Respuesta :
Answer:
the function varies linearly with the radius of the disk, so the smallest period is zero for a radius of zero centimeters
Explanation:
This system performs a simple harmonic movement where the angular velocity is given by
w = √ k / I
Where k is the constant recovered from the axis of rotation and I is the moment of inertia of the disk
The expression for the moment of inertia is
I = 1/2 m r²
Angular velocity, frequency and period are related
w = 2π f = 2π / T
Substituting
2π / T = √ k / I
T = 2π √ I / k
T = 2π √ (½ m r² / k)
T = (2π √m / 2k) r
We can see that the function varies linearly with the radius of the disk, so the smallest period is zero for a radius of zero centimeters
The minimum period of the pendulum Tmin is [tex]\mathbf{T_{min} = 2 \pi \sqrt{\dfrac{a^2}{ \Big ( \dfrac{ga}{\sqrt{2}}\Big)}} \ \ or \ \ 2 \pi\sqrt{ \dfrac{\sqrt{2a}}{g}}}[/tex]
The period of a pendulum refers to the time required to complete one single revolution or cycle.
Let consider the time period of the disk to that of bob of a pendulum, then, the period of the pendulum can be expressed as:
[tex]\mathbf{T(d) = 2 \pi \sqrt{\dfrac{a^2}{2gd}+\dfrac{d}{g} }}[/tex]
At minimum, the period of the pendulum needs to obey the condition that:
[tex]\mathbf{\dfrac{dT}{d(d)}=0}[/tex]
∴
The above equation becomes:
[tex]\mathbf{\dfrac{d \Big( 2 \pi \sqrt{\dfrac{a^2}{2gd}+\dfrac{d}{g} }\Big )}{d(d)}=0}[/tex]
Taking the differentiation of the above equation with respect to d, we have:
[tex]\mathbf{d = \dfrac{a}{\sqrt{2}}}[/tex]
∴
[tex]\mathbf{T_{min} = 2 \pi \sqrt{\dfrac{a^2}{2g \Big ( \dfrac{a}{\sqrt{2}} \Big)} + \dfrac{a}{g \sqrt{2}}}}[/tex]
[tex]\mathbf{T_{min} = 2 \pi \sqrt{\dfrac{a^2}{ \Big ( \dfrac{ga}{\sqrt{2}}\Big)}} \ \ or \ \ 2 \pi\sqrt{ \dfrac{\sqrt{2a}}{g}}}[/tex]
Learn more about the period of a pendulum here:
https://brainly.com/question/24159297?referrer=searchResults
