# Rotational kinetic energy of a dumbell

1. Nov 8, 2009

### KBriggs

1. The problem statement, all variables and given/known data
A dumbell made of two point masses separated by a massless rod of fixed length has centre of mass velocity $$V$$ and rotates about its CM with angular velocity $$\omega$$. Two point masses $$m_1$$ and $$m_2$$ are located at $$r_1$$ and $$r_2$$ repsectivelt wrt the CM. Let $$T$$ be the kinetic energy of the dumbell in the lab frame and $$T'_{rot}$$ be the kinetic energ of rotation about the CM. Show that $$T = \frac{1}{2}MV^2+T'_{rot}$$and derive and expression for $$T'_{rot}$$

2. Relevant equations

3. The attempt at a solution

In the lab frame, the velocity of the point masses is $$v'_1=V+v_1$$ and $$v'_2=V+v_2$$ where the little v refers to velocity wrt the CM. So Kinetic energy is given by

$$\frac{1}{2}m_1v'_1\cdotp v'_1 + \frac{1}{2}m_2v'_2\cdotp v'_2$$
$$=\frac{1}{2}m_1(V^2+2V\cdotp v_1+v_1^2) + \frac{1}{2}m_2(V^2+2V\cdotp v_2+v_2^2)$$
let $$M = m_1+m_2$$
$$=\frac{1}{2}MV^2 + V\cdotp (m_1v_1+m_2v_2) + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$

The last two terms can be reduces to the expression for the rotational kinetic energy, which means that I have the extra term $$V\cdotp(m_1v_1+m_2v_2)$$ in my answer. Now, either I did something wrong, or this is identically 0 for some reason, but I can't work out which. Can someone help me?

Last edited: Nov 8, 2009