An annular ring of internal and outer radii `r` and `R` respectively oscillates in a vertical plane about a horizontal axis perpendicular to its plane and passing through a point on its outer edge. Calculate its time period and show that the length of an equivalent simple pendulum is `(3R)/(2)` as `r rarr 0` and `2 R` as `r rarr R`.

Answer

B, C

Explanation

Mass per unit area `= (m)/(pi(R^(2) - r^(2))) = sigma ("say")`

Whole mass `m_(1) = (pi R^(2))sigma`
` = ((R^(2))/(R^(2) - r^(2)))m`
mass of cavity `m_(2)= (pir^(2))sigma`
`= ((r^(2))/(R^(2) - r^(2)))m`
`I = (3)/(2)m_(1)R^(2) - [(1)/(2)m_(2)r^(2) + m_(2)R^(2)]`
` = (3)/(2)[(R^(2))/(R^(2) - r^(2))]mR^(2) - [(1)/(2)((r^(2))/(R^(2) - r^(2)))mr^(2) + ((r^(2))/(R^(2) - r^(2)))mR^(2)]`
`= (m)/(2(R^(2) - r^(2)))[3R^(4) - r^(4) - 2r^(2)R^(2)]`
`= m((3R^(2) + r^(2)))/(2)`
Now, `T = 2pi sqrt((I)/(mgl))` ...(i)
Here, `l = R`
`:. (I)/(mR) = (3R)/(2) + (r^(2))/(2R)`
Substituting in Eq. (i) we have,
`T = 2pi sqrt (((3R)/(2) + (r^(2))/(2R))/(g))`
Comparing with `T = 2pi sqrt ((l)/(g))`
`l` of pendulum `= (3R)/(2) + (r^(2))/(2R)`
`l = (3R)/(2)` as `r rarr 0`
and `l = 2R` as `r rarrR`.