wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

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).

Open in App
Solution

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:

I

= × (10) × (0.15)2

= 0.1125 kg m2

Time period,

α is the torsional constant.

= 1.972 Nm/rad

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


flag
Suggest Corrections
thumbs-up
2
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Simple Harmonic Oscillation
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon