Question

A 25 Kg solid sphere with a 20 cm radius is suspended by a vertical wire. A torque of 0.10 N-m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its period of the oscillations will be :

• A. πsecond
• B. √2π second
• C. 2π second
• D. 4π second

Answer 1

Answer: (D)

4π second

Answer 2

τ = - kθ for torisional harmonic motion.

0.1 = - k(1.0) where k is torisional constant of the wire.

k = $\frac { 1 } { 10 }$

T = 2π$\sqrt { \frac { \mathrm { I } } { \mathrm { k } } } = 2 \pi \sqrt { \frac { \frac { 2 } { 5 } \times 25 \times ( 2 ) ^ { 2 } } { 1 / 10 } }$

= 2π$\sqrt { 10 \times .2 \times .2 \times 10 }$ = 4πsecond