A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude `2^@` and time period 2s. Find a the radius of the circular wire. b. the speed of the particle farthest away from the point of suspension as it goes though its mean position c. the aceleration of this particle as ilt goes through its mean position and extreme position. Take `g=pi^2 sms^-2`

Answer

A, B, C, D

Explanation

For a compound pendulum
a. `T=2pi sqrt(I/(mgl))=2pi sqrt(I/(mgr))`
Ml of the circular wire about the poinnt of suspension is givne by
`I=mr^2+mr^2=2mr^2` is
Moment of inertia about A
`2=2pisqrt((2mr^2)/(mgr))=2pi sqrt((2r)/g)`
`rarr(2r)/g=1/pi^2`
`rarr r=g/(2pi^2)`
`=0.5m=50cm.`
`b`. `(1/2)omega^2-0=mgr(1-cos2^@)`
`rarr (1/2)2mr^2.omega^2=mgr(1-cos2^@)`
`rarr w^2-g/r(1-cos2^@)`
`w^2=0.11rad/sec`
[putting the values of g and r]
`rarr v=wxx2r=11cmsec^-1`
`c`. Acceleration at the end position will be centripetal
`a_n=omega(2r)=(0.11)xx100`
`1.2cm/s^2`
The directionof `a_n` is towards the point of suspension.
d.At the extreme position the centripetal acceleration will be zero. But the particle will still have acceleration due to the SHM. because `t=2sec`
angualr frequency.
`omega=(2pi)/T=pi=3.14`
So, angular acceleration at the extreme position,
`alpha=pi^2 2^@=pi^2xx(2pi)/180`
`=(2pi)^3/18001^@=pi/180radian]`
So, tangential acceleration
`=alpha(2r)=((2pi^3)/180)xx100`
`=34cm/s^2`