Question

give correct answer only​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

• Here the concept Centripetal Force, Young's Modulus and Tension has been used. We see, we are given the dimensions of the rod by which we can apply the values in the formulas. First we can consider an element in the rod from axis of rotation. Using that we can find the Tension, then using Young's Modulus we can find the increase in its length due to axis of rotation.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{F_{c_{(angular)}}\;=\;\bf{\dfrac{m\omega^{2}}{r}}}}$

$\\\;\boxed{\sf{Y\;=\;\bf{\dfrac{F\:/\:A}{\Delta l\:/\:l}}}}$

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★ Solution :-

» Mass of the rod = m

» Length of the rod = L

» Angular Centripetal Force on Rod = Fc(angular)

» Young's Modulus of rod = Y

» Area of cross section of rod = A

» Angular Velocity of the rotation = ω

• Let there me a small element of length dx which is at a distance of x from the axis of rotation.

Mass of the element = dm

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For mass of the element ::

Mass of the element, μ is given as,

$\\\;\sf{\rightarrow\;\;Mass\;of\;element\;dm\;=\;\bf{\dfrac{m}{L}\;dx}}$

Now let, μ = m/L

Then,

$\\\;\bf{\rightarrow\;\;Mass\;of\;element\;dm\;=\;\bf{\mu\;dx}}$

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For centripetal force Fc and Tension on the rod ::

The centripetal force is given as,

$\\\;\;\sf{:\Rightarrow\;\;F_{c_{(angular)}}\;=\;\bf{\dfrac{m\omega^{2}}{r}}}$

Now the centripetal force acting on the element is gives as,

$\\\;\;\sf{:\Rightarrow\;\;dT\;=\;\bf{dm\omega^{2}\:x}}$

Now using the above relationship of mass of element, we get,

$\\\;\;\sf{:\Rightarrow\;\;dT\;=\;\bf{\omega^{2}\:\mu\omega^{2}x\:dx}}$

We know that total force is sum of all external force acting on the body. This centripetal force is provided by the Tension produced in the rod due to its elasticity.

This means the tension in the rod at a distance of x from the axis of rotation due to all centripetal forces which is due to all the elements between x = x to x = L. This is the distance from rhe axis of rotation to x and then from x to the length (L) of the rod which contains all the elements.

This will give us the equation that,

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;Tension,\;T\;=\;\bf{\int_{x} ^{L}\;\;\mu\omega^{2}x\:dx}}}$

On solving this, we get,

$\\\;\;\sf{:\Longrightarrow\;\;Tension,\;T\;=\;\bf{\dfrac{\mu\omega^{2}}{2}\;(L^{2}\;-\;r^{2})}}$

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★ For the increase in length (L) of rod ::

Let the increase in the length of the element be dl. Then Young's Modulus is given as,

$\\\;\;\sf{:\mapsto\;\;Y\;=\;\bf{\dfrac{F\:/\:A}{\Delta l\:/\:l}}}$

By replacing Force with Tension, l by dx and ∆l by dl in the above formula, we get,

$\\\;\;\sf{:\mapsto\;\;Y\;=\;\bf{\dfrac{T\:/\:A}{dl\:/\:l}}}$

$\\\;\;\sf{:\mapsto\;\;Y\;=\;\bf{\dfrac{\bigg(\dfrac{T}{A}\bigg)}{\bigg(\dfrac{dl}{l}\bigg)}}}$

$\\\;\;\sf{:\mapsto\;\;Y\;=\;\bf{\dfrac{T\;\times\;dx}{dl\;\times\;A}}}$

$\\\;\;\sf{:\mapsto\;\;dl\;=\;\bf{\dfrac{T\;\times\;dx}{Y\;\times\;A}}}$

Now using the relationship of T, we get,

$\\\;\;\sf{:\mapsto\;\;dl\;=\;\bf{\dfrac{\bigg(\dfrac{\mu\omega^{2}}{2}\;(L^{2}\;-\;x^{2})\:dx\bigg)}{Y\;\times\;A}}}$

$\\\;\;\sf{:\mapsto\;\;dl\;=\;\bf{\dfrac{\mu\omega^{2}}{2\;\times\;Y\;\times\;A}\;(L^{2}\;-\;x^{2})\:dx}}$

This is the elongation for elements between x to L. Now total elongation in the rod is given as,

$\\\;\;\displaystyle{\sf{:\mapsto\;\;l\;=\;\bf{\int_{0} ^{L}\;\;\dfrac{\mu\omega^{2}}{2\;\times\;Y\;\times\;A}\;(L^{2}\;-\;x^{2})\:dx}}}$

On solving this, we get:-

$\\\;\;\sf{:\mapsto\;\;l\;=\;\bf{\dfrac{\mu\omega^{2}}{2\:Y\:A}\;\:\bigg[L^{2}x\;-\;\dfrac{x^{3}}{3}\bigg]_{0} ^{L}}}$

$\\\;\;\sf{:\mapsto\;\;l\;=\;\bf{\dfrac{1}{3}\;\times\;\dfrac{\mu\omega^{2}\:L^{2}}{Y\;\times\;A}\;}}$

$\\\;\;\sf{:\mapsto\;\;l\;=\;\bf{\dfrac{1}{3}\;\times\;\dfrac{m\omega^{2}\:L^{2}}{Y\;\times\;A}}}$

$\\\;\;\sf{:\mapsto\;\;l\;=\;\bf{\dfrac{1}{3}\;\times\;\dfrac{m\omega^{2}\:L^{2}}{3Y\;\times\;A}}}$

∴ So, option c.) (m ω² L²) / (3AY) is the correct option.

$\\\;\underline{\boxed{\tt{Increase\;\;in\;\;Length\;\;of\;\;rod\;=\;\bf{\dfrac{m\omega^{2}\:L^{2}}{3AY}}}}}$

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★ More to know :-

• $\\\;\sf{\leadsto\;\;Coefficient\;of\;Elasticity\;=\;\dfrac{Stress}{Strain}}$

• $\\\;\sf{\leadsto\;\;Surface\;Tension\;=\;\dfrac{Force}{Length}}$

• $\\\;\sf{\leadsto\;\;Surface\;Energy\;=\;\dfrac{Work}{Area}}$

• $\\\;\sf{\leadsto\;\;Force\;Constant\;=\;\dfrac{Force}{Displacement}}$

• $\\\;\sf{\leadsto\;\;Stress\;=\;\dfrac{Force}{Area}}$