Calculating frequency for small torsional oscillation

[lzh]

Homework Statement

A thin, uniform, rigid disk of mass M, radius R is welded to a light, elastic shaft of radius r, length L, shear modulus G. Phi is the torsional oscillation.
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Homework Equations

Phi=TL/GI
I=pi/2*r^4(polar area moment of inertia)

The Attempt at a Solution

Since the shaft is "light", I assumed it to be massless and considered it a spring instead. I'm not sure if this is the right train of thought. I'm trying the Newtonian approach of equating torque:

I(total)*phi"=torque of shaft+torque of disk

somehow though, I couldn't figure out the torque of the disk. If the shaft is a spring, it's torque would be k*phi.

Am I even on the right track here? Should I do conservation of energy instead?

 

[tiny-tim]

hi lzhlzh1

A thin, uniform, rigid disk of mass M, radius R is welded to a light, elastic shaft of radius r, length L, shear modulus G …

do you know what shear modulus is?

if so, write an equation for it ​

 

[lzh]

it's the ratio of shear stress to shear strain.

G=(F/A)/tan(theta)
but truthfully I don't understand what role G plays in this question except for in the phi equation i posted

 

[tiny-tim]

you'll need to use conservation of energy (ie kinetic energy of the disc plus torsional potential energy of the shaft)

but i don't know how the shear modulus in the question is related to the torsional modulus

 

[DeeNos]

Your approach of equating torque is correct. To calculate the torque of the disk, you can use the formula T = I * alpha, where T is the torque, I is the moment of inertia, and alpha is the angular acceleration. In this case, the moment of inertia is given by I = (1/2)*M*R^2, and the angular acceleration alpha is equal to phi" (double derivative of phi with respect to time).

So the equation becomes (1/2)*M*R^2*phi" = k*phi, where k is the spring constant of the shaft.

To solve for phi, you can use the small angle approximation, which states that for small angles, sin(phi) is approximately equal to phi. This simplifies the equation to M*R^2*phi" = 2k*phi.

Using this equation, you can solve for the frequency of the torsional oscillation, which is given by f = 1/(2*pi)*sqrt(k/(M*R^2)).

In this case, since the shaft is considered to be a spring, you can use the formula for the spring constant k = (G*pi*r^4)/(2*L).

Substituting this into the frequency equation, you get f = 1/(2*pi)*sqrt((G*pi*r^4)/(2*M*L*R^2)).

Note that this is the frequency for small torsional oscillations, so it is valid only for small values of phi. If you need to calculate the frequency for larger values of phi, you will need to use a different approach, such as conservation of energy.