Question

A uniform ring of mass M and radius R is spinning at angular velocity '$\omega$' about axis passing through center and perpendicular to plane. Area of cross-section of wire of ring is A (that is very small). Breaking stress of material of ring is 'S' and young's modulus is Y. Then select the correct statements.

• A. Tension in ring due to rotation is $\frac{M{w}^{2}R}{2\pi }$.
• B. Tension in ring due to rotation is $\frac{M{w}^{2}R}{4\pi }$.
• C. Maximum angular velocity of ring without referring is $\sqrt{\frac{2\pi SA}{MR}}$.
• D. Extension in length or circumference of ring is $\frac{M{w}^2{R}^2}{AY}$.

Answer 1

Answer: (A), (C), (D)

The correct options are
A Tension in ring due to rotation is $\frac{M{w}^{2}R}{2\pi }$.
C Maximum angular velocity of ring without referring is $\sqrt{\frac{2\pi SA}{MR}}$.
D Extension in length or circumference of ring is $\frac{M{w}^2{R}^2}{AY}$.


$2T\sin \theta =dm{\omega }^{2}R$

$2T\theta =dm{\omega }^{2}R$
$2T\theta =\frac{M}{2\pi }2\theta {\omega }^{2}R$

$T=\frac{M{w}^{2}R}{2\pi }\dots \left(1\right)$

$Y=\frac{Tl}{A\Delta l}\Rightarrow \Delta l=\frac{T2\pi R}{AY}=\frac{M{w}^{2}R}{\text{/}2\pi }\times \frac{\text{/}2\pi R}{AY}$

$\Delta l=\frac{M{\omega }^2{R}^2}{AY}$

By $S=\frac{T}{A}=\frac{M{\omega }^{2}R}{2\pi A}$ is ${\omega }^2=\frac{SA2\pi }{MR}$

$\omega =\sqrt{\frac{2\pi SA}{MR}}$