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A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ=cx2, where x is measured from the center of the rod and c is a constant.A. What are the units of c?Mastering physics says the answer is: kg/m^3 but i don't understand where they got this answer.B. Find the expression for c.Mastering physics says the answer is 12M/L^3 --> No clue where they got this answer!C. Find the expression for the moment of inertia of the rod for rotation about an axis through the center.

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Answer:

Explanation:

λ=c x²

c = λ / x²

λ is mass / length

so its dimensional formula is ML⁻¹

x is length so its dimensional formula is L

c = λ / x²

= ML⁻¹ / L²

= ML⁻³

B )

We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M

The mass in the rod is symmetrically distributed on both side of middle point.

we consider a small strip of rod of length dx at x distance away from middle point

its mass dm = λdx = cx² dx

By integrating it from -L to +L we can calculate mass of whole rod , that is

M = ∫cx² dx

= [c x³ / 3] from -L/2 to +L/2

= c/3 [ L³/8 + L³/8]

M = c L³/12

c = 12 M L⁻³

C ) Moment of inertia of rod

∫dmx²

= ∫λdxx²

= ∫cx²dxx²

= ∫cx⁴dx

= c x⁵ / 5 from - L/2 to L/2

= c / 5 ( L⁵/ 32 +L⁵/ 32)

= (2c / 160)L⁵

= (c / 80) L⁵

= (12 M L⁻³/80)L⁵

= 3/20 ML²

=

=

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