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Question

A circular ring of radius R and mass m made of a uniform wire of cross-sectional area A is rotated about a stationary vertical axis passing through its center and perpendicular to the plane of the ring. The breaking stress of the material of the ring is σb. Determine the maximum angular speed ωmax at which the ring may be rotated without failure.

A
πAσbmR
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B
πAσb2mR
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C
2πAσbmR
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D
4πAσbmR
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Solution

The correct option is C 2πAσbmR
Let's consider a small portion of the ring subtending an angle dθ.

The ring, as a result of rotational motion, experiences a tension F along its circumference.

The net tangential component of this tension will be 0, since tensions of Fcos(dθ2) will act in opposite directions and cancel each other out, as shown in the figure.


Therefore, the tension on the ring comes entirely from the radial components of the tension, given by

Tnet=Fsin(dθ2)+Fsin(dθ2)

Tnet=2Fsin(dθ2)2F(dθ2)

since dθ is very small.

Tnet=Fdθ

Therefore, the mass of the ring over this small component will be: dm=m×dθ2π.

The centripetal force experienced by this component of the ring on rotating it about its vertical axis with the maximum angular velocity ωmax will be:

Fcentripetal=(dm)ω2maxR

=(m×dθ2π)ω2maxR

Also, given that the breaking stress of the wire is σb.

σb=FAF=σbA

Tnet=Fdθ=(σbA)dθ

Now, the centripetal force is nothing but the radial tension of the ring:

Fcentripetal=Tnet

(m×dθ2π)ω2maxR=(σbA)dθ

mω2maxR2π=σbA

ω2max=2πσbAmR

ωmax=2πσbAmR

Hence, option (C) is correct.
Key concepts:

Breaking stress is the maximum force that can be applied on a cross-sectional area of a material in such a way that the material is unable to withstand any additional amount of stress before breaking.
Breaking stress may also be known as ultimate tensile stress or breaking strength.

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