A uniform rod of mass M and length L is freely pivoted at one End.what is the period of its oscillation?

Answer 1

Using just the centre of gravity as "effective for all purposes" is too simplistic!

For rotating objects one must use the MOMENT OF INERTIA, I, which is the dynamical quantity defined by:

I = Sum of all bits of mass times the square of their distance from the relevant axis of rotation.

In general, this involves doing an integral. Fortunately, there are various standard forms for this, in terms of the total mass, the shape of the body (a rod, a homogeneous sphere or spherical shell, a circular cylinder, etc.) AND where the pivot is placed. (Other significant words for what you must do on changing the pivot's position are "the parallel axis theorem.")

For a uniform rod, the Moment of Inertia about one end is:

1/3 ML^2, where M is the mass, and L is the total length.

When this rod oscillates under gravity, the restoring force is indeed effectively the weight due to the total mass acting through the centre of gravity, but the expression for what that's accelerating involves the Moment of Inertia, as shown in this particular case, below.

For the rod pivoted at one end, if theta is the angular displacement from the verical, the resulting equation is:

1/3 (M L^2) d^2(theta)/dt^2 = -1/2 Mg L sin (theta),

or, in the usual small amplitude approximation, replacing sin (theta) just by theta, and re-arranging the differential equation into the standard form for simple harmonic motion,

d^2(theta)/dt^2 + 3g/2L theta = 0.

Thus the conventional angular velocity constant omega is given by

(omega)^2 = 3g/2L, so the Period, P, or 2 pi / omega is:

P = 2 pi sqrt [2L/(3g)].

In general, the effect of taking the Moment of Inertia properly into account in problems of this nature is to alter the factor(s) that multiply the usual simple pendulum's L/g (inside the square root factor) by simple numerical fractions, here 2/3.

Live long and prosper.

Answer 2

Uniform rod...Center of Gravity is in the middle so the effective length of the pendulum is 1/2 L. I don't believe the mass matters. There is a formula for the period of an ideal pendulum - I don't have it!

I'm back. 'Googled "period of Pendulum" and got Time for one occelation = 2 x pi x sqrt of (Length over accel. of gravity ). I think it will work for you... In your case Length would be 1/2L. 'Suggest you google it and see if it makes sense.

Answer 3

time period of the road is given by:
t=2*3.14*underroot of( I/mg)

where i is the moment of inerta of the rod about the point through which it is hanging(if x be the length of rod, I=m*x^2/3)

m is the mass of the rod

g is the acceleration due to gravity

Answer 4

T=2*pi*sqrt(intensity/(mass*g*Distance from CM) intensity=a million/3*M*L^2 for this reason, T=2*pi*sqrt((a million/3*M*L^2)/M*g*L) Distance from pivot= L this can provide us: T=2*pi*sqrt(2/(3*9.8))= a million.sixty 4 seconds.