A sphere of radius 0.1 m and mass `8pi` kg is attached to the lower end of a steel wire of length Smand diameter `10^(-3)` m. The wire is suspended from 5.22 m high ceiling of a room. When the sphere is made to swing as a simple pendulum, it just grazes the floor at its lowest point. Calculate the speed of the sphere at the lowest position.
`Y_("steel")= 1.994 xx 10^(11)" Nm^(-2)`?

Options

A. `8.5 ms^(-1)`
B. `7.7 ms^(-1)`
C. `8.0 ms^(-1)`
D. `6.6 ms^(-1)`

Answer

A

Explanation

Let `Delta l` = extension of wire when the sphere is at mean position.
r= 0.1 m (of sphere )
`l +Delta l + 2r = 5.22 m implies Deltal = 5.22 -5 -2 xx 0.1 = 0.2 `
Let F be the tension in the wire at mean position during oscillations then
`F = (YA Deltal)/(l) = (Y pi r^(2) Deltal)/(l) = (1.994 xx 10^(11) xx3.14 xx (0.5 xx10^(-3))^(2)xx 0.02)/(5.22)` F= 600 N
At mean position `F-mg = (mv^(2))/(R ) ` ( Hint R = 5.22 -r = 5.12 m )`
600 - 8 pi (9.8 ) = (8 pi xx v^(2))/(5.12) implies v = 8.5 ms^(-1)`