Answer to Question #102375 in Mechanics | Relativity for Promise Omiponle

Question

A rod of length L is pivoted from its top end about an axis perpendicular to the rod to form a physical pendulum. The rod has a non-uniform linear mass density λ(y) = A y2 where y is the distance from the pivoted top end and A is a positive constant. (In practice, this could be realized approximately by having the rod actually be conical in shape with a small cone angle.)
(a) What are the (SI) units of A?
(b) Find the period T of the rod’s oscillations (for small oscillations) in terms of L and perhaps, constants (e.g., π, ⅓, g, ...).

Answer 1

1.) $λ(y) = A y^2$

here dimensional formula of $\lambda$ is [M¹L^-1T⁰]

and dimensional formula of $y^2$ is [M⁰L²T⁰]

hence dimensional formula of A = [M¹L^-3T⁰]

2.) Here the mass of the rod is variable so we will find out the moment of inertial of the rod about the axis of rotation by taking a small element of mass 'dm' at a 'x' distance from the axis of rotation

$dI = dm\times x^2$

$dI=\lambda dx\times x^2 \newline dI=Ax^2dx\times x^2\newline dI=Ax^4dx$

$\int dI=A\intop x^4dx \newline I = A\intop {^L} x^4dx \newline I=\dfrac{AL^5}{5}$

now calculating the mass of the rod , lets take a small element dm and integrate it

$dm=\lambda dx \newline \int dm= \int ^L Ax^2dx m=\dfrac{AL^3}{3}$

Time period is given by $T=2\pi\sqrt{\dfrac{I}{mgd}}$ where I is the moment of inertia , m is the mass of the rod and d is distance between the axis of rotation and the end point of the rod

$T=2\pi\sqrt{\dfrac{AL^5/5}{(AL^3/3)\times g\times L}}=2\pi\sqrt{\dfrac{5L}{3g}}$