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Asked by Bindu Last Modified
Answer As an experienced tutor registered on UrbanPro, I'd be glad to assist you with this physics problem.
Given that a circular disc of mass 10 kg is suspended by a wire attached to its center, and the wire is twisted by rotating the disc and released, resulting in torsional oscillations with a period of 1.5 seconds. The radius of the disc is 15 cm.
To determine the torsional spring constant (α) of the wire, we can use the equation provided:
J=−αθJ=−αθ
Where:
The restoring couple JJ can be related to the torque acting on the disc, which is given by:
J=I⋅αJ=I⋅α
Where:
The moment of inertia of a circular disc about its center is given by:
I=12mr2I=21mr2
Where:
Given that m=10m=10 kg and r=15r=15 cm, we can calculate the moment of inertia II.
I=12×10×(0.15)2I=21×10×(0.15)2 I=0.1125 kg m2I=0.1125 kg m2
Now, substituting II into the equation for the restoring couple:
J=0.1125×αJ=0.1125×α
Since the period of torsional oscillations (TT) is related to the angular frequency (ωω) by T=2πωT=ω2π, we can find ωω:
T=2πωT=ω2π ω=2πTω=T2π ω=2π1.5ω=1.52π ω≈4.19 rad/sω≈4.19 rad/s
Now, the relation between angular frequency (ωω) and torsional spring constant (αα) is:
ω=αIω=Iα
Substituting the known values:
4.19=α0.11254.19=0.1125α
Solving for αα:
α=(4.19)2×0.1125α=(4.19)2×0.1125 α≈1.86 Nm/radα≈1.86 Nm/rad
So, the torsional spring constant of the wire is approximately 1.86 Nm/rad1.86 Nm/rad.
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