- #1
etotheipi
Near the end of this paper, Sherwood presents a rotational analogue of the centre of mass work equation. The derivation is as follows (##\tau_{i, \text{CM}}## is the torque of the ith force about the centre of mass): $$\sum_i \tau_{i, \text{CM}} = I_{\text{CM}} \alpha$$ $$\int \left( \sum_i \tau_{i, \text{CM}} d\theta \right) = \int I_{\text{CM}} \frac{d\omega}{dt} d\theta = \Delta \left(\frac{1}{2}I_{\text{CM}}\omega^2 \right)$$ I have no issues with this derivation. However, he then goes on to say that the actual work done in the centre of mass frame is $$W = \sum_i \left( \int \tau_{i, \text{CM}} d\theta_i \right)$$ The first derivation is only valid in the case that the body is rigid, since otherwise ##\omega## and ##\alpha## are not defined. This means that the power of any given torque is ##P = \mathbf{\tau}_i \cdot \mathbf{\omega}##, where ##\omega## is the angular velocity of the rigid body (regardless of the angular velocity of the force). And if the body is no longer rigid, then the first derivation doesn't apply anyway.
The second equation also relies on the assumption that the material being acted upon by the force also undergo angular displacements of ##d\theta_i##, since that is how work is defined. In the case of a rigid body, then all the angular displacements ##d\theta_i## must be equal.
So I struggle to see why these two equations would ever be different? Would you agree with my interpretation above? Thank you!
The second equation also relies on the assumption that the material being acted upon by the force also undergo angular displacements of ##d\theta_i##, since that is how work is defined. In the case of a rigid body, then all the angular displacements ##d\theta_i## must be equal.
So I struggle to see why these two equations would ever be different? Would you agree with my interpretation above? Thank you!