Question

A slender, uniform metal rod of mass M and length lis pivoted without friction about an axis through its midpoint andperpendicular to the rod. A horizontal spring, assumed massless andwith force constant k,is attached to the lower end of the rod, with the other end of thespring attached to a rigid support.
Find the torque tau due to the spring. Assume that theta is small enough that the spring remains effectivelyhorizontal and you can approximate \sin{(\theta)}\approx \theta (and \cos(\theta)\approx 1 ).
Express the torque as a function oftheta and other parameters of the problem.
What is the angular frequency omega of oscillations of the rod?
Express the angular frequency interms of parameters given in the introduction.

Answer 1

Answer:

Explanation:

Moment of inertia of the metal rod pivoted in the middle

= M l² / 12

If the spring is compressed by small distance x twisting the rod by angle θ

restoring force by spring

= k x

moment of torque  about axis

= k x l /2

= k θ( l /2 )²     ( x / .5 l = θ )

=

moment of torque = moment of inertia of  rod  x angular acceleration

k θ( l /2 )²   = M l² / 12 d²θ/dt²

d²θ/dt² = 3 k/M  θ

acceleration =  ω² θ

ω² = 3 k/M

ω = √ 3 k / M

Answer 2

Answer:

τ = FR = (−kRθ)R = −kR²θ

ω=√(3k/M)

Explanation:

the rod rotates by a small angle θ.  

The spring stretches by amount x = Rθ, and to exert a force

F = −kx = −kRθ on the rod.

Torque equal to

τ = FR = (−kRθ)R = −kR²θ. .........................1

Now, we know that τ = Iα (torque = moment of inertia times angular acceleration);

equsating 1 with the value of torque

Iα = −kR²θ .....................ii

moment of inertia of a rod is given by

I = ML²/12 = M(2R)²/12 = MR²/3 ...................................2

Substituting 2 into the equ 2

(MR²/3)α = −kR²θ

or:

Mα/3 = −kθ

Now, since α is the 2nd derivative of θ (with respect to time), this is just a differential equation:

(M/3)θ′′ = −kθ

from simple harmonic equation

we know that

T=2π√(m/k)

, "mx′′ = −kx"; except that "M/3" takes the place of "m".  That means, where the period of the standard SHM equation comes out to T=2π√(m/k),

we substitute M/3 for m

T = 2π√(M/(3k))..............3

ω=2π/T........4

T=2π/ω

putting angular frequency into 3

ω=√(3k/M)