Question

A small body of mass $m=0.1\text{Kg}$ moves in a frame, rotating about a stationary axis with a constant angular velocity $\omega =5.0\text{rad/s}$. What work (in $\text{J}$) does the centrifugal force perform during the transfer of the body along an arbitrary path from point $1$ to $2,$ which are located at distances $r_1=30\text{cm}$ and $r_2=50\text{cm}$ from the rotation axes?

Answer 1

We know that, Centrifugal force $=m{\omega }^{2}r$
Work done to displace it by a small distance $dr=m{\omega }^{2}rdr$
$\therefore$ The total work done due to centrifugal force as the particle moves from $r_1=30\text{cm}$ and $r_2=50\text{cm}\text{is}W=\underset{r_1}{\overset{r_2}{\int }}m{\omega }^{2}rdr={\left[\frac{m{\omega }^2{r}^2}{2}\right]}_{r_1}^{r_2}=\frac{m{\omega }^2}{2}[r_2^2-r_1^2]$
$=0.1\times \frac{25}{2}[{\left(\frac{50}{100}\right)}^2-{\left(\frac{30}{100}\right)}^2]=\frac{2.5}{2}[0.25-0.09]=\frac{2.5}{2}\times 0.16$
$=2.5\times 0.08=0.2\text{J}$