Mechanical vibration: single degree of freedom model of wheel mounted on a spring

Question

I think the author of my book has this solution wrong and would like some feedback on my thoughts (see Example 1.4.1).

The solution states that the total energy of the system is represented by the spring, rotation of the wheel, and translation of the wheel, therefore constant total energy E is given by

\begin{align} E = (1/2)mv^2 + (1/2)J\omega^2 + (1/2)kx^2 \end{align}

where $J$ is the inertia of the wheel, equal to $mr^2$, and omega is the angular velocity equal to $v/r$.

I think in this scenario the translational energy and rotational energy are the same quantity -- that is, there cannot be rotation without translation (since he defined "no slipping occurs and no energy is lost at contact"), and vice-versa; or, any time the object is rotating the energy of rotation will be consumed to produce translation, and vice-versa. Thus the equation should be written

\begin{align} E = (1/2)mv^2 + (1/2)kx^2 \end{align}

And the maximum kinetic energy will be

\begin{align} T_{max} = (1/2)m(\omega_n)^2A^2 \end{align}

yielding

\begin{align} \omega_n^2 = k/m \end{align}

Thank you in advance for all responses. This is my first time posting on stack exchange.

Answer 1

The energy of the system at a configuration $(x,\dot{x}, \dot{\theta})$ is given by $E(x,\dot{x}, \dot{\theta}) = {1 \over 2} m \dot{x}^2 + {1 \over 2} J \dot{\theta}^2 + {1 \over 2} k x$.

In this particular example, there is an additional constraint which is that $\dot{x} = r \dot{\theta}$.

When the object rolls, it will 'see' an effective mass of $m+ {J \over r^2}$. That is, it will behave as a single point mass with no inertial component of mass $m+ {J \over r^2}$.

The formula for the resonant frequency follows from the usual formula.

Aside:

Consider two systems of mass $m$, (i) is just a frictionless sliding mass and (ii) is an ideal, non slipping hoop with the mass at any positive radius.

The kinetic energies are (i) ${1 \over 2} m \dot{x}^2$ and (ii) $m \dot{x}^2$ (regardless of radius).

Answer 2

This is more a physics problem then math. But here are some points that you should consider:

1. The moment of inertia of a filled disk is not $J=mr^2$, but $J=\frac 12 mr^2$, so the kinetic energy of rotation would be half of the kinetic energy of translation. $$K.E.R.=\frac 12 \frac12 mr^2\frac{v^2}{r^2}=\frac 14 mv^2$$ You have $J=mr^2$ only if the mass of the wheel is uniformly distributed on the periphery of the disk. So bodies with the same mass can have different moments of inertia. If the kinetic energy of translation is the same as the kinetic energy of rotation, which formula would you use? Why is the other formula wrong?
2. Let's look at a simpler case, a wheel rolling with constant velocity $v$. The mass of the disk is concentrated in two points, on opposite sides of the disk. We have at each of these points a mass $m/2$. The kinetic energy of the center of the mass is $$K.E._{CM}=\frac12(m/2+m/2)v^2=\frac{mv^2}2$$ But this must be equal to the sum of kinetic energies of the two particles. When one of the particle is at the bottom, the velocity of that particle is $v_b=0$. At the same time, the velocity of the particle at the top is $v_t=2v$. The sum of kinetic energies is then $$\frac 12 \frac m2 0^2+\frac 12\frac m2 (2v)^2=mv^2\ne \frac{mv^2}2$$ So if the sum is $mv^2$ but the kinetic energy of the center of mass is only half, where did the other half go? The answer is in the kinetic energy of rotation.