Hi, I'm having trouble with a question. Can someone please help me out?(adsbygoogle = window.adsbygoogle || []).push({});

The scenario is I've got a uniform rod off mass M and a ball (particle) of mass m. They are connected together (so there's a rod with a ball at one end) to form a link called "2." The link 2 is of length L.

The link 2 is hinged to a circular platform (the shape of the platform seems irrelevant, I think it's just to make the situation easier to visualise) at the other end (the end of the rod without the particle attached).

The circular platform rotates at a constant angular velocity of w (omega). The angular motion of the link 2 is determined by the angular displacement [itex]\beta(t)[/itex], where beta is the angle between the link 2 and the vertical.

I need to determine the kinetic energy stored in the link 2.

Answer: [tex]T = \frac{1}{2}m\left[ {\left( { - R\omega \sin \left( {\omega t} \right) + L\mathop \beta \limits^ \bullet \cos \left( {\omega t} \right)} \right)^2 + \left( {R\omega \cos \left( {\omega t} \right) + L\mathop \beta \limits^ \bullet \sin \left( {\omega t} \right)} \right)^2 } \right]

[/tex]

In my first attempt I obtained the expression above except that in the terms with (d/dt)(beta) I had beta in the argument of the trig function instead of wt.

Anyway, what I'm having problems with are the energy considerations. I am looking for an expression for the kinetic energy of the link 2. So I start off by isolating it from the rest of the set up and consider it's motion alone.

From the given information I would say that there is only rotational kinetic energy stored in the link 2. It doesn't seem to be 'translating' so I'd say that there is no (1/2)mv^2 component in the kinetic energy, only a (1/2)Iw^2 component.

That's about all I've got to work with right now. I'm fairly sure I've missed a lot of things because the expression given in the answer looks like something of the form (1/2)mv^2. Also, the the mass of the rod (M) in the link doesn't appear in the answer whereas the mass of the particle (m) does.

I'm really unsure about what to do in this question. Can someone please help me out?

Edit: I think that the kinetic energy T will be:

[tex]

T = \frac{1}{2}m\left| {\mathop {v_G }\limits^ \to } \right|^2 + \frac{1}{2}I_G \mathop \beta \limits^ \bullet ^2

[/tex]

So I need to determine the COM of the link 2 (which consists of a rod of mass M and a particle of mass m). This seems rather difficult because the particle is just a dot. It's not like having two blocks where I can determine their individual COMs and then find the common COM.

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# Energy - Rotation

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