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:
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)
