# Torsional oscillator with angular displacement

1. Apr 24, 2013

### Robertoalva

1. A torsional oscillator of rotational inertia 2.1 kg·m2 and torsional constant 3.4 N·m/rad has a total energy of 5.4 J.
What is its maximum angular displacement?
What is its maximum angular speed?

2. Relevant equations
θ(t)=Acosωt

3. The attempt at a solution
still trying to think on how to use the energy given, I can't relate kinetic energy...

2. Apr 24, 2013

### mhsd91

I'm not sure if i completely understand the question, so please feel free to correct me.

Now, consider a torsional oscillator: f.ex. a mass at the end of a wire: The mass has the angular inertia, $I$, and the wire it's torsional constant, $\tau_{c}$, with values as specified. Assuming the total energy is conserved (no friction or outside forces), the total energy will be the sum of: (I) the potential energy, $E_{pot}$, corresponding to the work needed to rotate the mass to some angular displacement. (II) the kinetic energy, $E_{kin}$, due to the mass' velocity and angular inertia.

If you study the units of the torsional constant, $\tau_{c}$, you see that [N*m / rad] = [J / rad], applying that work is given from force times distance (newton meters). Hence, if the energy you submitted is the total energy, $E$, then the maximum angular displacement, $\theta_{max}$, will be when,
$E_{kin}=0 \\ E_{pot}= E = 5.4 J$

corresponding to the potential energy stored in the wire by 'twinning it up',
$\tau_{c} \cdot \theta_{max} = E_{pot}$

As you'll easily calculate yourself, inserting correct values and solving for $\theta_{max}$.

The maximum angular velocity, $\omega$, will then be at the point of rotation where,
$E_{pot}=0 \\ E_{kin}= E = 5.4 J \\ E_{kin} = \frac{1}{2} \cdot I \cdot \omega ^2$

Also there are all necessary values given: just solve for $\omega$. That should be it. Good Luck!

Last edited: Apr 24, 2013