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# Physics

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

Note by Cody Martin
3 years, 1 month ago

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Hi , 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 = $$\frac{T}{AY} = \frac{3m \omega^2 l }{8AY}$$ · 3 years, 1 month ago

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=$$\displaystyle \frac{T}{AY}=\frac{3m\omega^2l}{8AY}$$ · 3 years, 1 month ago