xand y.

The component force F_Tsin$\mathit{\theta }$equals the centrifugal force that rotates the pendulum in the horizontal component xand is given by

${\mathit{F}}_{\mathbf{T}}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}$ ………. (1)

Also, the upward component force F_Tcos$\mathit{\theta }$in the vertical component yis in the opposite direction of the gravitational force F_g, hence it is given by

$\begin{array}{rcl}{\mathit{F}}_{\mathbf{T}}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }& \mathbf{=}& \mathit{m}\mathit{g}\\ {\mathit{F}}_{\mathbf{T}}& \mathbf{=}& \frac{\mathbf{m}\mathbf{g}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}\end{array}$

$\begin{array}{rcl}{\mathit{F}}_{\mathbf{T}}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }& \mathbf{=}& \frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}\\ \left(\frac{mg}{cos\theta }\right)\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }& \mathbf{=}& \frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}\\ {\mathit{v}}^{\mathbf{2}}& \mathbf{=}& \mathit{g}\mathit{R}\left(\frac{sin\theta }{cos\theta }\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(R=Lsin\theta \right)\\ {\mathit{v}}^{\mathbf{2}}& \mathbf{=}& \mathit{g}\left(Lsin\theta \right)\left(\frac{sin\theta }{cos\theta }\right)\end{array}$

Solve further

$\mathit{v}\mathbf{=}\sqrt{\mathbf{g}\mathbf{L}\frac{\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}}$ …….. (2)

To get the value of v, substitute of the values in equation (2)

$\begin{array}{rcl}v& =& \sqrt{\frac{gL{\sin }^2\theta }{\cos \theta }}\\ & =& \sqrt{\frac{\left(9.8m/s^2\right)\left(1.1m\right){\sin }^2{28}^{\circ }}{\cos {28}^{\circ }}}\\ & =& 1.64m/s\end{array}$

The change in distance over time is the speed. As a result of v = d / t , the block moves over the circumference of a circle when it completes one circuit. The circumference is calculated using the formula $d=2\mathrm{\pi R}$ , where R is the circle's radius. As a result, the block's speed is determined by

$v=\frac{2\mathrm{\pi R}}{T}$ ………… (3)

Where T denotes the length of one period. Let's plug the $R=L\sin \theta$expression into equation (3) and solve it for T.

$\begin{array}{rcl}T& =& \frac{2\mathrm{\pi Lsin\theta }}{v}\\ & =& \frac{2\pi (1.1m)\sin {28}^{\circ }}{1.64m/s}\\ & =& 1.97s\end{array}$

Therefore, the speed of the mass at the end of the string is 1.64 m/s and the time is 1.97 s.