Question

A small block of mass $\mathit{m}$ is attached to a spring with stiffness ${\mathit{k}}_{\mathbf{s}}$ and relaxed length$\mathit{L}$. The other end of the spring is fastened to a fixed point on a low-friction table. The block slides on the table in a circular path of radius$\mathit{R}\mathbf{>}\mathit{L}$. How long does it take for the block to go around once?

Short answer

Time taken for the block to go around once is $2\pi \sqrt{\frac{mR}{k_s(R-L)}}$.

Step-by-step solution

Identification of the given data

A small block of mass is$m$

Stiffness of spring is $k_s$

Relaxed length is$L$

Understanding the concept

The block's speed is v. The spring exerts a tension force on the block with the tension force${\mathit{F}}_{\mathbf{T}}$ in the spring being determined by

${\mathit{F}}_{\mathbf{T}}\mathbf{=}{\mathit{k}}_{\mathbf{s}}\mathit{x}$ …(i)

Where, ${\mathit{k}}_{\mathbf{s}}$denotes the spring's stiffness andxdenotes the spring's stretched length or elongation. It's equal to the difference between the spring's original length and its final length after extension, which we can figure out by

$\mathit{x}\mathbf{=}\mathit{R}\mathbf{-}\mathit{L}$

The expression will be entered into equation (i) and it will take the form

${\mathit{F}}_{\mathbf{T}}\mathbf{=}{\mathit{k}}_{\mathbf{s}}\mathbf{(}\mathit{R}\mathbf{-}\mathit{L}\mathbf{)}$ …(ii)

The rate of change is related to the perpendicular rate of change and equals the centrifugal force${\mathit{F}}_{\mathbf{c}}$ that exists due to the tension force. Thus, the tension force is provided by

${\mathit{F}}_{\mathbf{T}}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}$

Determining the block’s speed

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 $d=2\pi R$, where is the circle's radius. As a result, the block's speed is determined by

$v=\frac{2\pi R}{t}$ …(iii)

The expression of $\nu$have to be put into equation (iii), so the time is:

$$\begin{gathered} F_T=\frac{m{v}^2}{R} \\ {F}_T=\frac{m}{R}{\left(\frac{2\pi R}{t}\right)}^2 \\ t=\sqrt{\frac{{\left(2\pi \right)}^{2}mR}{F_T}}\left(\text{Solve for}t\right) \\ t=2\pi \sqrt{\frac{mR}{k_s(R-L)}}\left(F_T=k_s(R-L)\right) \end{gathered}$$

Time taken for the block to go around once is $2\pi \sqrt{\frac{mR}{k_s(R-L)}}$.