A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 2° and time period 2 s. Find (a) the radius of the circular wire, (b) the speed of the particle farthest away from the point of suspension as it goes through its mean position, (c) the acceleration of this particle as it goes through its mean position and (d) the acceleration of this particle when it is at an extreme position. Take g = π^{2} m s^{-1}.

(a) 50 cm

(b) 11 cm s^{-1}

(c) 1.2 cm s^{-2} towards the point of suspension

(d) 34 cm s^{-2} towards the mean position

Given θ = 2°

Timeperiod = 2sec

a) Time period T= 2π

I =

=

= 2

Substitute I value in time period formula

T= 2π

2 = 2π

R =

b) From law of conservation of energy

Δ KE = Δ PE

- 0 = mgr(1-cosθ)

=

= = 0.11 m/sec or 11cm/sec

c) Accelaration of the particle is due to centripetal force on the circle.

a= = = 0.012m/ here distance between point of suspension and that particle is diameter of the circle 2r and v is obtained in the previous problem.

d) Accelaration of particle when it is at extreme position,

Given g=

T = 2sec

T = therefore ω = π /sec

Angular accelaration α = = × 2× π /180 =

Linear accelaration a = α(2r) = × 200 = 34 cm/

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