What is a "Torsional Constant"?

[Gourab_chill]

Homework Statement

I don't have much idea on what torsional constant means.

Relevant Equations

ω=√(C/I); here C is the torsional constant
ω'=ωβ; this formula which I saw in the solutions is related to the amplitude of the system.

The question was:

I will also include the solution:

So, what is the justification of the first formula [ω=√(C/I)]? I know how to derive simple harmonic equations, this one as I guess is probably similar? But I cannot connect as to how C is used exactly.

And the second formula [ω'=ωβ], I don't know why is it used; I'm clueless about it. Can you throw some light on these topics?

 

[haruspex]

Homework Statement:: I don't have much idea on what torsional constant means.
Relevant Equations:: ω=√(C/I); here C is the torsional constant
ω'=ωβ; this formula which I saw in the solutions is related to the amplitude of the system.

The question was:

View attachment 266508

I will also include the solution:

View attachment 266509

So, what is the justification of the first formula [ω=√(C/I)]? I know how to derive simple harmonic equations, this one as I guess is probably similar? But I cannot connect as to how C is used exactly.

And the second formula [ω'=ωβ], I don't know why is it used; I'm clueless about it. Can you throw some light on these topics?

I can explain the $\beta, \omega, \omega_0$ part.
The two $\omega$s represent quite different things. $\omega$ is the angular frequency, i.e. as in $\sin(\omega t)$. For a rotational oscillation, as here, there is also the rate of change of angular position. So if $\beta$ is the angular amplitude then $\theta=\beta\sin(\omega t)$ and $\dot\theta=\beta\omega\cos(\omega t)$.
They have, confusingly, used $\omega_0$ for the amplitude of $\dot\theta$.

 

[Gourab_chill]

I can explain the $\beta, \omega, \omega_0$ part.
The two $\omega$s represent quite different things. $\omega$ is the angular frequency, i.e. as in $\sin(\omega t)$. For a rotational oscillation, as here, there is also the rate of change of angular position. So if $\beta$ is the angular amplitude then $\theta=\beta\sin(\omega t)$ and $\dot\theta=\beta\omega\cos(\omega t)$.
They have, confusingly, used $\omega_0$ for the amplitude of $\dot\theta$.

Thanks, now I understand it! What are your views on the first equation [ ω=√(C/I) ]? That equation seems familiar to the other ones like ω=√(mgd/I), I'm confused on what exactly C means.

 

[haruspex]

Thanks, now I understand it! What are your views on the first equation [ ω=√(C/I) ]? That equation seems familiar to the other ones like ω=√(mgd/I), I'm confused on what exactly C means.

I can't make sense of that one. All the references I see on the web , like https://en.m.wikipedia.org/wiki/Torsion_constant, give it dimension $L^4$. It is similar to second moment of area, but is less easily calculated and is specific to the way a cross section reacts to torque.
The use of C above gives it the same dimensions as energy (!).

 

[haruspex]

After a bit more research, looks like there's a confusion over terminology.
Wikipedia use "torsion coefficient" for what the question above calls torsional constant.
See https://en.m.wikipedia.org/wiki/Torsion_spring.
It represents the energy stored in the structure per radian of twist.

 

[Gourab_chill]

After a bit more research, looks like there's a confusion over terminology.
Wikipedia use "torsion coefficient" for what the question above calls torsional constant.
See https://en.m.wikipedia.org/wiki/Torsion_spring.
It represents the energy stored in the structure per radian of twist.

Thanks again!
So Cθ = Iα is the equation, and therefore ω^2=C/I.

 

[haruspex]

First, a correction. As with force in a linear spring, torque rises in proportion to the twist, so energy rises as the square of the twist.

Thanks again!
So Cθ = Iα is the equation, and therefore ω^2=C/I.

What is $\alpha$ there?

In the question, the max angular speed is $\beta\omega$, so the max KE is $\frac 12I\beta^2\omega^2$.
The max twist of the fibre is $\beta$, so the max torsional energy is $\frac 12C\beta^2$. These must be equal, so we have $C=I\omega^2$.

 

[Gourab_chill]

First, a correction. As with force in a linear spring, torque rises in proportion to the twist, so energy rises as the square of the twist.

What is $\alpha$ there?

In the question, the max angular speed is $\beta\omega$, so the max KE is $\frac 12I\beta^2\omega^2$.
The max twist of the fibre is $\beta$, so the max torsional energy is $\frac 12C\beta^2$. These must be equal, so we have $C=I\omega^2$.

Since this pendulum is spinning it is coming to rest and then going the opposite way(harmonic motion i mean) so there is some some force in the wire which causes it to accelerate in the opposite direction, and therefore there must be some angular acceleration, right?

 

[haruspex]

Since this pendulum is spinning it is coming to rest and then going the opposite way(harmonic motion i mean) so there is some some force in the wire which causes it to accelerate in the opposite direction, and therefore there must be some angular acceleration, right?

Ok, but you had not mentioned acceleration before.
Yes, and $\theta$ is the angular displacement so $C\theta$ is the torque.
So to complete my correction, think of C as like the k for a spring: it's the torque per unit twist, and the stored energy is $\frac 12C\theta^2$.

 

[Gourab_chill]

Ok, but you had not mentioned acceleration before.
Yes, and $\theta$ is the angular displacement so $C\theta$ is the torque.
So to complete my correction, think of C as like the k for a spring: it's the torque per unit twist, and the stored energy is $\frac 12C\theta^2$.

Yes, I agree to this analogy too.

 

[Gourab_chill]

First, a correction. As with force in a linear spring, torque rises in proportion to the twist, so energy rises as the square of the twist.

What is $\alpha$ there?

In the question, the max angular speed is $\beta\omega$, so the max KE is $\frac 12I\beta^2\omega^2$.
The max twist of the fibre is $\beta$, so the max torsional energy is $\frac 12C\beta^2$. These must be equal, so we have $C=I\omega^2$.

I have a bit confusion as I overlooked a few things(my bad):
I thought the given angular velocity(ω0) shouldn't change but here as in the solutions I see they're considering ω0 to change with ω = √(C/I) [as ω0= ωβsin(ωt)]! Please explain me a bit here.

 

[haruspex]

as ω0= ωβsin(ωt)

Where are you getting that from?
In the extract you quoted in post #1 it says ω₀= ωβ, all constants.