MHB How Is Total Kinetic Energy Calculated in This Mechanical System?

[Dustinsfl]

For the following system, what is the total KE?

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I have the following
\[
\frac{1}{2}m_1\dot{x}_1^2 + \frac{1}{2}J_1\dot{\theta}_1 + \frac{1}{2}J_0\dot{\theta}_0^2 + \frac{1}{2}m_s\dot{x}_s^2 + \frac{1}{2}J_s\dot{\theta}_s^2
\]
\(J\) is the mass moment of inertia.

 

[Ackbach]

For the following system, what is the total KE?

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I have the following
\[
\frac{1}{2}m_1\dot{x}_1^2 + \frac{1}{2}J_1\dot{\theta}_1 + \frac{1}{2}J_0\dot{\theta}_0^2 + \frac{1}{2}m_s\dot{x}_s^2 + \frac{1}{2}J_s\dot{\theta}_s^2
\]
\(J\) is the mass moment of inertia.

I'm guessing you meant
$$\frac{1}{2}m\dot{x}_1^2 + \frac{1}{2}J_1\dot{\theta}_1^2 + \frac{1}{2}J_0\dot{\theta}_0^2 + \frac{1}{2}m_s\dot{x}_s^2 + \frac{1}{2}J_s\dot{\theta}_s^2,$$
right? If so, I'd agree. I'm assuming here that you meant $J_0$ to be the MoI of the vertical part of the crank lever, $J_1$ to be the MoI of the horizontal part, and $J_s$ to be the MoI of the sphere. Is that right?

I'm also guessing that you're meant to look up the necessary MoI's. I'm not sure how you can finish this problem without knowing the mass of the crank lever. The MoI of a rod of length $\ell$ and mass $m$ rotated about an end is $m\ell^2/3$. The MoI of a solid sphere rotating about its center is $mr^2/5$.

 

[I like Serena]

\[
\frac{1}{2}m_1\dot{x}_1^2 + \frac{1}{2}J_1\dot{\theta}_1 + \frac{1}{2}J_0\dot{\theta}_0^2 + \frac{1}{2}m_s\dot{x}_s^2 + \frac{1}{2}J_s\dot{\theta}_s^2
\]
\(J\) is the mass moment of inertia.

What are $m_1, x_1, J_1, \theta_1$?
I can't find them in the drawing.
It seems to me that at least $J_1, \theta_1$ are not given, nor relevant.

I'm not sure how you can finish this problem without knowing the mass of the crank lever.

We would not need the mass of the crank lever if the MoI is given with respect to the axis of rotation.

 

[Dustinsfl]

I'm guessing you meant
$$\frac{1}{2}m\dot{x}_1^2 + \frac{1}{2}J_1\dot{\theta}_1^2 + \frac{1}{2}J_0\dot{\theta}_0^2 + \frac{1}{2}m_s\dot{x}_s^2 + \frac{1}{2}J_s\dot{\theta}_s^2,$$
right? If so, I'd agree. I'm assuming here that you meant $J_0$ to be the MoI of the vertical part of the crank lever, $J_1$ to be the MoI of the horizontal part, and $J_s$ to be the MoI of the sphere. Is that right?

I'm also guessing that you're meant to look up the necessary MoI's. I'm not sure how you can finish this problem without knowing the mass of the crank lever. The MoI of a rod of length $\ell$ and mass $m$ rotated about an end is $m\ell^2/3$. The MoI of a solid sphere rotating about its center is $mr^2/5$.

My end goal is to determine the equivalent mass of the system. If we say the mass connected to spring 1 is displaced \(x_1\), the bell crank level will displace \(x_1 = \ell_1\theta_1\) (small angle approximaitons). Would that imply that the vertical arm of the bell crank would also displace at the same angle and distance? That is, would we have \(x_0 = x_1 = \ell_2\theta_0 = \ell_2\theta_1\)?

This would then cause the distance diplaced by \(x_s = x_1\)?

Using this idea, I have
\[
\frac{1}{2}\dot{x}^2\Big[m + m_s + \frac{m_0 + m_1}{3} + \frac{m_s}{5}\Big]
\]
Therefore, \(m_{eq} = m + m_s + \frac{m_0 + m_1}{3} + \frac{m_s}{5}\).