An electorn moves in a circle of radius 10cm with a constant speed of `4.0xx10^(6)ms^(-1)`.Find the electric current at a point on the circle.

Explanation

Let the position vectors of `m_(1)` and `m_(2)` be `r_(1)` and `r_(2)`.
The position vector of the center of mass, measured from the same origin, is
`R=(m_(1)r_(1)+m_(2)r_(2))/m_(1)+m_(2)`
where we have neglected the mass of the thin rod. The center of mass lies on the line joining `m_(1)` and `m_(2)`. To show this, suppose first that the tip of R dows not lie on the line, and consider the vectors `r_(1) r_(2)` from the tip of R to `m_(1)` and `m_(2)`
From the sketch we see that `r_(1) = r_(1) - R`
`r_(2) = r_(2) - R`
Using equation (1) gives `r_(1) = r_(1) - (m_(1)r_(1))/(m_(1) + m_(2)) - (m_(2)r_(2))/(m_(1)+m_(2)) = m_(2)/(m_(1) + m_(2))(r_(1) - r_(2))`
`r_(2) = r_(2) - (m_(1)r_(1))/(m_(1) + m_(2)) - (m_(2)r_ (2))/(m_(1) + m_(2)) = -(m_(1)/(m_(1) + m_(2)))(r_(1) - r_(2))`
`r_(1)` and `r_(2)` are propotional to `r_(1)-r_(2)`, the vector from `m_(1)` to `m_(2)`. Hence `r_(1)` and `r_(2)` lie along the line jpining `m_(1)` and `m_(2)` as shown.

Furthermore, `r_(1) = m_(2)/(m_(1)+m_(2))(r_(1)-r_(2)) = m_(2)/(m_(1) + m_(2))`
`r_(2) = m_(1)/(m_(1) + m_(2))(r_(1) -r_(2)) = m_(1)/(m_(1) + m_(2))l`
Assuming that friction is negligible, the external force on the baton is `F=m_(1)g + m_(2)g`
The equation of motion of the center of mass is `(m_(1) + m_(2)) R = (m_(1) + m_(2))g` or `R = g`
The center of mass follows the parabolic trajectory of single mass in a uniform gravitational field.