# Roller-spring system (Generalized mass)

Just a few weeks into a physics class on waves/oscillating systems here and I'm a bit stumped.

System is straight forward: Cylinder resting on a floor attached to a wall via a spring. Connection is at cylinder center of mass. Cylinder rolls without skidding, no air resistance.

Cylinder has radius 'a', mass 'm' and the spring constant is k.

I've already used the energy method to derive $\omega ^2 =\frac{2k}{3m}$

Where I'm stumped is the second part where we're instructed to find the generalized mass.

From my notes, this would take the initial form $\frac 1\mu=\frac1{m_1} + \frac 1{m_2}$

My initial attempt was to assume m1 was the translational mass and m2 was the rotational mass, I, giving:

$\frac 1\mu=\frac1m + \frac 1I=\frac1m + \frac 2{ma^2}$

$\mu=\frac{a^2}{a^2+2}m$

Given that the radius is irrelevant in the derivation for the first part, I'm assuming I've forgotten something or misapplied something else, seeing as this looks nothing like what I could use to plug into $\frac km$ to get something that resembles $\omega^2$ above... Any ideas what I've missed?

• You cannot just put the moment of inertia ("rotational mass") in an equation for the mass. It does not even match dimensions. Feb 5, 2017 at 21:43
• Also I presume that your equation for generalized mass applied to a specific physical system (coupled masses?). It is not anything like the definition for generalized mass. Feb 5, 2017 at 21:52
• I believe that the idea is to compare this to an oscillation without rotation. Something like saying: "the system oscillates like a point (generalized) mass (of value=...) on a spring" Feb 5, 2017 at 21:57
• @ja72: What is $c$ in this problem? Feb 5, 2017 at 22:14
• See my answer what I finished posting it. Feb 5, 2017 at 22:23

I think I understand the problem. Do a free body diagram and see that there are two forces acting on the cylinder in the horizontal direction

• The spring force $F_S = - k x$.
• The friction force $F_R$ keeping the cylinder under pure rolling.
• The equations of motion of the center of mass is $$m \ddot{x} = F_S + F_R$$

The contact point must not have velocity so the motion of the center of mass must obey $$\dot{x} + \dot{\theta} r = 0$$ or $$\ddot{x} + \ddot{\theta} r = 0$$ or $$\ddot{\theta} = -\frac{\ddot{x}}{r}$$

• The rotational equations of motion are $$\left. I \ddot{\theta} = r\, F_R \right\} \ddot{x} =- \frac{r^2}{I} F_R$$

Combined with the linear motion

$$\left. m \left(- \frac{r^2}{I} F_R \right) = -k x + F_R \right\} F_R = \frac{I}{I+m r^2} k x$$

So now the equations of motion are $$\boxed{ m \ddot{x} =-\left( \frac{m r^2}{I+m r^2} k \right) x }$$

You can proceed for here.

The effective mass $m^\star$ is defined as $$\ddot{x} = -\frac{k}{m^\star} x$$

or $$m^\star = m + \frac{I}{r^2}$$

• And with the moment of inertia $I=mr^2/2$ you end up with $m^*=3 m / 2$ which is what OP could have directly taken from the expression of the frequency (by comparing it with the frequency without rotation, $\omega^2=k/m$). Feb 5, 2017 at 23:21
• Ah, that's where I tripped. Using your equations, I'm supposed to be solving for m* and I'm missing the r squared under the I. Feb 5, 2017 at 23:26