The maximum tension that an inextensible ring of radius 1 m and mass density `0.1` kg `m^(-1)` can bear is 40 N. The maximum angular velocity with which it can be rotated in a circular path is

Options

A. 20 rad `s^(-1)`
B. 16 rad `s^(-1)`
C. 18 rad `s^(-1)`
D. 15 rad `s^(-1)`

Answer

A

Explanation

To find tension in the ring, let us take an arc which subtends angle `2(d theta)` at centre. Tangential components of T cancel out each other, while inward components provide the necessary centripetal force. Thus,

`2Tsin(d theta)=(dm)romega^(2)=(lambda2R.d theta)(Romega^(2))`
Here, `lambdararr` linear mass density.
For small angles, `sin d theta~~d theta`
`therefore" "2Td theta=2lambdaR^(2)omega^(2)d theta" or "T=lambdaR^(2)omega^(2)`
`omega_(max)=sqrt((T_(max))/(lambda)).(1)/(R)=sqrt((40)/(0.1))xx(1)/(1)=20"rad s"^(-1))`