A vertical brass rod of the circular section is loaded by placing a $5\;kg$ weight on top of it. If its length is $50\;cm$ and radius of the cross-section is $1\;cm$, find the contraction of rod and the energy stored in it.

Given,
Mass of brass $(M) = 5\;kg$
Length of rod $(l) = 50\;cm = 0.5=\,m$
Radius of brass rod $(r) = 1\;cm = 0.01\;m$
Contraction of rod $(e) = ?$
Energy stored $(W) = ?$

Young's modulus of brass $(Y) = 3.5 * 10^{10}\;N/m^2$
Load on the brass $(F) = m.g = 5 * 10 = 50\;N$
We have,

$Y = \frac{F/A}{e/l} = \frac{F \;*\; l}{e\;*\;A} = \frac{F \;*\; l}{e \;*\; \pi \; r^2}$

⇒ $e = \frac{F.l}{Y. \pi r^2} = \frac{50 \;*\; 0.5}{3.5 \;*\; 10^{10} \;*\; \pi \;*\; (0.01^2)} $ $ = 2.27 * 10^{-6}\;m$

∴  Contraction of rod $(e) = 2.27 * 10^{-6}\;m$

∴  Energy stored $(E) = \frac{1}{2}F.e = \frac{1}{2}* 50 * 2.27 * 10^{-6}$   $ = 5.7 * 10^{-5}\;J$

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