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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 is 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 α is Defined by the Relation J = –α θ, Where J is the Restoring Couple and θ the Angle of Twist). - Physics

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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 is 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 α is defined by the relation = –α θ, where is the restoring couple and θ the angle of twist).

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Solution 1

Mass of the circular disc, m = 10 kg

Radius of the disc, r = 15 cm = 0.15 m

The torsional oscillations of the disc has a time period, T = 1.5 s

The moment of inertia of the disc is:

`l = 1/2 mr^2`

`= 1/2 xx (10) xx (0.15)^2`

`=0.1125 kg m^2`

Time Period, `T = 2pi sqrt(1/alpha)`

α is the torsional constant.

`alpha = (4pi^2l)/T^2`

`= (4xx(pi)^2xx0.1125)/(1.5)^2`

= 1.972 Nm/rad

Hence, the torsional spring constant of the wire is 1.972 Nm rad–1.


Solution 2

`T = 2pi sqrt(I/alpha) or T^2 = (4pi^2I)/alpha`

or `alpha = (4pi^2I)/T^2 or alpha = (4pi^2)/T^2(1/2 MR^2)`

or `alpha = (2pi^2MR^2)/T^2`

or `alpha = (2(3.14)^2 xx 10xx (0.15)^2)/(1.5)^2 "Nm  rad"^(-1)`

= `1.97 " Nm rad"^(-1)`

Concept: Force Law for Simple Harmonic Motion
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NCERT Class 11 Physics
Chapter 14 Oscillations
Q 23 | Page 361
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