Suppose a uniform steel ring (mass M and radius R ) is kept on a smooth horizontal surface with its plane horizontal. It is hinged about a point O on the perimeter and is made to rotate in a horizonta

Question

Suppose a uniform steel ring (mass M and radius R ) is kept on a smooth horizontal surface with its plane horizontal. It is hinged about a point O on the perimeter and is made to rotate in a horizontal plane as shown: The angular speed ω is kept constant. As a result of rotation, the whole ring comes under tensile stress. Assuming the cross-sectional area of steel used to be a , the difference in tensile stress (absolute value) at location O and at the location farthest from O is

TestHub · Accepted Answer

Tensile stress at O is zero. Tensile at A:

$T d θ = (λ R d θ) ω^{2} (2 R)$

$T = \frac{m}{2 π R} 2 R^{2} ω^{2}$

$T = \frac{7 m ω ^{2} R}{22}$

Tensile stress at A: $T_{A} = \frac{7 m ω ^{2} R}{22 a _{o}}$

$T_{A} - T_{o} = \frac{7 m ω ^{2} R}{22 a _{o}}$

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