91Ó°ÊÓ

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

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

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

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

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

Solve further

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

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

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

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 $d=2\mathrm{Ï€¸é}$ , where R is the circle's radius. As a result, the block's speed is determined by

$v=\frac{2\mathrm{Ï€¸é}}{T}$ ………… (3)

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

$\begin{array}{rcl}T& =& \frac{2\mathrm{Ï€³¢²õ¾\pm ²Ôθ}}{v}\\ & =& \frac{2\mathrm{Ï€}(1.1m)\sin {28}^{∘}}{1.64m/s}\\ & =& 1.97s\end{array}$

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