Question #102375

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, ...).

(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, ...).

Expert's answer

1.) "\u03bb(y) = A y^2"

here dimensional formula of "\\lambda" is [M^{1}L^{-1}T^{0}]

and dimensional formula of "y^2" is [M^{0}L^{2}T^{0}]

hence dimensional formula of A = [M^{1}L^{-3}T^{0}]

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 \ndI=Ax^4dx"

"\\int dI=A\\intop x^4dx \n\\newline\nI = A\\intop {^L} x^4dx \\newline\nI=\\dfrac{AL^5}{5}"

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

"dm=\\lambda dx \\newline\n\\int dm= \\int ^L Ax^2dx\nm=\\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}}"

Learn more about our help with Assignments: MechanicsRelativity

## Comments

## Leave a comment