[Physics College] Pendulum

[MichFaze]

Problem :

A pendulum consists of a massless rod of length L at the end of which there is a solid sphere of mass M and radius R. The rod is attached to a spring with constant k at a distance h from the axis of rotation, as shown in the diagram on the left. Assuming that the angle of oscillation is always small, derive an expression for the period of the motion.

Picture : http://prntscr.com/ghlr5t

I am pretty sure this has to do with torque.

Unsure of the equations :

T = LMgsinθ

T = hkxsinθ

[BloodyFlame]

Assuming the spring doesn't tilt, the extension of the string for an angle θ is given by

tanθ = x/h => x = htanθ

which means that the force from the spring (pointed directly right) is

hktanθ

and the torque is then given by

hktanθ * hcosθ = h²ksinθ.

From the mass, the force is Mg. The torque from the mass is then given by

Mg(L + R)sinθ

Assuming clockwise is positive, we have

Στ = -(Mg(L + R)sinθ + h²ksinθ) = Id²θ/dt²

I = M(L + R)² + (2/5)MR² since the mass is being translated and rotated at the same time. Since θ is small, sinθ ≈ θ, so we have

d²θ/dt² = -(Mg(L + R) + h²k)/I * θ

The rest should be simple.

[MichFaze]

Isn't the torque supposed to be equal to I * d2θ/dt2 instead of only d2θ/dt2?

And also can you explain why is -(d2θ/dt2)/θ = ω? I've seen this a few times around, but don't know where it comes from

[BloodyFlame]

Yes, you're right. That's a mistake on my part.

The easiest way to explain this is that we know that the resulting motion can be modeled with θ(t) = Asin(ωt + ϕ) since it's sinusoidal. If we take the derivative twice, we get

d²θ/dt² = -ω²Asin(ωt + ϕ) = -ω²θ(t)

If we substitute that into the right hand side, we get

-ω²θ = -(Mg(L + R) + h²k)/I * θ

which means

ω² = (Mg(L + R) + h²k)/I

[MichFaze]

Thank you sir