×

# 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
4 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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 = $$\frac{T}{AY} = \frac{3m \omega^2 l }{8AY}$$

- 4 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=$$\displaystyle \frac{T}{AY}=\frac{3m\omega^2l}{8AY}$$

- 4 years, 1 month ago