A uniform rod of length L is rotated in a hor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Consider a small element of the rod of length $dx$ at a distance $x$ from the axis of rotation.The mass of this element is $dm = \frac{M}{L} dx =

Vedclass · Accepted Answer

$\frac{\rho \omega^2 L^3}{3 Y}$

Vedclass · Answer

(B) Consider a small element of the rod of length $dx$ at a distance $x$ from the axis of rotation.The mass of this element is $dm = \frac{M}{L} dx = \rho A dx$,where $A$ is the cross-sectional area.The centrifugal force $dF$ acting on this element is $dF = (dm) x \omega^2 = \rho A \omega^2 x dx$.The tension $T(x)$ at a distance $x$ from the axis is the sum of centrifugal forces on all elements from $x$ to $L$:$T(x) = \int_x^L \rho A \omega^2 x' dx' = \rho A \omega^2 \left[ \frac{x'^2}{2} \right]_x^L = \frac{\rho A \omega^2}{2} (L^2 - x^2)$.The elongation $d\Delta L$ of the element $dx$ is given by Hooke's Law:$d\Delta L = \frac{T(x) dx}{AY} = \frac{\rho A \omega^2 (L^2 - x^2) dx}{2 AY} = \frac{\rho \omega^2}{2Y} (L^2 - x^2) dx$.The total increase in length $\Delta L$ is the integral from $0$ to $L$:$\Delta L = \int_0^L \frac{\rho \omega^2}{2Y} (L^2 - x^2) dx = \frac{\rho \omega^2}{2Y} \left[ L^2 x - \frac{x^3}{3} \right]_0^L = \frac{\rho \omega^2}{2Y} \left( L^3 - \frac{L^3}{3} \right) = \frac{\rho \omega^2}{2Y} \left( \frac{2L^3}{3} \right) = \frac{\rho \omega^2 L^3}{3Y}$.

$A$ uniform rod of length $L$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $\omega$. Find the increase in length of the rod,if $\rho$ and $Y$ are the density and Young's modulus of the rod respectively.

Similar Questions

The following four wires are made of the same material. Which one will have the largest elongation when subjected to the same tension?

$A$ mild steel wire of length $1.0 \; m$ and cross-sectional area $0.50 \times 10^{-2} \; cm^{2}$ is stretched,well within its elastic limit,horizontally between two pillars. $A$ mass of $100 \; g$ is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

$A$ $5\, m$ long aluminium wire $(Y = 7 \times 10^{10}\, N/m^2)$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10}\, N/m^2)$ of the same length under the same weight,the diameter should now be,in $mm$.

The density of a rubber cord is $d$. $A$ thick rubber cord of length $L$ and cross-sectional area $A$ undergoes elongation under its own weight when suspended vertically. This elongation is proportional to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module