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Question

Find the time period of small oscillations of the following systems.

(p) A metre stick suspended through the 20 cm mark.

(q) A ring of mass m and radius r suspended through a point on its periphery.

(r) A uniform square plate of edge 'a' suspended through a corner.

(s) A uniform disc of mass m and radius r suspended through a point r2 away from the centre.

(i)2π22a3g (ii) 1.51 sec (iii) 2π3r3g (iv)T=2π2Rg


A

p - (i) ; q - (iii) ; r - (ii) ; s-(iv)

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B

p - (ii) ; q - (iv) ; r - (i) ; s-(iii)

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C

p - (ii) ; q - (i) ; r - (iii) ; s-(iv)

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D

p - (iv) ; q - (iii) ; r - (ii) ; s-(i)

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Solution

The correct option is B

p - (ii) ; q - (iv) ; r - (i) ; s-(iii)


The ruler is suspended at 20 cm mark.C.O.M is at 50cm so the C.O.M is going in a circle of 30 cm radius.

Circle of 30 cm radius so if ruler is displaced by an angleθ and left,then torque of mg will try to restore it.

Torque of mg=0.3 mg sin θ = -l α . . . (1)

Iruler=mL212+md2 (L=1m;d=30 cm)
Iruler=m[112+(30100)2]
= m (0.173)
eqn (1) becomes 0.3 mg θ=Irα
(θissmall,sinθθ)
α=0.3mgθIr(α=ω2θ)
T=2πω
=2πIr0.3mg
= 1.51 sec

(q)

If displaced by angle θ (small) restoring torque will be - mgR sin θ = lα
α=mgRθI(α=ω2θ)
ω2=mgRI
T=2π2Rg

(r)

Distance of Suspension = a2
mga2sinθ=Iα
mga2sinθ=(Iz+m(a2)2)α (Parallel axis theorem)

mga2sinθ=(Ix+Iy+m(a2)2)α (Perpendicular axis theorem)

mga2 sinθ=(ma212+ma212+m(a2)2)

mga2sinθ=2ma23

=3ga22θ

ω2=3ga22

T=2π22a3g

(s)

mgr2sinθ=I

For small θ

mgr2θ=I

mgr2θ=(Ic+m(r2)2)

mgr2θ=(mr22+m(r2)2)

mgr2θ=(3mr24)

=2g3rθ

ω2=2g3r

T=2π3r2g


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