A circular disc, of mass 10 kg, is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations of found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant) a is defined by the relation J = – αθ, where J is the restoring couple and θ the angle of twist).

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: JJ is the restoring couple, θθ is the angle of twist. The restoring couple JJ can be related to the torque acting on the disc, which is given by: J=I⋅αJ=I⋅α Where: II is the moment of inertia of the disc about its center. The moment of inertia of a circular disc about its center is given by: I=12mr2I=21mr2 Where: mm is the mass of the disc, rr is the radius of the disc. 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=0.1125kgm2I=0.1125kgm2 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≈4.19rad/sω≈4.19rad/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≈1.86Nm/radα≈1.86Nm/rad So, the torsional spring constant of the wire is approximately 1.86Nm/rad1.86Nm/rad. Feel free to ask if you have any questions or need further clarification! And remember, UrbanPro is a great platform for finding excellent tutors for your academic needs. read less