A uniform rod of mass m and length l is rotating with constant angular velocity w about its axis which passes through its one end and perpendicular to the length of rod. the area of cross section of rod is A and its Youngs Modulus is Y. The strain at the midpoint of the rod (neglect gravity) is

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TopNewestHi , let us say the rod is \(AB\) being rotated about \(A\) , and the center of rod being \(C\).

Consider force diagram on \(CB\), having mass \(\frac{m}{2}\) whose center of mass is rotating about \(A\) in a circle of radius \(\frac{l}{2} + \frac{l}{4} = \frac{3l}{4}\). If the tension (providing centripetal force) is \(T\),

\(T = \frac{m}{2} \omega^2 \frac{3l}{4}\)

Hence, strain = \(\frac{T}{AY} = \frac{3m \omega^2 l }{8AY}\) – Jatin Yadav · 3 years, 5 months ago

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Consider an infinitesimally small element of \(dr\) thickness at a distance \(r\) from the axis.

Let \(T(r)\) be the tension at \(r\).

Switching to a rotating reference frame and balancing the forces on this small element.

\(T(r)=T(r+dr)+(dm)\omega^2r\)

where \(dm=(m/l)dr\).

\(\Rightarrow -T'(r)dr=(m/l)\omega^2r\,dr \Rightarrow -dT=(m/l)\omega^2r\,dr\)

\(\displaystyle \Rightarrow \int_{0}^r \frac{m}{l}\omega^2r\,dr=\int_{T_0}^{T} -dT\)

Solving for T, we get:

\(\displaystyle T=T_0-\frac{m\omega^2r^2}{2l}\)

At \(r=l\), \(T=0 \Rightarrow T_0=\frac{m\omega^2l^2}{2l}\). Hence,

\(\displaystyle T=\frac{m\omega^2}{2l}(l^2-r^2)\)

Since we need tension at l/2,

\(\displaystyle T=\frac{m\omega^2}{2l}\left(l^2-\frac{l^2}{4}\right) =\frac{3m\omega^2l}{8}\)

Hence, strain=\(\displaystyle \frac{T}{AY}=\frac{3m\omega^2l}{8AY} \) – Pranav Arora · 3 years, 5 months ago

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