Question

A small object of mass m₁ moves in a circular path of radius r on a frictionless horizontal tabletop. It is attached to a string that passes through a frictionless hole in the center of the table. A second object with a mass of m₂ is attached to the other end of the string. Derive an expression for r in terms of m₁, m₂, and the time T for one revolution. (Use any variable or symbol stated above along with the following as necessary: g.)

r =

Answer 1

The acceleration of the \(\mathrm{m}\) ass \(\mathrm{m} 2\) along \(\mathrm{x}\) -axis is the centripetal acceleration \(\mathrm{a}_{\mathrm{c}}\)

$$ \begin{array}{l} \mathrm{m}_{2} \mathrm{~g}-\mathrm{F}_{2}=0 \\ \mathrm{~F}_{2}=\mathrm{m} \mathrm{g} \ldots \ldots \ldots(1) \\ \mathrm{F}_{1}=\mathrm{m}_{1} \mathrm{a}_{\mathrm{c}} \\ \quad=\mathrm{m}_{1} \mathrm{v}^{2} / \mathrm{r} \ldots \ldots \ldots(2) \end{array} $$

From the above equations

$$ \mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{1}\left(\frac{\mathrm{v}^{2}}{\mathrm{r}}\right) $$

$$ \mathrm{m}_{2} \mathrm{~g}-\mathrm{m}_{1}\left(\frac{\mathrm{v}^{2}}{\mathrm{r}}\right)=0 \ldots \ldots \ldots .(3) $$

Now the speed of the mass \(m_{1}\) is given by relation

$$ \mathrm{v}=\frac{2 \pi \mathrm{r}}{\mathrm{T}} \ldots \ldots \ldots \ldots(4) $$

From the above equations

$$ \begin{array}{l} m_{2} g-m_{1}\left(\frac{4 \pi^{2} r^{2}}{r^{2}}\right)=0 \\ m_{2} g=m_{1}\left(\frac{4 \pi^{2} r}{T^{2}}\right) \\ r=\frac{m_{2} g T^{2}}{4 \pi^{2} m_{1}} \end{array} $$